MATH10242 Sequences and Series

Mike Prest1

School of MathematicsAlan Turing Building

Room 1.120

mprest@manchester.ac.uk

January 26, 2018

1These notes are a slightly modified version of notes developed by Prof. J. T. Stafford and, beforehim, Prof. A. J. Wilkie.

Contents

0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 Before We Begin 41.1 Some Reminders about Mathematical Notation . . . . . . . . . . . . . . . . . . . 4

1.1.1 Special sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Set theory notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.3 Logical notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.4 Greek letters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.5 Structure of the course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.6 Where we’re headed and some things we’ll see on the way . . . . . . . . . 61.1.7 Reading outside the Course Notes . . . . . . . . . . . . . . . . . . . . . . 71.1.8 Basic properties of the real numbers . . . . . . . . . . . . . . . . . . . . . 71.1.9 The Integer Part (or ‘Floor’) Function . . . . . . . . . . . . . . . . . . . . 9

I Sequences 10

2 Convergence 112.1 What is a Sequence? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 The Triangle Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 The Definition of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 The Completeness Property for R . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Some General Theorems about Convergence . . . . . . . . . . . . . . . . . . . . . 182.6 Exponentiation - a digression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 The Calculation of Limits 213.1 The Sandwich Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 The Algebra of Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4 Some Special Sequences 274.1 Basic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 New Sequences from Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.3 Newton’s Method for Finding Roots of Equations - optional . . . . . . . . . . . . 34

5 Divergence 365.1 Sequences that Tend to Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6 Subsequences 396.1 The Subsequence Test for Non-Convergence . . . . . . . . . . . . . . . . . . . . . 396.2 Cauchy Sequences and the Bolzano-Weierstrass Theorem . . . . . . . . . . . . . . 41

6.2.1 Proofs for the section - optional . . . . . . . . . . . . . . . . . . . . . . . . 41

1

7 L’Hopital’s Rule 457.1 L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

II Series 48

8 Introduction to Series 498.1 The Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

9 Series with Non-Negative Terms 539.1 The Basic Theory of Series with Non-Negative Terms . . . . . . . . . . . . . . . 539.2 The Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

10 Series with Positive and Negative Terms 6210.1 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6210.2 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

11 Power Series 6711.1 The Radius of Convergence of a Power Series . . . . . . . . . . . . . . . . . . . . 68

12 Further Results on Power Series - further reading 7212.1 More General Taylor Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7212.2 Rearranging Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7312.3 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2

0.1 Introduction

Maybe you can see that

1 +1

2+

1

4+

1

8+ · · ·+ 1

2n+ · · · = 2

and even that

1 +1

3+

1

9+

1

27+ · · ·+ 1

3n+ · · · = 3

2.

But what exactly do these formulas mean? After all, we cannot really add infinitely manynumbers together. You might recognize them as geometric progressions and know the generalformula

1 + x+ x2 + x3 + · · ·+ xn + · · · = 1

1− x.

But how do we prove this? And what if x ≥ 1?In this course we shall answer these, and related, questions. In particular, we shall give a

rigorous definition of what it means to add up infinitely many numbers and then we shall findrules and procedures for finding the sum in a wide range of particular cases.

Here are some more, rather remarkable, such formulas:

1 +1

4+

1

9+

1

16+ · · ·+ 1

n2+ · · · = π2

6,

1− 1

2+

1

3− 1

4+ · · ·+ (−1)n+1

n+ · · · = log 2,

1 +1

2+

1

3+

1

4+ · · ·+ 1

n+ · · · =∞.

We shall prove the second and third of these formulas in this course unit, but the first one istoo difficult and will be done in your lectures on real and complex analysis in the second year.Here, “real analysis” means the study of functions from real numbers to real numbers, from thepoint of view of developing a rigorous foundation for calculus (differentiation and integration)and for other infinite processes.1 The study of sequences and series is the first step in thisprogramme.

This also means there are two contrasting sides to this course. On the one hand we willdevelop the machinery to produce formulas like the ones above. On the other hand it is alsocrucial to understand the theory that lies behind that machinery. This rigorous approach formsthe second aspect of the course—and is in turn the first step in providing the formal foundationof real analysis.

1and “complex analysis” refers to the complex numbers, not (necessarily) complexity in the sense of “compli-cated”!

3

Chapter 1

Before We Begin

1.1 Some Reminders about Mathematical Notation

1.1.1 Special sets

We use the following notation throughout the course.R - the set of real numbers;R+ - the set of strictly positive real numbers, i.e. R+ = {x ∈ R : x > 0};Q - the set of rational numbers;Z - the set of integers (positive, negative and 0);N (or Z+) - the set of natural numbers, or positive integers {x ∈ Z : x > 0}. (In this

course, we do not count 0 as a natural number. We can use some other notation like Z≥0 forthe set of integers greater than or equal to 0.)∅ - the empty set.

1.1.2 Set theory notation

The expression “x ∈ X” means x is an element (or member) of the set X. For sets A, B, wewrite A ⊆ B to mean that A is a subset of B (i.e. every element of A is also an element of B).Thus ∅ ⊆ N ⊆ Z ⊆ Q ⊆ R.

Standard intervals in R: if a, b ∈ R with a ≤ b, then

• (a, b) ={x ∈ R : a < x < b}; • [a, b) ={x ∈ R : a ≤ x < b};

• (a, b] ={x ∈ R : a < x ≤ b}; • [a, b] ={x ∈ R : a ≤ x ≤ b};

• (a,∞) ={x ∈ R : a < x}; • [a,∞) ={x ∈ R : a ≤ x};

• (−∞, b) ={x ∈ R : x < b}; • (−∞, b] ={x ∈ R : x ≤ b};

• (−∞,∞) = R.

1.1.3 Logical notation

The expression “∀x ...” means “for all x ...” and “∃x ...” means “there exists at least one x suchthat ...”. These are usually used in the context “∀x ∈ A ...” meaning “for all elements x of theset A ...”, and “∃x ∈ A ...” meaning “there exists at least one element x in the set A such that...”.

Thus, for example, “∀x ∈ R x > 1” means “for all real numbers x, x is greater than 1”(which happens to be false) and “∃x x > 1” means “there exists a real number x such that x isgreater than 1” (which happens to be true).

4

1.1.4 Greek letters

The two most commonly used Greek letters in this course are δ (delta) and ε (epsilon). Theyare reserved exclusively for (usually small) positive real numbers.

Others are α (alpha), β (beta), γ (gamma), λ (lambda), θ (theta-usually an angle), η (eta)and Σ (capital sigma - the summation sign which will be used when we come to study series inPart II).

1.1.5 Structure of the course

This course unit has two lectures each week; each student has one tutorial per week - you will beassigned to one of the four tutorial groups (this should be shown in your space on the studentsystem).

Lectures:

Tuesday 10:00–11:00, except for the first week of term, when it will be 11:00-12:00, inCrawford House, Theatre 1;

Thursday 12:00–1:00 in Simon Building, Theatre E.

Tutorials: These start on 5th February. You will be assigned to one of these:

Tutorial 1: Monday 11:00–12:00 in Simon Building, Room 3.40;

Tutorial 2: Monday 15:00–16:00 in Samuel Alexander Building, SG.16;

Tutorial 3: Tuesday 11:00–12:00 in University Place, 1.219;

Tutorial 4: Monday 16:00–17:00 in Samuel Alexander Building, SG.16.

The tutorials start in Week 2 and, typically, Week n tutorials concentrate on the subject matterintroduced in Week n− 1.

I will put the weekly exercise sheets on my teaching webpage website. (Complete coursenotes and some links are already there.) It is very important that you work at these sheets beforethe weekly tutorials; try to have a serious attempt at all the problems. But don’t waste timegoing round in circles: try to recognise when you’re doing that, have a break/do something elseand, maybe, when you come back to the question you’ll see your way around what seemed to bea problem. In general, for each lecture or tutorial hour, you should expect to spend two to threehours working at understanding the material and testing/strengthening your understanding bydoing problems. It is not enough to read and summarise the notes; it is only by testing yourunderstanding by doing examples that you will really understand the material. Moreover, theexam questions will be similar to these problems!

When you get really stuck on something, discuss it with someone! As well as me and thepeople who help in the tutorials, there are your fellow-students - they are an excellent resourceand bear in mind that one of the most useful exercises you can do is to try to explain somethingto someone else.

Office Hour: see my teaching website http://www.maths.manchester.ac.uk/∼mprest/teaching.html

Assessment:

1. Final exam carrying 80% of the course weight.

2. Coursework carries 20% of the course weight and will consist of:

an in-class test on Thursday 15th March.

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1.1.6 Where we’re headed and some things we’ll see on the way

In Part I we aim to understand the behaviour of infinite sequences of real numbers, meaningwhat happens to the terms as we go further and further on in the sequence. Do the terms allgradually get as close as we like to a limiting value (then the sequence is said to converge to thatvalue) or not? The “conceptual” aim here is to really understand what this means. To do that,we have to be precise and avoid some plausible but misleading ideas. It’s worthwhile trying todevelop, and refine, your own “pictures” of what’s going on. We also have to understand theprecise definition well enough to be able to use it when we calculate examples, though we willgradually build up a stock of general results (the “Algebra of Limits”), general techniques andparticular cases, so that we don’t have to think so hard when faced with the next example.

Part II is about “infinite sums” of real numbers: how we can make a sensible definition ofthat vague idea and then how we can calculate the value of an infinite sum - if it exists. We alsoneed to be able to tell whether a particular “infinite sum” does or doesn’t make sense/exist.Sequences appear here in two ways: first as the sequence of numbers to be “added up” (and theorder of adding up does matter, as we shall see); second as a crucial ingredient in the actualdefinition of an “infinite sum” (“infinite series” is the official term). What we actually do is addup just the first n terms of such an infinite series - call this value the n-th partial sum - andthen see what happens to this sequence (note) of partial sums as n gets bigger and bigger. Ifthat sequence of partial sums converges to a limit then that limit is what we define to be thesum of the infinite series. Hopefully that makes sense to you and seems like it should be theright definition to use. Anyway, it works and, again, we have the conceptual aspect to get togrips with as well as various techniques that we can (try to) use in computations of examples.

Here are some of the things we prove about our concept of limit: a sequence can have atmost one limit; if a sequence is increasing but never gets beyond a certain value, then it has alimit; if a sequence is squeezed between two other sequences which have the same limit l, thenit has limit l. These properties help clarify the concept and are frequently used in argumentsand calculations. We also show arithmetic properties like: if we have two sequences, each witha limit, and produce a new sequence by adding corresponding terms, then this new sequencehas a limit which is, as you might expect, the sum of the limits of the two sequences we startedwith.

Then we turn to methods of calculating limits. We compare standard functions (polynomials,logs, exponentials, factorials, ...) according to how quickly they grow, but according to a verycoarse measure - their “order” of growth, rather than rate of growth (i.e. derivative). That letsus see which functions in a complicated expression for the n-th term of a sequence are mostimportant in calculating the limit of the sequence. There will be lots of examples, so that youcan gain some facility in computing limits, and there are various helpful results, L’Hopital’sRule being particularly useful.

While the properties of sequences are, at least once you’ve absorbed the concept, quitenatural, infinite series hold quite a few surprises and really illustrate the need to be carefulabout definitions in mathematics (many mathematicians made errors by mishandling infiniteseries, especially before the concepts were properly worked out in the 1800s). Given an infiniteseries, there are two questions: does it have a sum? (then we say that it “converges”, meaningthat the sequence of partial sums has a limit - the value of that infinite sum) and, if so, whatis the value of the sum? There are a few series (e.g. a geometric series with ratio < 1) wherewe can quite easily compute the value but, in general this is hard. It is considerably easier todetermine whether a series has a sum or not by comparing it with a series we already knowabout. Indeed, the main test for convergence (that is, having a sum) that we will use, the RatioTest, is basically comparison with a geometric series.

6

Many infinite series that turn up naturally are “alternating”, meaning that the terms arealternately positive and negative. So, in contrast with the corresponding series where all theterms are made positive, there’s more chance of an alternating sequence converging, becausethe terms partly cancel each other out. Indeed, remarkably, it’s certain provided the absolutevalues of the individual terms shrink to 0!

We’ll finish with power series: infinite series where each term involves some variable x - youcould think of these as “infinite polynomials in x”. Whether or not a power series converges(i.e. has a sum), depends very much on the value of x, convergence being more likely for smallervalues of x. In fact, the picture is as good as it could be: there’s a certain “radius of convergence”R (which could be 0 or ∞ or anything in between, depending on the series) such that, withinthat radius we have convergence, outside we have divergence and, on the boundary (x = ±R)it could go either way for each point (so we have to do some more work there).

1.1.7 Reading outside the Course Notes

You should do reading outside the course notes.If you read only the course notes, do only the exercises set and look just at past exams

for this course unit, then you should pass and you might even get a very good mark. Butyour understanding and enjoyment of mathematics will be enhanced if you become more of an“independent learner” by reading around a bit, asking your own questions and looking beyondthe course notes for answers. You might see connections with other courses and you’re likelyto develop a stronger sense of “ownership” of the mathematics you’re learning. It takes moreeffort, but more engagement is likely to result in more enjoyment and, in the long run, bettermarks. In particular, you’re less likely to feel overwhelmed by all the new material in front ofyou because you will have had some practice in extracting what’s most useful from a variety ofsources.

The time to start reading outside the course notes is now! Not when exams are close -at that time you will probably want to focus your efforts on what is most directly relevant toexams but that is the time when the perspective and understanding that you have developedduring the semester will pay off and make revision easier.

Notation in this course is largely consistent with the Foundations of Pure Mathematics courseunit text:

[PJE] Peter J Eccles, ’An Introduction to Mathematical Reasoning.’

1.1.8 Basic properties of the real numbers

It will be assumed that you are familiar with the elementary properties of N, Z and Q that werecovered last semester in MATH10101/10111. These include, in particular, the basic facts aboutthe arithmetic of the integers and a familiarity with the Principle of Mathematical Induction.One may then proceed to construct the set R of real numbers. There are many ways of doingthis which, remarkably, all turn out to be equivalent in a sense that can be made mathematicallyprecise. One method with which you should be familiar is to use infinite decimal expansions asdescribed in Section 13.3 of [PJE].

Here we extract some of the basic arithmetic and order properties of the reals.First, just as for the set of rational numbers, R is a field. That means that it satisfies the

following conditions.

(A0) ∀a, b ∈ R one can form the sum a+ b and the product a · b (also written as just ab). Wehave that a+ b ∈ R and a · b ∈ R;

7

(A1) ∀a, b, c ∈ R, a+ (b+ c) = (a+ b) + c (associativity of +);

(A2) ∀a, b ∈ R, a+ b = b+ a (commutativity of +);

(A3) ∀a ∈ R, a+ 0 = 0 + a = a;

(A4) ∀a ∈ R, there is a unique element in R (denoted −a) suchthat a+ (−a) = 0 = (−a) + a;

(A5) ∀a, b, c ∈ R, a · (b · c) = (a · b) · c (associativity of ·);

(A6) ∀a, b ∈ R, a · b = b · a (commutativity of ·);

(A7) ∀a ∈ R, a · 1 = 1 · a = a;

(A8) ∀a ∈ R, if a 6= 0 then there is a unique element in R (denoted a−1

or 1a) such that a · (a−1) = 1 = a−1 · a;

(A9) ∀a, b, c ∈ R, a · (b+ c) = a · b+ a · c (the distributive law).

These axioms (A0)-(A9) list the basic arithmetic/algebraic properties that hold in R andfrom which all the other such properties may be deduced. Further identities, such as x ·0 = 0,x · (−y) = −(x · y), −(−x) = x, (x+ y)(x− y) = x2 − y2, (x+ y)2 = x2 + 2xy+ y2, ... followfrom these axioms.

Example 1.1.1. For example, let us prove the last identity above. First, however, note the usualnotation: for a ∈ R, a2 is an abbreviation for a · a (similarly a3 is an abbreviation for a · (a · a),etc.).

So, we have

(x+ y)2 = (x+ y)(x+ y) (by definition),

= (x+ y) · x+ (x+ y) · y (by A9),

= x · (x+ y) + y · (x+ y) (by A6),

= x · x+ x · y + y · x+ y · y (by A9),

= x · x+ x · y + x · y + y · y (by A6),

= x · x+ 1 · (x · y) + 1 · (x · y) + y · y (by A7),

= x · x+ (1 + 1) · (x · y) + y · y (by A9 and A6),

= x · x+ 2 · (x · y) + y · y (by definition),

= x2 + 2xy + y2 (by definition).

Actually we have also used A1 many times here. This allowed us to ignore brackets inexpressions involving many + symbols.

Note that if we replace R in these axioms by Q then they do hold; that is, Q is also a field(but Z is not since it fails (A8)). The point of giving a name (“field”) to an “arithmetic system”where these hold is that many more examples appear in mathematics, so it proved to be worthisolating these properties and investigating their general implications. Other fields that you willhave encountered by now are the complex numbers and the integers modulo 5 (more generally,modulo any prime). But the reals and rationals have some further properties not shared bythese latter examples.

Namely, the reals form an ordered field. This means that we have a total order relation< on R. All the properties of this relation and how it interacts with the arithmetic operationsfollow from these axioms:

8

(Ord 1) ∀a, b ∈ R, exactly one of the following is true: a < b or b < a or a = b;

(Ord 2) ∀a, b ∈ R, if a < b and b < c, then a < c;

(Ord 3) ∀a, b ∈ R, if a < b then ∀c ∈ R, a+ c < b+ c;

(Ord 4) ∀a, b ∈ R, if a < b then ∀d ∈ R+, a · d < b · d.

As usual, we write a ≤ b to mean either a = b or a < b.You do not need to remember exactly which rules have been written down here. The impor-

tant point is just that this short list of properties is all that one needs in order to be able toprove all the other standard facts about <. For example, we have:

Example 1.1.2. Let us prove that if x is positive (i.e. 0 < x) then −x is negative (i.e. −x <0). So suppose that 0 < x. Then by (Ord 3), 0 + (−x) < x + (−x). Simplifying we get−x = 0 + (−x) < x+ (−x) = 0, as required.

Other facts that follow from these properties include:∀x ∈ R x2 ≥ 0,∀x, y ∈ R x ≤ y ⇔ −x ≥ −y, et cetera.It left as an exercise to prove these facts using just the axioms (Ord 1–4). Also see the first

exercise sheet.

However, there are more subtle facts about the real numbers that cannot be deduced fromthe axioms discussed so far. For example, consider the theorem that

√2 is irrational. This

really contains two statements: first that√

2 6∈ Q (that is, no rational number squares to 2; thiswas proved in MATH10101/10111); second that there really is a such real number - there is a(positive) solution to the equation X2 − 2 = 0 in R. And this definitely is an extra propertyfor the reals—simply because it not true for the rationals (which satisfies all the algebraic andorder axioms that we listed above).

So, we need to formulate a property of R that says there are no “missing numbers” (like the“missing number” in Q where

√2 should be). Of course, we have to say what “numbers” should

be there, in order to make sense of saying that some of there are “missing”. The example of√2 might suggest that we should have “enough” numbers so as to be able to take n-th roots

of positive numbers and perhaps to solve other polynomial equations and following that ideadoes lead to another field - the field of real algebraic numbers - but we have a much strongercondition (“completeness”) in mind here. We will introduce it in Chapter 2, just before we needit.

1.1.9 The Integer Part (or ‘Floor’) Function

From any of the standard constructions of the real numbers one has the fact that any realnumber is sandwiched between two successive integers in the following precise sense:∀x ∈ R, ∃n ∈ Z such that n ≤ x < n+ 1.

The integer n that appears here is unique and is denoted [x]; this is called the integer part ofx. The function x 7→ [x] is called the integer part, or floor, function. Note that 0 ≤ x− [x] < 1.

Example 1.1.3. [1.47] = 1, [π] = 3, [−1.47] = [−2].

9

Part I

Sequences

10

Chapter 2

Convergence

2.1 What is a Sequence?

Definition 2.1.1. A sequence is a list a1, a2, a3, . . . , an, . . . of real numbers labelled (or in-dexed) by natural numbers. We usually write such a sequence as (an)n∈N, or as (an)n or justas (an). Then an is called the nth term of the sequence.

The word “sequence” suggests time, with the numbers occurring in temporal sequence:first a1, then a2, et cetera. Indeed, some sequences arise this way, for instance as successiveapproximations to some quantity we want to compute. Formally, a sequence is simply a functionf :N→ R, where we write an for f(n).

We shall be interested in the long term behaviour of sequences, i.e. the behaviour of thenumbers an when n is very large (e.g. do the approximations converge on some value?).

Example 2.1.2. Consider the sequence 1, 4, 9, 16, . . . n2, . . .. Here, the nth term is n2. So wewrite the sequence as (n2)n∈N or (n2)n or just (n2). What is the nth term of the sequence4, 9, 16, 25, . . .? What are the first few terms of the sequence

((n− 1)2

)n?

Example 2.1.3. Consider the sequence 2, 32 , 4

3 , 54 , . . . . Here, an =

n+ 1

n. The sequence is(

n+ 1

n

)n∈N

.

Example 2.1.4. Consider the sequence −1, 1,−1, 1,−1, . . .. A precise and succinct way ofwriting it is

((−1)n

)n∈N. The nth term is 1 if n is even and −1 if n is odd.

Example 2.1.5. Consider the sequence

((−1)n

3n

)n∈N

. The 5th term, for example, is−1

243.

Example 2.1.6. Sometimes we might not have a precise formula for the nth term but rathera rule for generating the sequence, E.g. consider the sequence 1, 1, 2, 3, 5, 8, 13, . . ., which isspecified by the rule a1 = a2 = 1 and, for n ≥ 3, an = an−1 + an−2. (This is the Fibonaccisequence.)

Long term behaviour: In Examples 2.1.2 and 2.1.6 the terms get huge, with no bound ontheir size (we shall say that they tend to ∞).

However, for 2.1.3, the 100th term is101

100= 1 +

1

100, the 1000th term is

1001

1000= 1 +

1

1000. It

looks as though the terms are getting closer and closer to 1. (Later we shall express this by

saying that

(n+ 1

n

)n∈N

converges to 1.)

In Example 2.1.4 , the terms alternate between −1 and 1.

11

In Example 2.1.5, the terms alternate between being positive and negative, but they are alsogetting very small in absolute value (i.e. in their size when we ignore the minus sign): so(

(−1)n

3n

)n∈N

converges to 0.

Before giving the precise definition of the word “convergence” we require some technical prop-

erties of the modulus (i.e. absolute value) function.

We now make the convention that unless otherwise stated, all variables (x, y, l, ε,δ ...) range over the set R of real numbers.

2.2 The Triangle Inequality

We define the modulus, |x| of x, also called the absolute value of x.

|x| =

{x if x ≥ 0,

−x if x < 0.

Clearly |x| = max{x,−x}, so x ≤ |x| and −x ≤ |x|.

Theorem 2.2.1 (The Triangle Inequality). For all x, y, we have

|x+ y| ≤ |x|+ |y|.

Proof. We have x ≤ |x| and y ≤ |y|. So by adding we get x+ y ≤ |x|+ |y|.Also, −x ≤ |x| and −y ≤ |y|, so −(x+ y) = (−x) + (−y) ≤ |x|+ |y|.It follows that max{x+ y,−(x+ y)} ≤ |x|+ |y|, i.e. |x+ y| ≤ |x|+ |y|, as required. 2

Remark: For x, y ∈ R, |x − y| is the distance from x to y along the “line” R. The triangleinequality is saying that the sum of the lengths of any two sides of a triangle is at least as bigas the length of the third side, as is made explicit in the following corollary. (Of course we aredealing here with rather degenerate triangles: the name really comes from the fact that in thisform, the triangle inequality is also true for points in the plane.)

Corollary 2.2.2 (Also called the Triangle Inequality). For all a, b, c, we have

|a− c| ≤ |a− b|+ |b− c|

.

Proof. |a− c| = |(a− b) + (b− c)| ≤ |a− b|+ |b− c| (by Theorem 2.2.1 with x = (a− b) andy = (b− c))). 2

Lemma 2.2.3. Some further properties of the modulus function:

(a) ∀x, y |x · y| = |x| · |y| and, if y 6= 0,∣∣xy

∣∣ =|x||y|

;

(b) ∀x, y |x− y| ≥∣∣|x| − |y|∣∣;

(c) ∀x, l and ∀ε > 0, |x− l| < ε ⇐⇒ l − ε < x < l + ε.

Proof. Exercises: (a) - consider the cases; (b), (c) - see the Problems sheet for Week 3. 2

We now come to the definition of what it means for a sequence (an)n∈N to converge to a realnumber l. We want a precise mathematical way of saying that “as n gets bigger and bigger, angets closer and closer to l”.

12

2.3 The Definition of Convergence

Definition 2.3.1. We say that a sequence (an)n∈N converges to the real number l if thefollowing holds:∀ε > 0 ∃N ∈ N such that for all n ∈ N with n ≥ N we have |an − l| < ε.

We use various expressions and notations for when this holds:

• an tends to l as n tends to infinity, written an → l as n→∞.

• the limit of (an)n∈N as n tends to infinity equals l, written limn→∞ an = l.

Example 2.3.2. Consider the sequence

(n+ 1

n

)n∈N

of 2.1.3.

Claim:n+ 1

n→ 1 as n→∞.

Proof Let ε > 0 be given. The definition requires us to show that there is a natural number N

such that for all n ≥ N ,

∣∣∣∣n+ 1

n− 1

∣∣∣∣ < ε. So we look for N so that ∀n ≥ N ,

∣∣∣∣(1 +1

n

)− 1

∣∣∣∣ < ε,

which is equivalent to requiring that ∀n ≥ N ,

∣∣∣∣ 1n∣∣∣∣ < ε.

The N here will depend on ε (in fact in this example, as in most, no choice will work forall ε, so any choice of value for N will be in terms of ε). Since the last inequality is equivalent

to ∀n ≥ N ,1

ε< n we take N to be any natural number greater than

1

ε, say for definiteness

[ε−1] + 1.

Example 2.3.3. Now consider the sequence

((−1)n

3n

)n∈N

of 2.1.5. The claim is that

(−1)n

3n→ 0 as n→∞.

Proof: Let ε > 0 be given. To prove the claim we must find N ∈ N so that ∀n ≥ N ,∣∣∣∣(−1)n

3n− 0

∣∣∣∣ < ε. In fact N = [ε−1] + 1 (which implies 1N < ε) works here. For suppose

that n ≥ N . Then

∣∣∣∣(−1)n

3n− 0

∣∣∣∣ =

∣∣∣∣(−1)n

3n

∣∣∣∣ =|(−1)n||3n|

(by Lemma 2.2.3(a)) =1

3n≤ 1

n(since it

is easy to show (by induction) that ∀n ∈ N, 3n ≥ n, and hence that1

3n≤ 1

n). But

1

n≤ 1

N< ε,

and we are done.

The definition of convergence is notoriously difficult for students to take in, and rather fewgrasp it straight away. So here’s a different, but equivalent, way of saying the same thing.Maybe one of the definitions might make more sense to you than the other - it can be usefulto look at something from different angles to understand it. And, of course, the more examplesyou do, the more quickly you will get the picture.

By Lemma 2.2.3(c) the condition |an − l| < ε is equivalent to saying that l− ε < an < l+ ε.This in turn is equivalent to saying that an ∈ (l − ε, l + ε). So, to say that an → l as n → ∞is saying that no matter how small an interval we take around l, the terms of the sequence(an)n∈N will eventually lie in it, meaning that, from some point on, every term lies in thatinterval. You might like to look at the formula for N (in terms of ε) in Example 2.3.2 and

check that (taking ε = 110),

n+ 1

n∈ (1 − 1

10 , 1 + 110) for all n ≥ 11, and that (taking ε = 3

500)

13

n+ 1

n∈ (1 − 3

500 , 1 + 3500) for all n ≥ 167 (so, for ε = 3

500 we can take N = 167 or any integer

≥ 167 - the definition of convergence doesn’t require us to choose the least N that will work).

Example 2.3.4. This is more of a nonexample really. Consider the sequence ((−1)n)n∈N (of2.1.4). Then there is no l such that the terms will eventually all lie in the interval (l− 1

2 , l+12).

This is because if l ≤ 12 then the interval does not contain the number 1, and infinitely many

of the terms of the sequence are equal to 1, and if l ≥ 12 then the same argument applies to the

number −1. Hence there is no number l such that (−1)n → l as n→∞. In general, if there isno l such that an → l as n→∞ then we say that the sequence (an)n∈N does not converge.

Example 2.3.5. Another nonexample. Consider the sequence (n2)n∈N of Example 2.1.2. Thisdoes not converge either. Here is a rigorous proof of this fact using the original Definition 2.3.1.

Suppose, for a contradiction, that there is some l such that an → l as n→∞. Choose ε = 1in Definition 2.3.1. So there must exist some N ∈ N such that |n2 − l| < 1 for all n ≥ N . Inparticular |N2−l| < 1 and |(N+1)2−l| < 1. Therefore |(N+1)2−N2| = |(N+1)2−l+l−N2| ≤|(N +1)2− l|+ |l−N2| (by the triangle-inequality). But each of the terms in the last expressionhere is less than 1, so we get that |(N+1)2−N2| < 1+1 = 2. However, |(N+1)2−N2| = 2N+1,so 2N + 1 < 2, which is absurd since N ≥ 1.

This last example is a particular case of a general theorem. Namely, if the set of terms ofa sequence is not bounded (as is certainly the case for the sequence (n2)n∈N) then it cannotconverge. We develop this remark precisely now.

Definition 2.3.6. Let S be any non-empty subset of R.

• Then we say that S is bounded above if there is a real number M such that ∀x ∈ S,x ≤M . Such an M is an upper bound for S.

• Similarly, S is bounded below if there is a real number m such that ∀x ∈ S, x ≥ m.Such an m is a lower bound for S.

• If S is both bounded above and bounded below, then S is bounded.

Example 2.3.7. (1) Let S = {17,−6,−25, 25, 0}. Then S is bounded: an upper bound is 25(or any larger number) and a lower bound is −25 (or any smaller number). In fact one canshow easily by induction on the size of X that if X is a non-empty, finite subset of R then Xis bounded.(2) Intervals like (a, b] or [a, b], etc, for a < b are bounded above, and below, by a, respectivelyb. However an interval like (a,∞) is only bounded below.(3) Let S = {x ∈ R : x2 < 2}. Since 1.52 > 2 it follows that if x2 < 2 then −1.5 < x < 1.5. Ofcourse there are better bounds, but this is certainly sufficient to show that S is bounded.

Applying this to the set {an : n ∈ N} of terms of a sequence we make the following definition.

Definition 2.3.8. A sequence (an)n∈N is bounded if there exists M ∈ R+ such that for alln ∈ N, |an| ≤M .

Theorem 2.3.9 (Convergent implies Bounded). Suppose that (an)n∈N is a convergent sequence(i.e. for some l, an → l as n→∞). Then (an)n∈N is a bounded sequence.

Proof. Choose l so that an → l as n → ∞. Now take ε = 1 in the definition of convergence.Then there is some N ∈ N such that for all n ≥ N , |an − l| < 1.

14

But |an| = |(an − l) + l| ≤ |an − l| + |l| (by the triangle inequality). Thus for all n ≥ N ,|an| ≤ 1 + |l|. So if we take M= max{|a1|, |a2|, . . . , |aN−1|, 1 + |l|} we have that |an| ≤ M forall n ∈ N, as required. 2

We give one last example before we prove some more general theorems about convergence.

Example 2.3.10. Let an =8n

13

n3 +√n

. Then an → 0 as n→∞.

Proof: Let ε > 0 be given. We must find N (which will depend on the given ε) such that for

all n ≥ N ,

∣∣∣∣∣ 8n13

n3 +√n− 0

∣∣∣∣∣ < ε, i.e. so that8n

13

n3 +√n< ε (since the terms are positive, we may

remove the modulus signs).

So the game is to find a decent looking upper bound for8n

13

n3 +√n

so that one can easily read

off what N must be. Well, we certainly have n3 ≥√n (for all n), so

8n13

n3 +√n≤ 8n

13

√n+√n

=

4n13

√n

.

(You must get used to tricks like this: replacing an expression in the denominator by asmaller expression and/or replacing the numerator by a larger term always increases the size ofthe fraction, provided everything is positive.)

Now4 · n

13

√n

=4 · n

13

n12

=4

n12− 1

3

=4

n16

.

So we have that8n

13

n3 +√n≤ 4

n16

.

So we will have the desired inequality, namely

∣∣∣∣∣ 8n13

n3 +√n− 0

∣∣∣∣∣ < ε, provided that4

n16

< ε

which, after rearrangement, is provided that

(4

ε

)6

< n. So to complete the argument we just

take N to be any natural number greater than

(4

ε

)6

, say N =

[(4

ε

)6]

+ 1.

2.4 The Completeness Property for R

If we consider the set S = {x ∈ Q : x2 < 2} of rational numbers with square less than 2, thenwe can find upper bounds for S: 10 or, better, 1.5 or, better still, 1.4142135623731, ... but thereis no optimal (in the sense of least) upper bound in Q.

Definition 2.4.1. Let S be a non-empty subset of R and assume that S is bounded above. Thena real number M is a supremum or least upper bound of S if the following two conditionshold:

(i) M is an upper bound for S, and

(ii) if M ′ ∈ R satisfies M ′ < M then M ′ is not an upper bound for S.

It is easy to see that a set S cannot have more than one supremum. For if M1 and M2 wereboth suprema and M1 6= M2, then either M1 < M2 or M2 < M1 (this is the sort of situation

15

where one automatically uses rule (Ord 1)). But, in either case, this contradicts condition2.4.1(ii) above.

Thus we have proved:

Lemma 2.4.2. The supremum of a set S, if it exists, is unique and we then denote it bysup(S).

Lemma 2.4.3. Let S be a non-empty subset of R and assume that M = sup(S) exists. Then∀ε ∈ R+, ∃x ∈ S such that M − ε < x ≤M .

Proof. Let ε > 0 be given. Let M ′ = M − ε. Then M ′ < M , so by 2.4.1(ii), M ′ is not an upperbound for S. So there must be some x ∈ S such that x > M ′. Since M is an upper bound forS (by 2.4.1(i) ), we also have x ≤M . So M ′ < x ≤M , i.e. M − ε < x ≤M , as required. 2

Example 2.4.4. For S = {17,−6,−25, 25, 0} we clearly have that sup(S) = 25. Indeed, if Sis any non-empty, finite subset of R then it contains a greatest element and this element isnecessarily its supremum.

Example 2.4.5. However, it need not be the case that a set contains its supremum as a member.Indeed, if a < b then we claim that sup(a, b) = b despite the fact that b /∈ (a, b).

Proof. Let us prove this. Certainly b is an upper bound for (a, b). To see that 2.4.1(ii) is alsosatisfied, suppose that M ′ < b. Let c = max{a,M ′}. Then certainly c < b and so the average,

d =c+ b

2of c and b satisfies M ′ ≤ c < d < b. Similarly, a < d < b. So d is an element of the set

(a, b) which is strictly greater than M ′. Hence M ′ is not an upper bound for (a, b), as required.2

It is now time to introduce the final property that characterises the reals. It is a succinct,but sweeping property of R. It is strong enough to imply facts as diverse as “the number 2 hasa square root” on the one hand and apparently easy things like “N has no upper bound in R”on the other.

Property 2.4.6. (The Completeness property of R) (A14) Any non-empty subset of R whichis bounded above has a supremum.

Remark. We will not prove the completeness property. After all, we’ve not actually definedwhat real numbers are, so we can’t begin to prove that they have such-and-such a property.The usual route is to carefully define the real numbers in terms of so-called Dedekind cuts (analternative approach is in terms of Cauchy sequences). The completeness axiom is then almostpart of the definition. We don’t have the time to give the details in this course but you can findthe construction in most analysis text books.

Let us list some consequences of the completeness property.

Example 2.4.7. Consider now the set S = {x ∈ R : x2 < 2} of Example 2.3.7. We have alreadynoted that it is bounded above, for example by 1.5. Thus by Property 2.4.6 it has a supremum,say s. We will see below that s2 = 2 and so s does indeed equal

√2. (Clearly s is the positive

square root of 2 since 1 ∈ S, so 1 ≤ s).So suppose, for a contradiction, that s2 6= 2. Then either s2 > 2 or s2 < 2. Assume first

that s2 > 2. What we want to do is find some (positive) number t that is small enough so that

(s− t) is still an upper bound for S. It might be tempting to take t to be the average t = s2−22 ,

but this does not quite work. However the idea is good:

16

Let t =s2 − 2

4s. Then t > 0. Also t <

s2

4s=s

4, so t < s and s− t > 0. Then we check that

s − t is also an upper bound for S, which contradicts the fact that s is the least upper boundfor S. The key point is to compute

(s− t)2 = s2 − 2st+ t2 ≥ s2 − 2st = s2 − 2s

(s2 − 2

4s

)= s2 −

(s2 − 2

2

)=s2

2+ 1 >

2

2+ 1 = 2.

This does it, since if x ∈ S then x2 < 2 < (s − t)2. As (s − t) > 0, it follows (see the firstexample sheet) that x < (s− t) contradicting the fact that s is a supremum for S.

We should also dispose of the case s2 < 2; that is left for the exercise sheet.

In fact, by using a similar method (though the details are rather more complicated), onecan also show that

• for any x ∈ R with x > 0, and any natural number n, there exists y ∈ R, with y > 0, suchthat yn = x. Such a y is unique and is denoted x

1n or n

√x (the positive real nth root of

x). Furthermore,

• for any x ∈ R, and any odd natural number n, there exists a unique y ∈ R (of the same

sign as x) such that yn = x. Again, we write this y as x1n .

Finally, here is a property of the reals that you probably took for granted. However let’sprove it using the completeness property.

Example 2.4.8. N is not bounded above.

Proof. To see this suppose, for a contradiction, that N is bounded above. Then the complete-ness property says that N has a supremum, say s = sup(S). Apply Lemma 2.4.3 with ε = 1

2 .Then the lemma guarantees that there is some x ∈ N such that s − 1

2 < x. Adding 1 to bothsides we get s + 1

2 < x + 1. But then we have x + 1 ∈ N and s < x + 1 which contradicts ourassumption that s was the supremum of, and hence an upper bound for, S. 2

Using similar ideas one can prove:

Example 2.4.9. (i) (See Problems for Week 2) ∀x, y ∈ R, if x < y then ∃q ∈ Q such thatx < q < y.

We should also make the obvious corresponding definition for lower bounds for sets of realnumbers.

Definition 2.4.10. Suppose that S is a non-empty subset of R and that S is bounded below.Then a real number m is an infimum (or greatest lower bound) of S if

(i) m is a lower bound for S, and

(ii) if m′ ∈ R and m < m′, then m′ is not a lower bound for S.

Again, one can show that the infimum of a set S is unique if it exists, and we write inf(S)for it when it does exist. One might think that we need a result like Property 2.4.6 to postulatethe existence of infima. But we can directly prove from that proposition.

Theorem 2.4.11 (Existence of infima). Every non-empty subset T of R which is bounded belowhas an infimum.

17

Proof. See the Problem Sheet for Week 1. As a hint, define T− = {−x : x ∈ T}. Thenyou should prove that T− has a supremum, M say. Now show that −M is the infimum of theoriginal set T . 2

So, to summarise, we are not going to define the real numbers from first principles - you canfind their construction in various sources and, in any case, you are used to dealing with them- but it is the case that everything we need about the reals follows from the axioms that welisted above for an ordered field, together with the completeness property. Indeed, these axiomscompletely characterise the reals. That is, and this is a rather remarkable fact: although thereare many different fields, and even many different ordered fields, if we add the completenessaxiom then there is just one mathematical structure which satisfies all these conditions - namelythe reals R.

2.5 Some General Theorems about Convergence

Theorem 2.5.1 (Uniqueness of Limits). A sequence can converge to at most one limit.

Proof. What we have to show is that if an → l1 as n→∞ and an → l2 as n→∞, then l1 = l2.So suppose that l1 6= l2. Say, wlog (“without loss of generality”), that l1 < l2. Let ε =

l2 − l12

, so ε > 0.

Since an → l1 as n→∞, there exists N ∈ N such that for all n ≥ N , |an − l1| < ε.Also, since an → l2 as n→∞, there exists N ′ ∈ N such that for all n ≥ N ′, |an − l2| < ε.Now, let m = max{N,N ′}.Then |am − l1| < ε and |am − l2| < ε.But l2 − l1 = |l2 − l1| = |(l2 − am) + (am − l1)| ≤ |l2 − am| + |am − l1| (by the triangle

inequality). But the right hand side here is strictly less than ε+ ε. That is, substituting in our

value ε =l2 − l1

2, we obtain l2 − l1 < l2 − l1, a contradiction! 2

So a sequence can have at most one limit. But when does it actually have a limit? The nextpiece of theory provides us with a criterion that covers many particular cases.

Definition 2.5.2. A sequence (an)n∈N is said to be

(i) increasing if for all n ∈ N, an ≤ an+1;

(ii) decreasing if for all n ∈ N, an+1 ≤ an;

(iii) strictly increasing if for all n ∈ N, an < an+1;

(iv) strictly decreasing if for all n ∈ N, an+1 < an.

A sequence satisfying any of these four conditions is called monotone or monotonic. Ifit satisfies (iii) or (iv) it is called strictly monotonic.

Theorem 2.5.3 (Monotone Convergence Theorem). Let (an)n∈N be a sequence that is bothincreasing and bounded. Then (an)n∈N converges.

Proof. Since the set {an : n ∈ N} is bounded, it has a supremum, l say. We show thatan → l as n→∞.

So let ε > 0 be given. We now use Lemma 2.4.3 which tells us that there is some x ∈ {an :n ∈ N} such that l − ε < x ≤ l.

Obviously x = aN for some N ∈ N. So l − ε < aN ≤ l.

18

Now for any n ≥ N we have that an ≥ aN (since the sequence (an)n∈N is increasing) and ofcourse we also have that an ≤ l (since l is certainly an upper bound for the set {an : n ∈ N},being its supremum).

So, for all n ≥ N , we have l − ε < an ≤ l, and hence l − ε < an ≤ l + ε. But (by 2.2.3(c))this is equivalent to saying that for all n ≥ N , we have |l − an| < ε. Thus an → l as n→∞ asrequired. 2

Example 2.5.4. Let an =n2 − 1

n2. Then (an)n∈N is an increasing sequence since an = 1− 1

n2≤

1 − 1

(n+ 1)2= an+1. Further, 0 ≤ an < 1 for all n, so (an)n∈N is also a bounded sequence.

Hence (an)n∈N converges.

In fact it is fairly easy to show directly from the definition of convergence that, with an as inthe example above, an → 1 as n→∞. However, Theorem 2.5.3 comes into its own in situationswhere it is far from easy to guess what the limit is:

Example 2.5.5. Let an =

(n+ 1

n

)n=

(1 +

1

n

)n. Then one can show (though it’s not partic-

ularly easy) that (an)n∈N is an increasing, bounded sequence. So it converges. But what is itslimit? It turns out to be e (the base for natural logarithms).

2.6 Exponentiation - a digression

This subsection is really a digression, so you do not need to remember the details but it uses theideas we have developed in an interesting way.

There are also some theoretical applications of Theorem 2.5.3 which allow one to give rig-orous definitions of some classical functions of analysis. For example, if x and y are positivereal numbers what does xy, or “x to the power y”, exactly mean? The answer “x multipliedby itself y times” doesn’t really make sense if y is not a natural number! Here we see the ideaof how to define this rigorously using Monotone Convergence although, since this is a bit of adigression, some steps will be skipped.

Let x, y be positive real numbers. We aim to give a rigorous definition of the real number

xy. If y ∈ Q, say y =n

mwith n,m ∈ N, then we may use the existence of roots (see Section

2.4) to define xy to be m√xn. But if y /∈ Q how should we proceed? We shall use Question 5 of

the Exercise sheet for Week 2, which tells us that we can approximate y arbitrarily closely byrational numbers q. We then show that the xq converge and we call the limit xy. The precisedetails are as follows.

Lemma 2.6.1. (a) Let y ∈ R. Then there exists a sequence (qn)n∈N which (i) is strictlyincreasing, (ii) has rational terms (i.e. qn ∈ Q for each n ∈ N) and (iii) converges to y.

(b) If q and s are positive rational numbers such that q < s, and if x ≥ 1, then xq < xs.

Proof. (a) Firstly, let q1 be any rational number strictly less than y (e.g. q1 = [y]− 1.)Let q2 be any rational number satisfying max{q1, y − 1

2} < q2 < y (using the result fromthe problem sheet). Now let q3 be any rational number satisfying max{q2, y − 1

3} < q3 < y (asabove).

We continue: once q1 < q2 < · · · < qn < y have been constructed, we choose a rationalnumber qn+1 satisfying max{qn, y − 1

n+1} < qn+1 < y.Clearly our construction has ensured that the sequence (qn)n∈N is (strictly) increasing and is

bounded above (by y). It therefore converges by the Monotone Convergence Theorem. However,

19

we have also guaranteed that for all n ∈ N, y− 1n < qn < y, from which it follows that the limit

is, in fact, y.1

(b) This is left as an exercise; see the solution to Question 4 from the Exercise sheet forWeek 3. 2

Construction of xy (outline).We first use the lemma to construct an increasing sequence (qn)n∈N of rational numbers that

converges to y. Part (b) of the lemma implies that (xqn)n∈N is a strictly increasing sequencewhich is bounded above by xN where N is any natural number greater than y (e.g. [y] + 1).Hence, by the Monotone Convergence Theorem, there is some l such that xqn → l as n→∞.We define xy to be this l.

We can extend this definition to negative y by the formula x−y =1

xy, and we also set x0 =

1. Finally, if 0 < x < 1, so x−1 ≥ 1, then we define xy to be (x−1)−y. We do not give any valueto xy for negative x.

With these definitions all the usual laws for exponentiation can now be established. Onefirst proves them for rational exponents and then shows that the laws carry over to arbitraryreal exponents upon taking the limit described above. This latter process is made easier bydeveloping an “algebra of limits” which we shall do in the next chapter.

The laws of exponentiation being referred to above are as follows (where x is assumedpositive throughout).

(E1) xy+z = xy · xz;

(E2) (x · y)z = xz · yz when x, y > 0;

(E3) (xy)z = xy·z;

(E4) if 0 < x < y and 0 < z < t, then xz < yz.

Similarly, if x ≥ 1 and 0 < z < t, then xz < xt.

Since E1-E4 encapsulate all we will need to know about exponention, we don’t need to referback to our particular construction. In particular, we will use (E4) a number of times, withoutparticular comment.

1Can you see this final step? As a hint, note that for any ε > 0 there exists n ∈ N with 0 <1

n+ 1< ε. Now

use Lemma 2.2.3(c) as usual.

20

Chapter 3

The Calculation of Limits

We now develop a variety of methods and general results that will allow us to calculate limitsof particular sequences without always having to go back to the original definition.

3.1 The Sandwich Rule

Roughly stated, this rule states that if a sequence is sandwiched between two sequences eachof which converges to the same limit, then the sandwiched sequence converges to that limit aswell. More precisely:

Theorem 3.1.1 (The Sandwich Rule). Let (an)n∈N and (bn)n∈N be two sequences and supposethat they converge to the same real number l. Let (cn)n∈N be another sequence such that for allsufficiently large n, an ≤ cn ≤ bn. Then cn → l as n→∞.

Proof. The hypotheses state that for some N ∈ N we have

an ≤ cn ≤ bn for all n ≥ N .

Now we write down what it means for limn→∞

an = ` = limn→∞

bn. So, fix ε > 0. Then there

exists N1 such that if n ≥ N1 then |`− an| < ε. Equivalently

`− ε < an < `+ ε for all n ≥ N1.

Similarly, there exists N2 such that if n > N2 then

`− ε < bn < `+ ε for all n ≥ N2.

Now, let M = max{N,N1, N2}. Then the displayed inequalities combine to show that:

for all n ≥M we have `− ε < an ≤ cn ≤ bn < `+ ε.

In particular, ` − ε < an ≤ cn ≤ bn < ` + ε, which is what we needed to prove to show thatlimn→∞

cn = `. 2

Remark 3.1.2. The Sandwich Rule is often applied when either (an)n∈N or (bn)n∈N is a constantsequence: if a ≤ cn ≤ bn for sufficiently large n and bn → a as n→∞, then cn → a as n→∞.(Just take an = a for all n ∈ N in 3.1.1.) Similarly, if an ≤ cn ≤ b for sufficiently large n andan → b as n→∞, then cn → b as n→∞.

Definition 3.1.3. A sequence (an)n∈N is a null sequence if an → 0 as n→∞.

21

Theorem 3.1.4. (i) If (an)n∈N is a null sequence then so are the sequences (|an|)n∈N and(−|an|)n∈N.(ii) (Sandwich rule for null sequences.) Suppose that (an)n∈N is a null sequence. Let (bn)n∈Nbe any sequence such that for all sufficiently large n, |bn| ≤ |an|. Then (bn)n∈N is also a nullsequence.

Proof. (i) This follows from the definitions. For any ε > 0, by definition there exists N suchthat |an| < ε for all n ≥ N . But then | |an| | = |an| < ε and so, by definition lim

n→0|an| = 0. The

same works for −|an| since | − |an| | = |an|.(ii) We are given that for some N ∈ N we have |bn| ≤ |an| for all n ≥ N . It follows that for

all n ≥ N , −|an| ≤ bn ≤ |an|. But by (i) both (|an|)n∈N and (−|an|)n∈N are null sequences.Hence by the Sandwich Rule 3.1.1 (with l = 0) it follows that (bn)n∈N is null. 2

Of course, this is really what we were doing in explicit cases like Question 2 of the Week 3Exercise Sheet.

Example 3.1.5. (i)(1n

)n∈N is null.

(ii)

(1

n2 + n32

)n∈N

is null.

Proof (i) We did this as part of Example 2.3.2.

(ii) For all n ∈ N,

∣∣∣∣ 1

n2 + n32

∣∣∣∣ =1

n2 + n32

≤ 1

n2=

∣∣∣∣ 1

n2

∣∣∣∣ ≤ ∣∣∣∣ 1n∣∣∣∣. So the result follows from

Theorem 3.1.4 and Part (i) (Or you could use Question 2(a) of the Week 3 Exercise Sheet.)

Lemma 3.1.6. (a) For all m ≥ 5 we have 2m > m2.(b) In particular, if an = 2−n, then limn→∞ an = 0.

Proof. (a) You may have seen this in the Foundations of Pure Mathematics course.Base case: For m = 5 we have 25 = 32 > 25 = 52.

So, suppose for some k ≥ 5 we have 2k > k2. Then in trying to relate 2k+1 and (k+ 1)2 onegets

2k+1 = 2 · 2k > 2 · k2 = (k + 1)2 + k2 − 2k − 1.

Now the key point is that, as k ≥ 5 we have (k2 − 2k − 1) = (k − 1)2 − 2 > 2. So from thedisplayed equation we get

2k+1 > (k + 1)2 +(k2 − 2k − 1) > (k + 1)2 + 2.

Thus, by induction 2m > m2 for all m ≥ 5.(b) Of course, you can prove this more directly, but part (a) implies that 0 < an = 2−n ≤ n−2.Thus, by Theorem 3.1.4 and Question 2(a) (of Problem Sheet 2) limn→∞(an) = 0. 2

Example 3.1.7.

(1

2n + 3n

)n∈N

is null.

Proof: For all n ∈ N, we have

∣∣∣∣ 1

2n + 3n

∣∣∣∣ =1

2n + 3n≤ 1

2n. So the result follows from Theo-

rem 3.1.4 and Lemma 3.1.6.

Example 3.1.8.

(n3

4n

)n∈N

is null.

Proof: By the lemma, ∀n ≥ 5,∣∣∣∣n34n

∣∣∣∣ =n3

4n=

n3

2n · 2n≤ n3

n2 · n2=

1

n.

So the result follows from 3.1.4 (and the fact that ( 1n) is null).

22

3.2 The Algebra of Limits.

Hopefully you now have a feel for testing whether a given sequence is convergent or not. Whatshould also be apparent is that we tend to repeat the same sorts of tricks. When that happens,one should suspect that there are general rules in play. This is the case here and we can convertthe sorts of manipulations we have been doing into theorems telling us when certain types offunctions converge (or not). The next theorem can be paraphrased as saying that: Limits ofconvergent series satisfy the same rules of addition and multiplication as numbers.

Theorem 3.2.1. [The Algebra of Limits Theorem] Let (an)n∈N, (bn)n∈N be sequences and leta, b be real numbers. Suppose that limn→∞ an = a and that limn→∞ bn = b. Then:

(i) limn→∞ |an| = |a|;

(ii) for any k ∈ R, limn→∞ kan = ka;

(iii) limn→∞(an + bn) = a+ b;

(iv) limn→∞(an · bn) = a · b;

(v) if bn 6= 0 (for all n), and b 6= 0, then limn→∞anbn

=a

b.

(vi) in particular, if bn 6= 0 (for all n), and b 6= 0, then limn→∞1

bn=

1

b.

Remarks: (i) In many of the proofs we are going to want to change ε in the definition ofconvergence, so let’s rephrase the definition of convergence as: a sequence (cn) converges to `if, for all η > 0 there exists M such that,

for all m ≥M we have `− η < am < `+ η.

(ii) Secondly, we will frequently use the various versions of the triangle inequality from 2.2.1and 2.2.2, generally without explicit mention.

Proof. (i) Let ε > 0 be given. Choose N so that for all n ≥ N , |an − a| < ε. Now byLemma 2.2.3(b),

∣∣|an| − |a|∣∣ ≤ |an − a|. Hence, for all n ≥ N ,∣∣|an| − |a|∣∣ < ε, as required.

(ii) Firstly, if k = 0 the result is obvious since limn→∞ 0 = 0. So suppose that k 6= 0.

Let ε > 0 be given. Here we take η =ε

|k|. Then η > 0 so we may apply the Remark to

obtain M ∈ N such that for all m ≥ M , we have |am − a| < η. Multiplying by |k| (which is> 0) we get that |k| · |an − a| < |k| · η = ε for all n ≥ M . Therefore, for n ≥ M , we have|kan − ka| = |k(an − a)| < ε. This shows that kan → ka as n→∞, as required.

(iii) Let ε > 0 be given. We take (for reasons that will soon become apparent) η =ε

2in the

Remark applied to the two sequences. Thus we can find M1 such that

a− η ≤ an ≤ a+ η for all n ≥M1

and M2 such thatb− η ≤ bn ≤ b+ η for all n ≥M2.

Adding these equations (and using the triangle inequality as always) we find that, for N =max{M1,M2}

(a+ b)− 2η ≤ (an + bn) ≤ (a+ b) + 2η for all n ≥ N.

23

Since ε = 2η this means that, for N = max{M1,M2}

(a+ b)− ε ≤ (an + bn) ≤ (a+ b) + ε for all n ≥ N.

So an + bn → a+ b as n→∞, as required.

(iv) This is harder. We first use the fact (Theorem 2.3.9) that (an)n∈N and (bn)n∈N are boundedsequences. So we may choose B ∈ R+ such that:

for all n ∈ N, |an| ≤ B (1)

Now let ε > 0 be given. Let M = max{B, |b|} and take η =ε

2M; so η > 0. Now use the

Remark to choose M1 ∈ N such that

∀n ≥M1, |an − a| < η (2).

Similarly, choose M2 ∈ N such that

∀n ≥M2, |bn − b| < η (3).

Let N = max{M1,M2}. Then for all n ≥ N we have

|anbn − ab| = |anbn − anb + anb− ab|≤ |anbn − anb|+ |anb− ab| (by the triangle inequality)

= |an||bn − b|+ |an − a||b| (since |xy| = |x| · |y|, see Lemma 2.2.3(a))

≤ |B||bn − b|+ |an − a||b| (by (1))

≤ |M ||bn − b|+ |an − a||M | (by the definition of M)

< M · η + η ·M (by (3), (2))

= 2 ·M · η = ε.

Thus anbn → ab as n→∞, as required.

(v) This is immediate from parts (iv) and (vi). Indeed, if1

bn→ 1

bas n→∞, then we can then

apply part (iv) to conclude that an ·1

bn→ a · 1

bas n→∞.

(vi) We start with:

Claim: There exists N1 ∈ N such that ∀n ≥ N1, one has |bn| > 12 |b|.

Proof of Claim: Use the Remark with η = 12 |b| (which is strictly positive as |b| 6= 0). This gives

N1 ∈ N such that ∀n ≥ N1, we have |b| − 12 |b| < |bn| < |b|+

12 |b|, which is what we wanted.

We now complete the proof. So let ε > 0 be given.

Let η =|b|2

2· ε. Then η > 0 and hence by the Remark there exists N2 ∈ N such that, for all

n ≥ N2, we have |bn − b| < η. Now let N = max{N1, N2}. Then, for n ≥ N we have∣∣∣∣ 1

bn− 1

b

∣∣∣∣ =

∣∣∣∣b− bnbnb

∣∣∣∣ =|b− bn||bn||b|

However, |bn| >|b|2

(because n ≥ N1) and hence |bn| · |b| >|b|2

2. Therefore

|b− bn||bn||b|

<2|b− bn||b|2

<2 · η|b|2

(because n ≥ N2).

24

Since η =|b|2

2· ε we obtain that

∣∣∣ 1bn− 1b

∣∣∣ < ε for all n ≥ N , as required. 2

Theorem 3.2.2. Let (an)n∈N be a null sequence and let (bn)n∈N be a bounded sequence (notnecessarily convergent). Then (an · bn)n∈N is a null sequence.

Proof. Exercise. 2

Example 3.2.3. For any fixed positive real number p,1

np→ 0 as n→∞.

Proof: Let ε > 0 be given. We want to show that1

np< ε for all large n. But

1

np< ε ⇐⇒ np >

1

ε⇐⇒ n >

(1

ε

)1/p

,

(where the final equivalence uses E4 of Section 2.6). So, we take N to be [ε− 1

p ] + 1. Thus, if

n ≥ N then n > ε− 1

p and so the above computation shows that1

np< ε. Therefore

∣∣∣∣ 1

np− 0

∣∣∣∣ < ε

and hence limn→01

np= 0.

Aside: By the way, if you do not want to use E4 here, you could also use the argument fromthe solutions to the Exercise sheet for Week 3. So, (using that remark) if we take q = [p] thenq ∈ Q+ and n−p ≤ n−q. So (by the Sandwich Theorem again) it is enough to prove Example3.2.3 for q or, as is the same statement, for p rational. This is the case of E4 proved explicitlyon that solution sheet.

Example 3.2.4.n2 + n+ 1

n2 − n+ 1→ 1 as n→∞

Proof: Divide top and bottom of the nth term by n2. (This trick, and variations of it, is themain idea in all examples like this.) Thus

n2 + n+ 1

n2 − n+ 1=

1 + 1n + 1

n2

1− 1n + 1

n2

.

We now apply the above example with p = 1 and p = 2 to deduce that 1n → 0 and that

1n2 → 0 as n→∞. (Of course, these are also examples we have done several times before!)

So by the Algebra of Limits Theorem 3.2.1(ii, iii)

1 +1

n+

1

n2−→ 1 + 0 + 0 = 1 as n→∞

and

1− 1

n+

1

n2−→ 1 + (−1) · 0 + 0 = 1 as n→∞.

So by using Algebra of Limits Theorem 3.2.1(vi)) again we obtain that

1 + 1n + 1

n2

1− 1n + 1

n2

→ 1

1= 1,

that is, n2 + n+ 1n2 − n+ 1

→ 1 as n→∞.

25

For many functions given as polynomials or fractions of polynomials we can apply methodsas in the above example. It is however very important to use this result (and earlier results)only for convergent sequences.

For example, we have seen that (an) is not convergent when an = (−1)n. This is an exampleof a bounded sequence that is not convergent (so the converse of Theorem 2.5.3 fails).

There are however cases where we can deduce negative statements.

Example 3.2.5. If p > 0 then (np) is an unbounded sequence (and hence is not convergent byTheorem 2.3.9).Proof. Given any ` we can certainly find a natural number n > (`)1/p since the natural numbersare unbounded. However, using (E4) of Section 2.6 we have np > ` ⇐⇒ n > (`)1/p. Thus thisalso proves that (np) is unbounded.

It is worth emphasising that in proving unboundedness (or any counterexample) it’s notnecessary that every value of n is “bad”, just that we can always find a larger “bad” value of n.

Example 3.2.6. (i) If (an) is unbounded and (bn) is convergent, then (an + bn) is unbounded.Similarly, (kan) is unbounded whenever k 6= 0.Proof. Since (bn) is convergent, it is bounded, say |bn| < B for all n. But now given any ` thereexists some an with |an| > B + `. Hence by the triangle inequality, |an + bn| > `, as required.The proof for products is similar.

26

Chapter 4

Some Special Sequences

In the previous chapter we saw how to establish the convergence of certain sequences that werebuilt out of ones that we already knew about. In this chapter we build up a stock of basicconvergent sequences that will recur throughout your study of analysis.

4.1 Basic Sequences

Lemma 4.1.1. For any c > 0, c1/n → 1 as n→∞.

Proof. This proof has several steps, as follows:Step I: Prove that (1 + x)n ≥ 1 + nx for all x ≥ 1 and n ∈ N.

Step II: By taking x =y

nin (a), deduce that for all y > 0 and n ∈ N, (1 + y)

1n ≤ 1 +

y

n.

Step III: Hence show that for fixed c > 1, one has c1n → 1 as n→∞.

Step IV: Complete the proof.

Proofs of Steps:Step I: Just use the binomial theorem—which says that

(1 + x)n = 1 + nx+ (lots of positive terms).

Step II: Take nth roots in Part I and use (E4) of Section 2.6 to get (1 + x) ≥ (1 + nx)1/n.

Substituting x =y

ngives

(1 +

y

n

)≥ (1 + y)1/n, which is what we want.

Step III: Fix ε > 0. For any c > 1 we can write c = 1 + y for y > 0. Now we choose

N =[yε

]+ 1, so that y

N < ε. Then for any n ≥ N we have:∣∣c1/n − 1∣∣ = c1/n − 1 since clearly c1/n > 1 (or use E4 of Section 2.6)

= (1 + y)1/n − 1

≤(

1 +y

n

)− 1 by part II

= yn ≤

yN < ε.

Step IV: So it remains to consider the case when 0 < c < 1. But then d = c−1 > 1 so by Part

III, d1/n → 1 as n→∞. By the Algebra of Limits Theorem 3.2.1(vi),1

d1/n→ 1

1 = 1 as n→∞.

Finally, since1

d1/n= c1/n, we have shown that c1/n → 1 as n→∞. 2

For completeness, recall Question 4 from the Week 4 Example Sheet:

27

Lemma 4.1.2. For 0 < c < 1 we have cn → 0 as n→∞.

Proof. Here is a different (and much simpler) proof. Let d = 1/c > 1 and write d = 1 + x, forx > 0. Then dn = (1 + x)n = 1 + nx + · · ·xn by the binomial theorem. As all the terms arepositive, dn > nx. Thus for any number E, we have dn > E whenever n ≥ E/x.

Now suppose that ε is given. Then cn < ε ⇐⇒ dn =1

cn>

1

ε. So, using the computations

of the last paragraph, take N = 1 +

[1

xε

], where x = 1/c− 1. Then, for n ≥ N we have n >

1

xε

and so dn >1

εby the last paragraph. In other words, cn < ε, as required. 2

We now consider several sequences where it is harder to see what is going on; typically they

are of the form an =bncn

where both bn and cn get large (or small) as n gets large. What matters,

for the limit, is how quickly bn grows (or shrinks to 0) in comparison with cn. Roughly, if (bn)nhas a higher “order of growth”1 than cn then the modulus of the ratio will tend to infinity andthere will be no limit; if (bn)n and (cn)n have roughly the same order of growth, then we mightget a limiting value for (an)n; if (cn)n has the higher order of growth, then (an)n → 0 as n→∞.

Lemma 4.1.3. Suppose that (an) is a convergent sequence with limit `. For any integer M ,let bn = an+M (if M happens to be negative, we just take bn = 0 or any other number for1 ≤ n ≤ −M). Then lim

n→∞bn = `.

Proof. The point is that (apart from starting at different places) the two sequences are reallythe same, so ought to have the same limit. More formally, let ε > 0. Then we can findN ∈ N such that |` − an| < ε for all n ≥ N . Hence |` − bn| = |` − an+M | < ε for all for alln ≥ max{1, N −M}. So the sequence (bn) converges. 2

Lemma 4.1.4. For any c,cn

n!→ 0 as n→∞.

Proof. Set an =cn

n!. Since it suffices to prove that |an| → 0 (see Theorem 3.1.4), we can

replace c by |c| and assume that c > 0. In particular, an > 0 for all n.

Now, an+1 = an·c

n+ 1. For all n ≥ 2c, we have

c

n+ 1< 1

2 and hence an+1 = an·c

n+ 1<an2

.

So, fix an integer N ≥ c. Then for all m > 0 a simple induction says that 0 < am+N <aN2m

=

aN2−m.But we know that limm→0 (2−m) = 0 by Lemma 4.1.2. Thus by the Algebra of Limits The-

orem 3.2.1 limm→0

aN (2−m) = 0 and by the Sandwich Theorem limm→0

(am+N ) = 0. By Lemma 4.1.3,

limn→0

(an) = 0. 2

Next, two special sequences where the proofs are harder.

Lemma 4.1.5. The sequence (n1n )n∈N converges to 1 as n→∞.

Proof. This is not so obvious.Throughout the argument we consider only n ≥ 2.

1By “order of growth” we mean something coarser than “rate of growth”, that is, derivative. (The rate ofgrowth = derivative is, however, relevant to computing limits, see L’Hopital’s Rule which we discuss later - itsays that we can compute limits by replacing functions by their derivatives.

28

Let kn = n1/n − 1. Then E4 of Section 2.6 says that kn > 0 and clearly n = (1 + kn)n. Bythe binomial theorem

n = (1 + kn)n = 1 + nkn +n(n− 1)

2k2n + · · ·+ knn.

Since all the terms are positive, n >n(n− 1)

2k2n, and hence k2n <

2n

n(n− 1)=

2

n− 1. So, for all

n ≥ 2 we have that 0 < kn <

√2√

n− 1.

Now

√2√

n− 1→ 0 as n → ∞ (exercise), so by the Sandwich Rule, kn → 0 as n → ∞. But

n1n = 1 + kn, so by the Algebra of Limits Theorem, n

1n → 1 as n→∞. 2

Lemma 4.1.6. Fix c with 0 < c < 1 and fix k. Then limn→∞ nk · cn = 0.

Remark. This proof is quite subtle, and it is included for completeness. Once we haveL’Hopital’s Theorem, we can give a very quick proof.

Proof. If k = 0 then the result is Lemma 4.1.5. Hence the result is also true for k ≤ 0 by theSandwich Rule (as 0 < nk · cn ≤ cn if k ≤ 0).

So we may assume that k > 0.Let us first assume that k ∈ N. Now n

1n → 1 as n → ∞ (by 4.1.2). Hence, by the AoL

(Algebra of Limits Theorem), n1n · n

1n → 1 · 1 = 1 as n → ∞. By repeatedly multiplying by

n1n we see that (n

1n )m → 1 as n → ∞ for any positive integer m, and in particular for m = k.

Another use of AoL gives that (n1n )k ·c→ c as n→∞ or, to put this another way, (nk)

1n ·c→ c

as n→∞.

Now let ε =1− c

2so that ε > 0 by our assumption on c.

Choose N ∈ N so that for all n ≥ N we have that

|(nk)1n · c− c| < ε.

We may rewrite this as

c− ε < (nk)1n · c < c+ ε (∀n ≥ N).

Raising to the power n gives

0 < nk · cn < (c+ ε)n (∀n ≥ N).

However, c + ε = c +1− c

2=

1 + c

2and

∣∣∣∣1 + c

2

∣∣∣∣ =1 + c

2< 1 (as 0 < c < 1) and so by by

Lemma 4.1.2 we have that (c+ ε)n → 0 as n→∞. Thus by the Sandwich Rule nk · cn → 0 asn→∞ as required.

Now in the case that k is (positive and) not an integer we simply observe that 0 < nk · cn ≤n[k]+1 · cn and, since [k] + 1 is an integer we have that n[k]+1 · cn → 0 as n→∞ by what we’vejust shown.

Hence nk · cn → 0 as n→∞ by the Sandwich Rule. 2

Example 4.1.7. Find: (1) limn→∞

n√3n.

(2) limn→∞

(−12)n.

(3) limn→∞

n!

nn.

29

Proofs: (1) Here we can break it up to getn√

3n = 31/nn1/n. Now Lemmas 4.1.1 and 4.1.5 showthat lim

n→∞31/n = 1 = lim

n→∞n1/n. Thus by the Algebra of Limits Theorem, lim

n→∞

n√3n = 1.

(2) We know from Lemma 4.1.2 that limn→∞

(12)n = 0, and so Theorem 3.1.4 (or 3.2.2 implies that

limn→∞

(−1

2

)n= 0.

(3) This does not follow directly from our results, but think about individual terms:

n!

nn=

1 · 2 · 3 · · ·nn · n · n · · ·n

=1

n

2

n· · · n

n≤ 1

n.

Since all the terms are positive and limn→∞

1

n= 0, we can use the Sandwich Theorem to get

limn→∞

n!

nn= 0.

4.2 New Sequences from Old

Example 4.2.1.n100 + 2n

2n + n2→ 1 as n→∞.

The general method with sequences like this is to find the fastest growing term, and thendivide top and bottom by it. The fastest-growing term is usually found using the followingobservation.

If we have a fractionanbn

where we know thatanbn→ 0 as n→∞, then (bn) has higher order

of growth than (an).For example, if we just have powers of n, then the highest positive power grows fastest and

we can divide all terms by it. If n occurs in the exponent, as in cn, then provided that c > 1,such a term will always have higher order of growth than any fixed power of n (by 4.1.6). If thereare several terms of the form cn, then the one with the largest c will grow fastest. If, however,there are n!’s around (or nn terms) then they will grow faster than any cn term, no matter howlarge the constant c is. Also Example 4.1.7 shows that nn has higher order of growth than n!.

So in this example 2n is the fastest-growing term. Therefore we divide top and bottom byit:

n100 + 2n

2n + n2=

n100

2n + 1

1 + n2

2n

Nown100

2n= n100·(12)n → 0 by Lemma 4.1.6 (with k = 100 and c = 1

2), andn2

2n= n2·(12)n → 0

by 4.1.6 (with k = 2 and c = 12).

Hence, by AoL,n100 + 2n

2n + n2→ 0 + 1

1 + 0= 1 as n→∞.

Example 4.2.2.106n + n!

3 · n!− 2n→ 1

3as n→∞.

The fastest-growing term is n! so we divide top and bottom by it:

106n + n!

3 · n!− 2n=

106n

n! + 1

3− 2n

n!

→ 0 + 1

3− 0=

1

3as n→∞.

30

Here we applied Lemma 4.1.4 with c = 106 to deduce that106n

n!→ 0, and again with c = 2

to deduce that2n

n!→ 0, and finally we applied AoL.

Alternative viewpoint. Another way to think about “relative orders of growth” is to drawup a table.

We will say that an goes to zero faster than bn if limn→∞

anbn

= 0. The first table shows common

functions ordered by how quickly they slip or plunge to zero.

Goes to zero fastest (top to bottom) comments

1

nnSee 4.1.7

1

n!See 4.1.4

cn for 0 < c < 1 See 4.1.6 and think of e−n =1

en

1

np= n−p for p > 0 you can put any polynomial in the denominator

1

ln(n)see below

c1n and n

1n for c > 0 These come last as they tend to 1; see 4.1.1 and 4.1.5.

Taking reciprocals we can get a table with fastest-growing terms first.

Fastest-growing (top to bottom) comments

nn See 4.1.7

n! See 4.1.4

dn for 1 < d See 4.1.6 and think of en

np for p > 0 See 3.2.3; you can also take any polynomial here

ln(n) see below

c1n and n

1n for c > 0 These come last as they tend to 1; see 4.1.1 and 4.1.5.

These tables include ln(n), which we have yet to discuss but you should have the intuitionthat “Since exponentials grow faster than any polynomial, so logs grow slower.” We show thisin the next example.

31

Example 4.2.3. For any 0 < c, we have limn→∞

ln(n)

nc= 0.

Proof: Just like Lemma 4.1.6, this is best done using L’Hopital’s rule. You can prove it using(a variant of) Lemma 4.1.6, but it is sufficiently messy I will leave it to L’Hopital’s rule.

Examples 4.2.4. Find the limits of the sequences (an) where:

(a) an =n3

3n(b) an =

3n

n3(c) an =

3n + n!

n!(d) an =

3!

n3(e)

n28 + 5n7 + 1

2n.

Answers (a) By the table, exponentials grow faster than polynomials (or use 4.1.6), so thistends to zero.

(b) This is really something we deal with in the next chapter, but notice it is the reciprocal of(a). So we would expect that it tends to 1

0 =∞. This is indeed a true statement, but can yousee how to prove it?

(c)3n + n!

n!=

3n

n!+1. By the table, limn→∞

3n

n!= 0 and so by the AoL Theorem limn→∞

(3n

n!+ 1

)=

1.

(d) Careful! This is just a constant times 1n3 and so (by the AoL or directly) it tends to zero.

(e) Here the table says that the limit is 0. Note that to prove it using Lemma 4.1.6,, you should

write it asn28

2n+ 5

n7

2n+

1

2n. Then 4.1.6, says each fraction has limit zero and so the AoL says

that

limn→∞

n28 + 5n7 + 1

2n= 0 + 5 · 0 + 0 = 0.

In the next example we will use the following general result:

Lemma 4.2.5. Suppose that (an)n∈N is a convergent sequence, say with limn→∞ an = `. Let rand s be real numbers such that r ≤ an ≤ s for all sufficiently large n ∈ N, say for all n ≥ M .Then r ≤ ` ≤ s.

Proof. Suppose (for a contradiction) that, ` < r. Let ε = r − ` and choose N1 so that for alln ≥ N1, |an − `| < ε. Set N = max{N1,M}. Then for any n ≥ N we have

`− ε < an < `+ ε = `+ (r − `) = r.

This contradicts the hypothesis that an ≥ r. The same argument shows that ` ≤ s. 2

Here is a more complicated type of example which we will see quite a lot.

Example 4.2.6. Let the sequence (an)n∈N be defined inductively by

a1 = 2,

and for n ≥ 1

an+1 =a2n + 2

2an + 1.

We show that (an)n∈N converges and then find the limit.To do this we first show that

(A) ∀n ∈ N, an ≥ 1, and

32

(B) ∀n ∈ N, an+1 ≤ an.

Finally,

(C) This forces (an) to converge. Now use the AoL to solve an equation for the limit `.

Proof of A: Now obviously a1 ≥ 1. So, suppose, for some natural number n ≥ 1 we have

an ≥ 1. Then an+1 =a2n + 2

2an + 1, so an+1 ≥ 1 ⇐⇒ a2n + 2

2an + 1≥ 1 ⇐⇒ a2n + 2 ≥ 2an + 1 (here we

are using the fact that by induction 2an + 1 is positive because an is). But

a2n + 2 ≥ 2an + 1 ⇐⇒ a2n − 2an + 1 ≥ 0 ⇐⇒ (an − 1)2 ≥ 0,

which is certainly true. So working back through these equivalences we see that an+1 ≥ 1 asrequired.

Remark. It is very important in this sort of argument that one has ⇐⇒ or at least ⇐ sinceone is trying to prove the first statement using the validity of the final statement. If you justhave ⇒ you would not be justified in drawing these conclusions.

Proof of B: We have

an+1 ≤ an ⇐⇒a2n + 2

2an + 1≤ an ⇐⇒ a2n + 2 ≤ 2a2n + an ⇐⇒ 0 ≤ a2n + an − 2.

But an ≥ 1 (by A), so a2n ≥ 1, and hence a2n + an ≥ 2, so we are done.

Step C. So we have now shown that (an)n∈N is a strictly decreasing sequence which is bounded(above by 2 and below by 1). Hence, by the Monotone Convergence Theorem (or, rather, by itsvariant in the Exercises for Week 3) it converges. Let its limit be `. We calculate ` explicitlyas follows.

We first note that, by Lemma 4.2.5, as an ≥ 0 for all n, then ` ≥ 0 also holds.

By repeated use of the Algebra of Limits theorem, we have a2n → `2, a2n + 2 → `2 + 2,2an + 1→ 2`+ 1 and, finally,

a2n + 2

2an + 1→ `2 + 2

2`+ 1as n→∞

(note that the denominator is not zero since ` ≥ 0.) Thus limn→∞ an+1 =`2 + 2

2`+ 1.

However, Lemma 4.1.3 shows that limn→∞

an+1 = limn→∞

an. So putting these equations together

we see that`2 + 2

2`+ 1= `.

We can now solve this for `: rearranging the equation`2 + 2

2`+ 1= ` gives `2 + 2 = 2`2 + ` and

hence gives the quadratic equation `2 + `− 2 = 0, or (` + 2)(`− 1) = 0. Thus, either ` = 1 or` = −2. But we know that ` ≥ 0, so ` 6= −2 and hence ` = 1. Thus

limn→∞

an = 1.

Remarks 4.2.7. It is important to realise that the calculation of the limit in Example 4.2.3 wasvalid only because we already knew that the limit existed.

To bring this point home, reflect on the following (incorrect!) proof that (−1)n → 0 asn→∞ (which we know to be false from Example 2.1.4): Let an = (−1)n. Then the sequence(an)n∈N is defined inductively by a1 = −1 and an+1 = −an. Let l = limn→∞ an. Then by theAlgebra of Limits Theorem we have limn→∞(−an) = −l, i.e. limn→∞ an+1 = −l. But, as above,limn→∞ an = limn→∞ an+1, so l = −l, whence 2l = 0 and so l = 0 as required!

More subtle examples will appear on future Exercise sheets.

33

4.3 Newton’s Method for Finding Roots of Equations - optional

This sub-section is an aside that will not be covered in the class; as such, it is not examinable,but you might find it interesting and/or useful.

Suppose that we wish to find a solution to an equation of the form

f(x) = 0

where f : R→ R is some function (e.g. we shall take f(x) to be x2 − 2 below and thereby lookfor a square root of 2).

Newton’s method is as follows.Let x1 be a reasonable approximation to a solution of the equation. For n ≥ 1, let

xn+1 = xn −f(xn)

f ′(xn).

Then, under suitable assumptions on x1 and f , it can be shown that the sequence (xn)n∈Nwill converge to a solution and, further, is a very good way of finding closer and closer approx-imations to a solution.

Example 4.3.1. Let x1 = 1 and f(x) = x2 − 2. Then Newton’s method suggests looking at thesequence defined inductively by

xn+1 = xn −(x2n − 2

2xn

)Simplifying this gives:

xn+1 =xn2

+1

xn.

Now one can show that this sequences converges but let us just assume that here and let λdenote the limit. We check that λ =

√2.

Now one can easily show that λ 6= 0, so we may apply the Algebra of Limits Theorem toobtain

limn→∞

(xn2

+1

xn

)=λ

2+

1

λ.

But also xn+1 → λ as n→∞, so λ = λ2 + 1

λ and it is a simple matter to solve this equation

to obtain λ =√

2 as required.

Let us now evaluate a few terms to see how well they approximate√

2.We have

x1 = 1

x2 =1

2+

1

1=

3

2' 1 · 5000000

x3 =3

4+

2

3=

17

12' 1 · 4166667

x4 =17

24+

12

17=

577

408' 1 · 4142156

x5 =577

816+

408

577' 1 · 4142136.

In fact, to seven decimal places we have that√

2 is indeed equal to1 · 4142136.

34

Example 4.3.2. Let us calculate√

10 by Newton’s method. We take f(x) = x2− 10 and x1 = 3.The Newton sequence for this function is defined inductively by

xn+1 = xn −(x2n − 10

2xn

)which, upon simplification, becomes

xn+1 =xn2

+5

xn.

[Check: If xn → λ as n→∞, then by the usual argument, λ = λ2 + 5

λ , which solves to λ =√

10.

So the limit, if it exists (which we assume here), is√

10.]We have

x1 = 3

x2 =3

2+

5

3=

19

6' 3 · 1666667

x3 =19

12+

30

19=

721

228' 3 · 1622807

In fact, to seven decimal places,√

10 = 3 · 1622776.

35

Chapter 5

Divergence

Example: Recall from Theorem 2.3.9 that if a sequence is convergent then it is bounded. Wecan turn this around and see that certain sequences are not convergent simply because they are

not bounded; for example

(en

n2

)n∈N

(or some of the examples on the Exercise sheet for Week

6). To see this: we know from Lemma 4.1.6 that

(n2

en

)→ 0. So for any ε; in particular ε =

1

K

there exists N such thatn2

en<

1

Kfor n ≥ N . Taking reciprocals we get

en

n2> K for all n ≥ N.

So the sequence

(en

n2

)n∈N

is unbounded and hence not convergent. Of course, this argument

is completely general, so we will make a definition and a theorem out of it.

5.1 Sequences that Tend to Infinity

Definition 5.1.1. A sequence (an)n∈N is divergent if it is not convergent, i.e. if there is nol ∈ R such that an → l as n→∞.

Definition 5.1.2. We say that a sequence (an)n∈N tends to infinity as n tends to infinityif for each positive real number K, there exists an integer N (depending on K) such that for alln ≥ N , an > K. In this case we write an →∞ as n→∞, or limn→∞ an =∞.

Example 5.1.3. The sequence ((−1)n)n∈N is divergent since there is no l ∈ R such that (−1)n → las n → ∞. (see Example 2.3.4). But it does not tend to infinity as n → ∞ because one maysimply take K = 1: there is clearly no N such that for all n ≥ N , (−1)n > 1.

Example 5.1.4. The sequence (√n)n∈N is not bounded and so is divergent (by Theorem 2.3.9).

It does tend to infinity as n → ∞. For let K > 0 be given. Choose N = [K2] + 1. Then ifn ≥ N we have that n ≥ [K2] + 1 > K2. Hence

√n >√K2 = K, as required.

Example 5.1.5. Here is an example one needs to be careful about.Consider the sequence ((−1)n · n)n∈N. This is clearly not a bounded sequence, so is not

convergent. But it does not tend to infinity either. For if it did, taking K = 1 we would havean N such that for all n ≥ N, (−1)n ·n > 1. Which is absurd since half the time it is negative.

36

Theorem 5.1.6 (The Reciprocal Rule). (i) Let (an)n∈N be a sequence of non-zero real numbers

such that an →∞ as n→∞. Then1

an→ 0 as n→∞.

(ii) Let (an)n∈N be a sequence of non-zero real numbers such that for all sufficiently large n,

an > 0. Assume that the sequence (1

an)n∈N is null. Then an →∞ as n→∞.

Proof. (i) Suppose that an →∞ as n→∞ and let ε > 0 be given.

Then1

ε> 0, so taking K =

1

εin Definition 5.1.2, there exists N ∈ N such that for all

n ≥ N , an > K, i.e. for all n ≥ N , an >1

ε. Hence for all n ≥ N , 0 <

1

an< ε. Thus for all

n ≥ N we have that

∣∣∣∣ 1

an− 0

∣∣∣∣ < ε, which proves that1

an→ 0 as n→∞.

(ii) Now suppose that1

an→ 0 as n→∞. Let K > 0 be given.

Then1

K> 0, so taking ε =

1

Kwe see that there exists an N such that for all n ≥ N ,∣∣∣∣ 1

an− 0

∣∣∣∣ < ε. We may also take N large enough so that we have an > 0 for all n ≥ N . Thus

for all n ≥ N , 0 <1

an< ε =

1

K. Therefore for all n ≥ N , an > K, so limn→∞ an = ∞ as

required. 2

Examples 5.1.7. One may invert the special null sequences of Section 4.1. For example, we have

that for any c > 0, n! · cn → ∞ as n → ∞. Similarly, if c > 1, we havecn

nk→ ∞ as n → ∞.

(In both cases the proof is left as an exercise.)

Consider now the sequence (n!− 8n)n∈N. We have1

n!− 8n=

1

n!(1− 8n

n! )=

1

n!· 1

(1− 8n

n! )→

0 · 1

1− 0= 0 (by Lemma 4.1.4 with c = 8, and AoL). Also, the fact that 1− 8n

n!→ 1 (as n→∞),

ensures that for sufficiently large n, n! − 8n > 0. Thus, by The Reciprocal Rule 5.1.6(ii),n!− 8n →∞ as n→∞.

We now prove an Algebra of Limits Theorem for sequences tending to infinity.

Theorem 5.1.8. (i) Suppose that (an)n∈N and (bn)n∈N both tend to infinity. Then(a) an + bn →∞ as n→∞;(b) if c > 0, then c · an →∞ as n→∞;(c) an · bn →∞ as n→∞.

(ii) (The Infinite Sandwich Rule.) If (bn)n∈N → ∞ as n → ∞, and (an)n∈N is any sequencesuch that an ≥ bn for all sufficiently large n, then an →∞ as n→∞.

Proof. For (i)(b), let K > 0 be given. ThenK

c> 0, so there exists N ∈ N such that an >

K

cfor all n ≥ N . Hence c · an > K for all n ≥ N , and we are done.

The rest of the proofs are left as exercises. 2

Definition 5.1.9. We say that a sequence (an)n∈N tends to −∞ as n→∞, if for all K < 0,there exists N ∈ N such that for all n ≥ N , an < K. This is written: an → −∞ as n→∞.

[This is easily seen to be equivalent to saying that −an →∞ as n→∞.]

37

Example 5.1.10. The sequence ((−1)n · n)n∈N is unbounded (so does not converge) but neithertends to ∞ nor to −∞ as n→∞.

Similarly (8n − n!)→ −∞ as n→∞. Why? Because this is exactly the same statement assaying that −(8n − n!)→ +∞ as n→∞. Which we proved in Examples 5.1.7.

Questions 6A:(These will appear on the Exercise sheet for Week 7, but are also relevant to the Coursework

Exam.)Do the following sequences converge/diverge/tend to infinity or tend to minus infinity?

(a)(cos(nπ)

√n)n∈N (b)

(sin(nπ)

√n)n∈N

c)

(√n2 + 2√n

)n∈N

(d)

(n3 + 3n

n2 + 2n

)n∈N

e)

(n2 + 2n

n3 + 3n

)n∈N

(f)

(1

√n−√

2n

)n∈N

38

Chapter 6

Subsequences

Looking at subsequences of a given sequence gives a practical test, Theorem 6.1.3, for non-convergence. These also feature in two of the most important ideas and results in analysis,namely Cauchy sequences and the Bolzano-Weierstrass theorem.

6.1 The Subsequence Test for Non-Convergence

Definition 6.1.1. Suppose that 1 ≤ k1 < k2 < · · · < kn < · · · is a strictly increasing sequenceof natural numbers. Then, given any sequence (an)n∈N of real numbers we can form the sequence(akn)n∈N. Such a sequence is called a subsequence of the sequence (an)n∈N. In other words,a subsequence of (an)n∈N is any sequence obtained by leaving out terms (as long as infinitelymany are left in).

Example 6.1.2. The sequence (4n)n∈N is a subsequence of the sequence (n2)n∈N: we take kn = 2n.Then with an = n2, we have akn = (kn)2 = (2n)2 = 22n = 4n.

Theorem 6.1.3. Let (an)n∈N be a sequence. For any subsequence (akn)n∈N of the sequence(an)n∈N we have:(i) if an → l as n→∞, then akn → l as n→∞;(ii) if an →∞ as n→∞, then akn →∞ as n→∞;(iii) if an → −∞ as n→∞ then akn → −∞ as n→∞.

Proof. (i) Let ε > 0 be given. Choose N ∈ N so that for all n ≥ N , |an − l| < ε. Now aneasy induction shows that for all n ∈ N we have kn ≥ n. Hence, if n ≥ N then kn ≥ N , so|akn − l| < ε. So the same N works for the sequence (akn)n∈N.The proofs of (ii) and (iii) are similar. 2

As mentioned above, the practical importance of Theorem 6.1.3 is that it gives us a methodof proving that certain sequences do not converge. We can either find two subsequences of thegiven sequence that converge to different limits, or else one subsequence that converges to either∞ or −∞.

Example 6.1.4. The sequence

((−1)n +

1

n2

)n∈N

does not converge. For suppose it does, to l

say. Consider the subsequence with kn = 2n. This is the sequence

((−1)2n +

1

(2n)2

)n∈N

, which

is

(1 +

1

(2n)2

)n∈N

and this converges to 1. Hence, by Theorem 6.1.3(i), we must have l = 1.

39

But now consider the subsequence with kn = 2n + 1, i.e.

((−1)2n+1 +

1

(2n+ 1)2

)n∈N

, which

is

(−1 +

1

(2n+ 1)2

)n∈N

, and this converges to −1. Hence by 6.1.3(i) we must have l = −1, a

contradiction!

Example 6.1.5. The sequence(n

4−[n

4

])n∈N

does not converge. For suppose it does, to l say.

Consider the subsequence with kn = 4n, i.e.

(4n

4−[

4n

4

])n∈N

. This is the sequence with

all terms equal to 0 which of course converges to 0 and hence by 6.1.3(i), l = 0. However,

consider now the subsequence with kn = 4n + 1, i.e.

(4n+ 1

4−[

4n+ 1

4

])n∈N

. Note that[4n+ 1

4

]=

[n+

1

4

]= n. So this subsequence has all terms equal to

1

4, and therefore it

converges to1

4. So by 6.1.3(i), l =

1

4, a contradiction.

Example 6.1.6. The sequence(n sin

(nπ2

))n∈N

neither converges, nor tends to ∞ nor tends

to −∞. For consider the subsequence with kn = 4n + 1, i.e.

((4n+ 1) sin

((4n+ 1)π

2

))n∈N

.

Now (4n+ 1) sin

((4n+ 1)π

2

)= (4n+ 1) sin

(2nπ +

π

2

)= (4n+ 1) sin

(π2

)= 4n+ 1. So this

subsequence is (4n+ 1)n∈N which tends to ∞ as n→∞.

But now consider the subsequence with kn = 4n, i.e. the sequence

(4n sin

(4nπ

2

))n∈N

.

Every term here is 0 (since sin(2nπ) = 0 for all n ∈ N). So this subsequence converges to 0. Soby 6.1.3 (i), (ii) and (iii), the original sequence does not converge and nor does it tend to ∞ or−∞.

Example 6.1.7. Does limn→∞

( [√n]−√n)

exist?

Hint: you may assume that [√m2 +m] = m.

Answer: Set an = [√n]−√n. We should use subsequences. One subsequence is pretty obvious—

if we take kn = n2 then akn = [√n2]−

√n2 = n− n = 0, so this subsequence clearly has limit

0.For the second one we use the hint and take kn = n2 + n; thus

akn =[√

n2 + n]−√n2 + n = n−

√n2 + n.

This is now one of those cases where we use the trick (x− y) =(x− y)(x+ y)

(x+ y). So, in this

case

n−√n2 + n =

(n−√n2 + n)(n+

√n2 + n)

(n+√n2 + n)

=n2 − (n2 + n)

n+√n2 + n

=−n

n+√n2 + n

=−1

1 +√

1 + 1/n.

Now, for example from the last Homework set, we know that limn→∞(√

1 + 1/n) =√

1 = 1.

Hence by the AoL the last display has limit−1

1 + 1=−1

2.

Since we have two subsequences of the original sequence with different limits the originalsequence cannot have a limit.

40

(Here is the proof of the hint: Simply observe that

[√m2 +m] = m ⇐⇒

√m2 +m < m+ 1 ⇐⇒ (m2 +m) < (m+ 1)2 = m2 + 2m+ 1,

which is certainly true!)

6.2 Cauchy Sequences and the Bolzano-Weierstrass Theorem

We conclude this chapter with two fundamental results of analysis. You are not required to knowtheir proofs in this course. However, since the proofs do not require any further knowledge onyour part, and since they are such useful results, the detailed proofs are given in the Appendixto this chapter.

Theorem 6.2.1 (The Bolzano-Weierstrass Theorem (1817)). Every bounded sequence (an) hasa convergent subsequence.

Remarks: (1) It is important to note that the Theorem is not saying that (an) is convergent—which is false in general. For instance take our favourite bad example (an = (−1)n). So, forthis example one would have to take a genuine subsequence. Two obvious examples would be(a2 = 1, a4 = 1, a6 = 1, ...) and (a1 = −1, a3 = −1, a5 = −1, ...)(2) The way to think about the theorem is as a generalisation of the Monotone ConvergenceTheorem—which says that if one has an increasing bounded sequence then it is convergent. So,you can probably guess how one might try to prove this theorem: given a bounded sequence(an) then start with a1, and look for some ak2 ≥ a1 and induct. If this is not possible then tryto do the same with descending sequences. In fact the proof needs to be a bit more subtle, butit develops from this idea.

We now come to the problem of giving a necessary and sufficient condition on a sequence forit to converge, without actually knowing what the limit might be. The following definition iscrucial and will feature in many different guises in your future Analysis and Topology courses.

Definition 6.2.2. A sequence (an)n∈N is called a Cauchy sequence if for all ε > 0, thereexists N ∈ N such that for all i, j ∈ N with i, j ≥ N , we have |ai − aj | < ε.

So this is saying that the terms of the sequence (an)n∈N are getting closer and closer together.Then we have a central theorem.

Theorem 6.2.3. A sequence converges if and only if it is a Cauchy sequence.

It is important in the definition of a Cauchy sequence that we are saying that all the terms getclose together. For example if one defined a sequence inductively by a1 = 1 and an = an−1 + 1

n ,then certainly for all ε > 0 we can find N such that |an− an+1| < ε for n ≥ N . But, in fact thissequence is not convergent (we have seen an informal proof of this already and will see it againwhen we consider infinite series). So, the last theorem would not be true if we just assumedthat successive terms got close together.

6.2.1 Proofs for the section - optional

Here are detailed proofs of the Bolzano-Weierstrass Theorem and the resulting theorem onCauchy sequences. As mentioned above, these proofs are not required in this course. But theresults are so useful that, if you have the time do go through them, it will set you up nicely forwhen you study real and complex analysis and metric spaces next year.

41

Theorem 6.2.4. (The Bolzano-Weierstrass Theorem). Any bounded sequence (an) pos-sesses a convergent subsequence. In other words there exist a sequence of numbers

k1 < k2 · · · < kr < kr+1 < · · ·

such that the subsequence (ank)k∈N is convergent.

Proof. Since (an) is bounded, it has a supremum M = sup{an : n ∈ N} and an infimumN = inf{an : n ∈ N}. (see Chapter 2). The proof is quite sneaky; what we will do is constructa chain of real numbers

M1 = M ≥M2 · · · ≥Mr · · · ≥ N

and a second sequence

N1 = M1 − 1, N2 = M2 −1

2, N3 = M3 −

1

3, · · · , Nr = Mr −

1

r· · ·

such that some element akr (with kr > kr−1) is squeezed between them: Nr ≤ akr ≤ Mr. Thiswill do. The reason is that by the Monotone Convergence Theorem, the {Mr} have a limit, sayµ. Then one shows that lim

r→∞Nr = µ as well and hence by the Sandwich Theorem lim

r→∞akr = µ

as well.

OK, let’s see the details. As we said, we take M1 = M and N1 = M1 − 1. The point hereis that N1 is not an upper bound for {an} and so there exists some ak1 with N1 < ak1 ≤ M1.Now we let M2 = sup{an : n > k1}. We note here that any upper bound for the {an} mustalso be an upper bound for the smaller set {an : n > k1}. Thus M2 ≤ M1. Once again we setN2 = M2 − 1

2 . Then since N2 is not an upper bound for {an : n > k1}, there exists some ak2with k2 > k1 such that N2 < ak2 ≤M2.

Now we induct. Suppose that we have found (M1, N1, ak1 , · · ·Mr, Nr, akr) in this way; thusthey satisfy Ni = Mi − 1

i < aki ≤ Mi = sup{aj : j > ki−1} ≤ Mi−1 and ki > ki−1 for each1 < i ≤ r.

Then we set Mr+1 = sup{aj : j > kr} and Nr+1 = Mr+1 −1

r + 1. Then as N+1 is not an

upper bound of {aj : j > kr} there exists Nr+1 < akr+1 ≤Mr+1 with kr+1 > kr. This completesthe inductive step.

Now, the {M`} is a descending sequence and, as each M` is an upper bound of some collectionof the ap, we see that the M` ≥ ap for some such ap. Thus the M` are bounded below by N .Hence the Monotone Convergence Theorem from Q3 of the Examples for Week 3 implies thatthe limit lim

r→∞Mr exists, say lim

r→∞Mr = µ.

We next claim that limr→∞

Nr = µ as well. To see this pick ε > 0 and pick P such that if

p ≥ P then Mp − µ = |Mp − µ| < ε/2. We also know that (for q ≥ Q =

[2

ε

]+ 1), we have

Nq ≥Mq −1

q≥Mq −

ε

2. By the triangle inequality

|Nq − µ| ≤ |Nq −Mq|+ |Mq − µ| <ε

2+ε

2= ε for all q ≥ max{P,Q}.

Thus, limr→∞

Nr = µ, as claimed.

Now we are done: since Nr < akr ≤ Mr for each r, the Sandwich Theorem ensures thatlimr→∞

akr = µ, as well. 2

42

Examples 6.2.5. The Theorem tells us nothing about what is the subsequence {akr} or its limitµ. Some exercises might give you a feel for the vagaries inherent in the argument.

For the ascending chain an = 1− 1

nthen Mr = 1 for all r and the {akr} is just some (indeed

any) subsequence of the {an}. For a sequence like

(an = (−1)n

1

n

)there are lots of possible

subsequences— you could take the descending chain

(a2n =

1

2n

)n∈N

for all n or the ascending

chain

(a2n+1 = − 1

2n+ 1

)n∈N

for all n or some messier combination of the two. The limit is

also not uniquely determined; just think about the case of (an = (−1)n).

In fact one can extend the proof of this theorem a bit to prove the following:

Challenging Exercise: Prove that every sequence (an)n∈N has a subsequence (akn)n∈N whichis either increasing (i.e. for all n ∈ N, akn ≤ akn+1) or decreasing (i.e. for all n ∈ N,akn ≥ akn+1).

We now consider Cauchy sequences as defined in Definition 6.2.2. We begin with:

Lemma 6.2.6. Every Cauchy sequence (an) is bounded.

Proof. Take N such that, for i, j ≥ N we have |ai − aj | < 1. In particular, taking i = N wesee that

aN − 1 < aj < aN + 1 for all j ≥ N.

Thus an ≤ max{a1, a2, . . . , aN−1, aN + 1} for all n. The lower bound is similar. 2

We know that an increasing sequence (an) has a limit if and only if it is bounded (combinethe Monotone Convergence Theorem with Theorem 2.3.9). Using Cauchy sequences we can geta general analogue:

Theorem 6.2.7. (The Cauchy Sequence Theorem) A sequence (an) converges if and onlyif it is a Cauchy sequence.

Proof. Suppose first that (an) is a Cauchy sequence and let ε > 0 be given. As so often, we

will use our various rules for convergence using η =ε

2. By the Bolzano-Weierstrass Theorem

we can at least find a convergent subsequence; say (akr : r ∈ N) with limr→∞

akr = µ. If we just

had a bounded sequence then there would be no reason why µ would equal limn→∞

an; just think

of our favourite bad sequence ((−1)n). So somehow we have to use the Cauchy condition.

So, let ε > 0 be given. In particular, there exists N0 such that |akr − µ| < η =ε

2for all

kr ≥ N . Also, by the Cauchy condition pick N1 such that |ai − aj | < η =ε

2for all i, j ≥ N1.

Set N = max{N0, N1} Then there is some kr > N and so, for this kr we get |akr − aj | <ε

2for

all j ≥ N . But we also know that |akr − µ| <ε

2for this kr. By the triangle inequality these

combine to give

|aj − µ| ≤ |aj − akr |+ |akr − µ| <ε

2+ε

2= ε for all j ≥ N .

Thus limn→∞

an = µ as we wanted.

43

Conversely, suppose that (an) converges; say with limn→∞

an = ν. Let ε > 0 be given. Once

again we use η =ε

2. Thus there exists N such that |aj −µ| < ε

2 for all j ≥ N . But now we find

that for i, j ≥ N we have

|ai − aj | ≤ |ai − ν] + |aj − ν] <ε

2+ε

2= ε.

So we have a Cauchy sequence. 2

44

Chapter 7

L’Hopital’s Rule

We will use L’Hopital’s Rule, which some of you will have seen before. We can’t mathematicallyjustify the rule in this course unit, simply because it involves differentiation of functions andwe have not yet given the rigorous definition of when a function is differentiable nor of whatthe derivative is. That will be done in the second year course on Real Analysis. But the rule isvery useful, so we will use it.

We consider two sequences (an)n∈N and (bn)n∈N, both tending to infinity. L’Hopital’s Rule

gives a method for calculating limn→∞anbn

under certain circumstances.

7.1 L’Hopital’s Rule

Assume that f : [1,∞) → R and g : [1,∞) → R are two functions which can be differentiatedat least twice and for some N > 0 that g′(x) 6= 0 for x > N . Set an = f(n) and bn = g(n) forn ∈ N.

Assume that limn→∞

an =∞ = limn→∞

bn. (†)

Then limn→∞

anbn

= limn→∞

f ′(n)

g′(n), assuming that the Right Hand Side exists. Here g′ =

dg

dx.

Second Version: L’Hopital’s Rule also holds if you replace (†) by

Assume that limn→∞

an = 0 = limn→∞

bn. (††)

Since we do not yet have a rigorous definition of the derivative, we’re not in a positionto have a rigorous proof of the validity of L’Hopital’s Rule, though you may have seen somearguments for it when studying Calculus. Anyway, we will take it as correct (which it is) andsee some examples of its use.

Example 7.1.1. Say an = 2n + 3 and bn = 3n + 2. Take f(x) = 2x + 3 and g(x) = 3x + 2, so

that clearly anbn

= f(n)g(n) and the hypotheses of the rule are satisfied. Thus L’Hopital’s Rule tells

us that

limn→∞

anbn

= limn→∞

f ′(n)

g′(n)= lim

n→∞

2

3=

2

3.

Example 7.1.2. Let an = 3n2−4n+2 and bn = (n+1)2. Here we apply the rule to the functionsf(x) = 3x2 − 4x + 2 and g(x) = (x + 1)2 both of which, as one may easily check, are positivefor x ≥ 1, to obtain

limn→∞

3n2 − 4n+ 2

(n+ 1)2= lim

n→∞

6n− 4

2(n+ 1)

45

.Still top and bottom go to ∞ as n→∞. But we may apply the rule again to obtain

limn→∞

6n− 4

2(n+ 1)= lim

n→∞

6

2= 3.

Of course, in these examples one may also use the previously discussed method of dividingtop and bottom by the fastest-growing term. However, L’Hopital’s Rule really comes into itsown when applied to examples like the following.

Example 7.1.3. Consider the sequence

(log n

n

)n∈N

. Before applying L’Hopital’s Rule we should

note that log n→∞ as n→∞.

So we may indeed apply the Rule in this example (with f(x) = log x and g(x) = x) and wesee that

limn→∞

log n

n= lim

n→∞

1n

1= lim

n→∞

1

n= 0.

Example 7.1.4. Similarly (with f(x) = log x and g(x) = log(x+ 1)):

limn→∞

log n

log(n+ 1)= lim

n→∞

1n1

(n+1)

= limn→∞

n+ 1

n= lim

n→∞

(1 +

1

n

)= 1.

Example 7.1.5. And a more complicated example:

limn→∞

log(4n5 − 1)

log(n2 + 1)= lim

n→∞

(20n4

4n5 − 1· n

2 + 1

2n

)= lim

n→∞

10n3(n2 + 1)

4n5 − 1

= limn→∞

10n5 + 10n3

4n5 − 1= lim

n→∞

10 + 10n2

4− 1n5

=10 + 0

4− 0=

5

2.

However, before applying L’Hopital’s Rule you have to be careful to ensure that (†) holds,since otherwise you may get rubbish:

Example 7.1.6. Take an = 1− 1n and bn = 2− 1

n . then using the Rule we get

limn→∞

1− 1n

2− 1n

= limn→∞

n−2

n−2= 1.

But even more obviously

limn→∞

1− 1n

2− 1n

=1− 0

2− 0=

1

2.

What went wrong? The problem is that (†) does not hold, since limn→∞

an 6=∞. So, the rule

should not have been applied.

Finally, L’Hopital’s Rule gives easy proofs of Lemma 4.1.6 and the related result for logsmentioned after the Tables on the Supplement to Section 5:

Example 7.1.7. For any 0 < c < 1 and k we have limn→∞

nkcn = 0.

Proof: First, pick a natural number K ≥ k. Since nKcn ≥ nkcn, the Sandwich Rule says we

need only prove that limn→∞

nKcn = 0. Secondly, we rewrite this as limn→∞

nK

dn= 0 (for d = 1

c > 1).

This ensures that we are in a situation where we can apply L’Hopital’s Rule and gives

limn→∞

nK

dn= lim

n→∞

K

ln(d)

nK−1

dn.

46

By induction on K the RHS has limit 0 (where one starts the induction at K − 1 = 0).

The rule for logs is similar: We are discussing limn→∞

ln(n)

nc, for any c > 0. Again we can apply

the rule to get

limn→∞

ln(n)

nc= lim

n→∞

1n

cnc−1= lim

n→∞

1

c

1

nc→ 0.

47

Part II

Series

48

Chapter 8

Introduction to Series

In this chapter we give a rigorous definition of infinite sums of real numbers. We are interestedin expressions of the form

∞∑i=1

ai = a1 + a2 + · · ·+ an + · · ·

and need to make sense of such expressions. This is closely related to sequences since this“infinite sum” is (by definition! as you will see) the limit of the sequence of partial sums (sn)where

sn =

n∑i=1

ai = a1 + a2 + · · · an.

You have probably seen that∑∞

i=0

1

2i= 2. To see that: an easy induction shows that sn =∑n

i=0

1

2i= 2− 1

2n, then the limit of the sequence (sn)n of partial sums is 2.

However, there are some surprising subtleties in these series. One of the basic tricky ones is

the Harmonic Series:∑∞

i=1

1

nwhich equals ∞. That might seem counterintuitive but, to see

it, collect terms together to get:∑∞i=1

1

n= 1 +

(12

)+

(13 + 1

4

)+

(15 + 1

6 + 17 + 1

8

)+ · · ·

≥ 1 +(12

)+

(12

)+

(12

)+ · · · .

(8.1)

So this gives an infinite sum of halves, which is therefore infinite. In contrast, we will see that

∞∑i=1

1

n1.001<∞.

So, we are going to need some new techniques!

49

8.1 The Basic Definitions

For real numbers an, an infinite series is an expression of the form

∞∑n=1

an (also written a1 + a2 + a3 + · · ·+ an + · · · ).

This may also be written∑n≥1

an or even just∑an.

Given such a series, we form the sequence of partial sums (sn)n∈N defined by setting

sn = a1 + a2 + · · · an =n∑i=1

ai.

If sn → s as n→∞, we then say that the series∞∑n=1

an is convergent with sum s, and write

∞∑n=1

an = s. If there is no real number s with this property, that is if the sequence (sn) does

not have a limit, then we say that the series∞∑n=1

an is divergent. If the series is divergent but

limn→∞

sn =∞ then we say that∞∑n=1

an =∞ (similarly with −∞).

The Harmonic Series, revisited. Let’s first check that the harmonic series (8.1) really doeshave sum∞. So, what we have seen is that for any K there exists M (in fact M = 2K+1 works)

such that∑M

n=1

1

ngives (more than) a sum of 2K copies of 1

2 . In other words, if m ≥ M then

sm =∑m

n=1

1

n≥∑M

n=1

1

n≥ K. Which is just what one needs to prove to see that

∑∞i=1

1

n=∞.

Remark. Sometimes (as in the next example), it is more convenient to start the series/sequenceat 0, thus looking at series of the form

∑∞n=0 = a0 + a1 + · · · .

Example 8.1.1 (Geometric series). Let r be a real number with |r| < 1. Then the series∞∑n=0

rn

converges with sum1

1− r.

Proof. Here, an = rn and sn = a0 + a1 + · · ·+ an = 1 + r + · · ·+ rn. So

sn = 1 + r + · · ·+ rn ⇒ rsn = r + r2 + · · ·+ rn+1.

Subtracting we obtain(1− r)sn = 1− rn+1

because all the other terms cancel. Hence

sn =1− rn+1

1− r.

Now since −1 < r < 1, rn+1 → 0 as n → ∞. (Use 4.1.2.) Hence1− rn+1

1− r→ 1− 0

1− r=

1

1− r(by AoL). Thus sn →

1

1− ras n→∞. So, by definition,

∞∑n=0

rn =1

1− r. 2

50

Remark: This proof only works for |r| < 1. It might be tempting to put in other values ofr since the right hand side is well defined for any r with r 6= 1. But this gives nonsense, e.g.

“∞∑n=0

2n = 11−2 = −1” is definitely not correct.

Example 8.1.2 (Using partial fractions). Consider the series∞∑n=1

1

n(n+ 1).

Here an =1

n(n+ 1)and sn = a1 + a2 + · · ·+ an =

1

1 · 2+

1

2 · 3+

1

3 · 4+ · · ·+ 1

n(n+ 1).

A trick that sometimes works is to use partial fractions on the terms. Here we use theidentity

1

x(x+ 1)=

1

x− 1

x+ 1

valid for x 6= 0,−1. Hence

sn =

(1

1− 1

2

)+

(1

2− 1

3

)+

(1

3− 1

4

)+ · · ·+

(1

n− 1

n+ 1

).

Now all the terms cancel apart from the first and the last, so we obtain

sn = 1− 1

n+ 1→ 1− 0 = 1 as n→∞.

Thus∞∑n=1

1

n(n+ 1)= 1.

Remark: It is very important in Example 8.1.2 that one only rearranges terms in a finite sum

rather than rearranging terms in the infinite sum∑∞

n=1

1

n(n+ 1). The reason is that, with

infinite sums, you can get all sorts of strange different answers by arranging things in differentways—the details can be found in Section 12.2.

It is quite rare for one to be able to find an exact formula for the nth partial sum (i.e. forthe number sn) as we did in these two examples. So, just as we did with the study of sequences,we need to build up a stock of general theorems that will help us to tell when a series does anddoes not converge without explicitly computing the partial sums. Our first result along theselines states that a series cannot converge unless the terms (i.e. the an) form a null sequence.

Note, however, that the converse of this statement is false in general, as we saw with∞∑n=1

1

n.

Here is a similar type of example.

Example 8.1.3 (of non-convergence). Consider the series∞∑n=1

1√n

. Here, an =1√n

and we have

shown (see Example 3.2.3) that (an)n∈N is a null sequence. However, the series∞∑n=1

an does not

converge. For sn =1√1

+1√2

+1√3

+ · · · + 1√n

, from which we deduce that sn ≥ (number of

terms) · (smallest term) = n · 1√n

=√n.

Thus sn ≥√n for all n ∈ N, so sn → ∞ as n → ∞ by the Infinite Sandwich Rule (since

√n→∞ as n→∞). Hence, by definition, the series

∞∑n=1

1√n

does not converge.

51

Theorem 8.1.4. (The nth term Test.) If∞∑n=1

an is a convergent series, then (an)n∈N is a

null sequence.

Proof. Suppose that∞∑n=1

an = s. Let sn = a1 + · · ·+ an be the nth partial sum. So sn → s as

n → ∞. Hence limn→∞

sn−1 → s as well (see Lemma 4.1.3), and so by AoL, limn→∞

(sn − sn−1) =

s− s = 0. But sn − sn−1 = an. So an → 0 as n→∞, i.e. (an)n∈N is a null sequence. 2

Theorem 8.1.5. (Algebra of Infinite Sums)

(i) If∞∑n=1

an and∞∑n=1

bn are both convergent series, with sums A and B respectively, then the

series∞∑n=1

(an + bn) is also convergent with sum A+B.

(ii) If∞∑n=1

an is a convergent series with sum A, and λ is any real number, then the series

∞∑n=1

λan is convergent with sum λA.

Proof. (i) For n ∈ N, let sn =n∑k=1

ak and tn =n∑k=1

bk be the partial sums and similarly let

un =n∑k=1

(ak + bk) be the partial sums for the series∞∑n=1

(an + bn).

Then, rearranging terms,

un = (n∑k=1

ak) + (n∑k=1

ak) = sn + tn for n ∈ N.

But sn → A and tn → B as n→∞ by the definition of convergence of the original series. Thus,

by AoL for sequences, un → A+B as n→∞, i.e.∞∑n=1

(an + bn) = A+B, as required.

The proof of (ii) is left as an exercise. 2

52

Chapter 9

Series with Non-Negative Terms

In this chapter we establish some key facts about series with non-negative terms. The theoryof such series is very much easier than the general case where negative terms are allowed.

9.1 The Basic Theory of Series with Non-Negative Terms

Theorem 9.1.1. Suppose that an ≥ 0 for all n ∈ N and let sn =n∑k=1

ak. Then the series∞∑n=1

an

is convergent if and only if the sequence (sn)n∈N is bounded above.

Proof. Note that sn+1 = sn + an+1 ≥ sn, since an+1 ≥ 0. Hence the sequence (sn)n∈N isincreasing. So, by the Monotonic Convergence Theorem, if (sn)n∈N is bounded above, then it

converges and thus, by definition, the series∞∑n=1

an is convergent.

Conversely, if∞∑n=1

an is convergent, then (sn) is convergent (definition!) and so Theorem 2.3.9

says that (sn)n∈N is bounded above. 2

Theorem 9.1.2 (The Comparison Test). If 0 ≤ an ≤ bn for all n ∈ N and∞∑n=1

bn converges,

then∞∑n=1

an converges.

Equivalently, if 0 ≤ an ≤ bn for all n ∈ N and∞∑n=1

an diverges, then∞∑n=1

bn diverges.

Proof. Let sn = a1 + a2 + · · ·+ an and tn = b1 + b2 + · · ·+ bn. Since an ≤ bn for all n it follows

that sn ≤ tn for all n. But as∞∑n=1

bn is a convergent series of non-negative terms there is, by

Theorem 9.1.1, some M > 0 such that tn ≤M for all n. Hence sn ≤M for all n, and so, again

by 9.1.1,∞∑n=1

an is convergent. 2

Exercise 9.1.3. Let (an)n∈N be any sequence and let N be any positive integer. Show that the

series∞∑n=1

an is convergent if and only if the series∑

n≥N an (which is the same as the series

∞∑n=1

an+N−1) is convergent.

Remark: This exercise is done in the Exercise sheet for Week 9, and essentially means thatfor any given test, you need only apply the test for “large n”. Let’s make this precise with:

53

Slightly Improved Comparison Test: Let N ∈ N. If 0 ≤ an ≤ bn for all n ≥ N and∞∑n=1

bn

converges, then∞∑n=1

an converges.

Proof. By (9.1.3),∞∑n=1

bn converges⇒∑n≥N

bn =∑m≥1

bN+m−1 converges⇒∑n≥N

an =∑m≥1

aN+m−1

converges (by the comparison test) ⇒∑n≥1

an converges (by 9.1.3 again) 2

Example 9.1.4. The series

∞∑n=1

1

n2is convergent. For we have that

1

(n+ 1)2≤ 1

n(n+ 1)for all

n ∈ N, and hence by Example 8.1.2 and the Comparison Test,∞∑n=1

1

(n+ 1)2is convergent. Since

this is just the series∑

n≥21

n2, the convergence of the series

∞∑n=1

1

n2follows from Exercise 9.1.3.

This example shows the value of the Comparison Test: it was straightforward to calculate

the partial sums of the series∞∑n=1

1

n(n+ 1)and hence calculate explicitly the sum of this series

(thereby establishing its convergence). But it is impossible to give a neat formula for the

partial sums of the series∞∑n=1

1

n2, so we resort to comparing its terms with those of the series

∞∑n=1

1

n(n+ 1)in order to establish its convergence. In fact, as mentioned in the introduction to

this course,∞∑n=1

1

n2=π2

6

but the proof of this requires methods from next year’s complex analysis course.

Example 9.1.5. For any p ≥ 2, the series∞∑n=1

1

npis convergent. To see this we simply observe

that for all n ∈ N,1

np≤ 1

n2(since p ≥ 2) and apply the Comparison Test to the previous

example.

Similarly we have:

Example 9.1.6. For any p ≤ 1 the series∞∑n=1

1

npis divergent. For suppose that p ≤ 1 and, for

a contradiction, that∞∑n=1

1

npis convergent. Since

1

n≤ 1

npfor all n ∈ N we would have, by the

Comparison Test, that∞∑n=1

1

nis convergent. But this contradicts Equation 8.1.

Remark: We’ve left the case 1 < p < 2 open but, later, we will show that∞∑n=1

1

npconverges if

p > 1 but diverges if p ≤ 1.

We now come to an important test for the convergence of series.

Theorem 9.1.7 (The Ratio Test for series.). Suppose that an > 0 for all n ∈ N and assume

thatan+1

an→ l as n→∞.

(i) If l < 1, then∞∑n=1

an is convergent.

54

(ii) If l > 1, then∞∑n=1

an is divergent.

Remark: If l = 1, no conclusion can be drawn: For example, if an =1

n2or an =

1

n, then in

both cases we have thatan+1

an→ 1 as n → ∞. But in the first case

∞∑n=1

an converges, whereas

in the second case it diverges.

Proof. (i) Suppose that l < 1. We want to compare our series to the geometric series∑rn for

some r. We need a little room to manoeuvre and taking l = r wouldn’t give us that, so we willtake a slightly bigger number for r.

So, choose r so that l < r < 1 (e.g. take r =1 + l

2to be the average of l and 1) and set

ε = r − l. Then ε > 0 so we may choose N ∈ N so that l − ε < an+1

an< l + ε for all n ≥ N .

But ε = r − l says that l + ε = r and so, as an > 0 we get 0 < an+1 < r · an for n ≥ N . Inparticular

aN+1 < r · aN , and aN+2 < r · aN+1 < r2 · aNand so on. So (by induction) we obtain

∀n ≥ N, 0 < an < rn−N · aN

Collecting terms we see that∑n≥N

an ≤∑n≥N

rn−N · aN =∑n≥0

rn · aN = aN∑n≥0

rn. (∗)

But as 0 < r < 1 it follows from Example 8.1.1 that∑

n≥0 rn converges. Now we are basically

done: by The Algebra of Infinite Sums 8.1.5(ii), aN∑

n≥0 rn converges and so from (*) and the

Slightly Improved Comparison Test,∑

n≥1 an converges, as required.

(ii) Now suppose that l > 1. In this case we choose r so that 1 < r < l and, by following thesame procedure as above, we obtain an N ∈ N such that for all n ≥ N we have an > rn−N ·aN =

rn · (aNrN

). But, since r > 1, this implies that an → ∞ as n → ∞ and, in particular, it implies

that (an)n∈N is not a null sequence. So∞∑n=1

an diverges by Theorem 8.1.4. 2

Example 9.1.8. Consider the series∞∑n=1

n2

2n.

Let an =n2

2n. Then

an+1

an=

(n+ 1)2

2n+1· 2n

n2=

1

2·(

1 +1

n

)2

.

So, by AoL,an+1

an→ 1

2· (1 + 0)2 =

1

2as n→∞.

Since 12 < 1, it follows from the Ratio Test that

∞∑n=1

n2

2nconverges.

55

Example 9.1.9. Let x be any positive real number and consider the series∑

n≥0xn

n!.

Here, an =xn

n!, so

an+1

an=

xn+1

(n+ 1)!· n!

xn=

x

n+ 1.

Butx

n+ 1→ 0 as n→∞ and 0 < 1. So by the Ratio Test,

∑n≥0

xn

n!converges.

Aside: We can define ex to be the sum∑ xn

n!. One can then show (directly from this definition)

that ex+y = ex ·ey. This justifies the notation, and also implies its consistency with our previousdiscussion of exponentiation (see Section 2.6).

Example 9.1.10. (The Bouncing Ball.) Suppose that a rubber ball is dropped from a height of1 metre and that each time it bounces it rises to a height of (2/3) of the previous height. Howfar does it travel before it stops bouncing (and yes it does stop)?

Answer: First it drops 1m. Then it rises up and drops 2/3m. Then it rises up and drops(2/3)2m, etc etc. So the total distance travelled is

1 + 2× 23 + 2×

(23

)2+ 2×

(23

)3+ · · ·

= 1 + 2× 23

(1 + 2

3 +(23

)2+ · · ·

)= 1 + 4

3 ·1

1−2/3 = 5.

9.2 The Integral Test

We consider a function f : [1,∞)→ R which is:

(i) positive (at least, non-negative), so f(x) ≥ 0 ∀x ≥ 1;

(ii) decreasing, so f(x) ≤ f(y) ∀ x ≥ y ≥ 1;

(iii) continuous.

Here continuity (meaning being continuous) is a condition you may have seen before, but will bemade precise in your analysis course next year. For the record (although you need not rememberthis) the definition is as follows. A function f is continuous on X = [1,∞) if for all x ∈ Xand ε > 0 there exists δ > 0 such that if y ∈ R with |x − y| < δ then |f(x) − f(y)| < ε. Thatis, if we choose x ∈ X and a (small) interval I around f(x), then there is some (perhaps verysmall) interval around x which is sent inside I by f . Roughly speaking it means that “there areno breaks in the graph of f”. This condition rules out functions such as

f(x) =

{1 if x ≤ 2,12 if x > 2.

(In this example, try taking x = 2 and ε = 14 .) Examples of functions f(x) satisfying (i), (ii)

and (iii) are1

x,

1

x2,

1

x log(1 + x), 2−x

Theorem 9.2.1 (The Integral Test). Let f : [1,∞) → R be a function satisfying the threeconditions above. Then the series

∞∑n=1

f(n)

56

converges if and only if the sequence (∫ n

1f(x)dx

)n∈N

converges as n→∞.This in turn happens if and only if the sequence

(∫ n1 f(x)dx

)n∈N is bounded.

Before giving the proof here are some examples of how the test works.

Example 9.2.2. Consider the series∞∑n=1

1

n. We shall show that it diverges.

Let f : [1,∞)→ R be the function f(x) =1

x, which clearly satisfies our three conditions.

Now ∫ n

1f(x)dx =

∫ n

1

1

xdx = [log x]n1 = log n− log 1 = log n.

But (as in Example 7.1.3) (log n)n∈N is a divergent sequence, so by the Integral Test,∞∑n=1

1

nis a divergent series.

Here is the important example which was left partly unresolved in the previous chapter:

Example 9.2.3. If p is a real number with p > 1, then∞∑n=1

1

npis a convergent series.

Here we take f(x) =1

xp. This clearly satisfies the three conditions for the Integral Test to

be applicable. We have∫ n

1f(x)dx =

∫ n

1x−pdx =

[x−p+1

(−p+ 1)

]n1

=n−p+1

−p+ 1− 1

−p+ 1=

1

p− 1

(1− 1

np−1

).

Now since p > 1, we have that1

np−1→ 0 as n→∞. So∫ n

1

1

xpdx→ 1

p− 1(1− 0) =

1

p− 1as n→∞.

In particular, the sequence

(∫ n1

1

xpdx

)n∈N

is convergent (with limit1

p− 1). Hence, by the

Integral Test, the series∞∑n=1

1

npis convergent (for p > 1).

Exercise 9.2.4. Use the Integral test to prove that the series∞∑n=1

1

npis divergent if p < 1. (We

also proved this in Example 9.1.6.)

Proof of the Integral Test: First of all, it is clear that∫ n1 f(x)dx ≤

∫ n+11 f(x)dx and so the

sequences of integrals is an increasing sequence. Hence the second paragraph of the Theoremis nothing more than the Monotone Convergence test.

So we look at the first assertion. The basic idea is that the integral∫ nx=1 f(x)dx defines the

area under that curve for 1 ≤ x ≤ n. This we can approximate by summing the areas of aseries of rectangles with width 1. The point of the proof is that this is (close to) the partialsum we want. To make this precise we can either draw it (which I will do in the lectures sinceit is probably easier to understand) or a give more algebraic proof, as follows.

57

Let n ∈ N and consider the two rectangles A, B in the xy-plane given by

A = {< x, y >∈ R2 : n ≤ x ≤ n+ 1, 0 ≤ y ≤ f(n+ 1)},

B = {< x, y >∈ R2 : n ≤ x ≤ n+ 1, 0 ≤ y ≤ f(n)}.

If you draw the graph of f , bearing in mind that it is a positive and decreasing function,you will see that the rectangle A is contained in the region between the x-axis and the graphof f , and that B contains this region. So it follows that

area(A) ≤∫ n+1

nf(x)dx ≤ area(B).

But area(A) = 1 · f(n+ 1) and area(B) = 1 · f(n), so

f(n+ 1) ≤∫ n+1

nf(x)dx ≤ f(n) .....(∗)

and this holds for all n ∈ N.We now let N ∈ N, and we sum the inequalities (∗) for n = 1, . . . , N − 1 to obtain

N−1∑n=1

f(n+ 1) ≤∫ 2

1f(x)dx+

∫ 3

2f(x)dx+ · · ·+

∫ N

N−1f(x)dx ≤

N−1∑n=1

f(n)

which gives, for all N ∈ N,

N∑n=2

f(n) ≤∫ N

1f(x)dx ≤

N−1∑n=1

f(n) .....(∗∗).

After this preparatory work we now prove Theorem 9.2.1.

(⇒): Suppose that the series∞∑n=1

f(n) converges. Then the partial sums are bounded (by 9.1.1,

since the terms f(n) are positive). So there is a positive real number M such that for all N ∈ Nwe have that

∑N−1n=1 f(n) ≤ M . Hence by (∗∗),

∫ N1 f(x)dx ≤ M for all N ∈ N and so the

sequence(∫ n

1 f(x)dx)n∈N is a bounded sequence.

But since f is a positive function, an increase in the range over which we integrate f willresult in a larger value for the integral. This implies that the sequence

(∫ n1 f(x)dx

)n∈N is an

increasing sequence. Thus, by the Monotone Convergence Theorem, it is a convergent sequenceas required.

(⇐): Suppose that the sequence(∫ n

1 f(x)dx)n∈N converges. Then it is bounded (by Theo-

rem 2.3.9). Let L be a bound for it. So∫ N1 f(x)dx ≤ L for all N ∈ N. Hence, by (∗∗),∑N

n=2 f(n) ≤ L for all N ∈ N. This means that the partial sums of the series∑∞

n=2 f(n) are

bounded and so, being a series of positive terms, it converges (by 9.1.1). But then so does theseries

∑∞n=1 f(n) (by Exercise 9.1.3), as required.

This completes the proof of the Integral Test.

Remark 9.2.5. Sometimes the integral∫ nx=1 f(x)dx can be awkward to compute at the Left

Hand End, but in that case, since it is always OK to compute integrals of the form∫ nx=K f(x)dx

for some fixed 1 ≤ K (and then K ≤ n), it may be that there is a convenient choice for K.Indeed, in view of 9.1.3, one only has to verify that that the three conditions (i), (ii) and

(iii) hold for all sufficiently large x, i.e. for all x ≥ K (for some given K ∈ N). The same proofshows that

58

∞∑n=1

f(n) converges ⇐⇒∞∑n=K

f(n) converges ⇐⇒(∫ n

Kf(x)dx

)n≥K

converges.

Example 9.2.6. The series∑∞

n=2

n2

n3 − 1diverges.

Proof: Consider the function f : [2,∞)→ R defined by f(x) =x2

x3 − 1.

Clearly f(x) > 0 for x ≥ 2 and f is continuous. To see that f is decreasing, we can either

use calculus or algebra. Using calculus is easier: if f(x) =x2

x3 − 1then (after simplification)

f ′(x) = (−x4 − 2x)(x3 − 1)−2. Clearly this is negative for x > 1 and so our function f(x)decreases.

For an algebraic proof, suppose that 2 ≤ x ≤ y. Then

x2

x3 − 1≥ y2

y3 − 1⇐⇒ x2y3 − x2 ≥ y2x3 − y2

⇐⇒ y2 − x2 ≥ y2x3 − x2y3 ⇐⇒ (y − x)(y + x) ≥ y2x2(x− y).

Since the last statement here is true (for 2 ≤ x ≤ y) we can reverse the bi-implications. Thisshows that f is indeed decreasing.

So we may apply the Integral Test (in the form of 9.2.5). Now∫ n2 f(x)dx =

∫ n2

x2

x3 − 1dx.

In order to evaluate the integral we make the substitution u = x3 − 1. Thus du = 3x2dx andthe new limits are u = 7 to u = n3 − 1. Thus∫ n

2f(x)dx =

1

3

∫ n3−1

7

du

u=

1

3[lnu]n

3−17 =

1

3(ln(n3 − 1)− ln 7).

Since 13(ln(n3 − 1) − ln 7) → ∞ as n → ∞, the sequence

(∫ n2 f(x)

)n≥2 diverges and hence

so does the series∑∞

n=2

n2

n3 − 1by the Integral Test.

Remark 9.2.7. Assume that conditions (i)–(iii) from Theorem 9.2.1 hold. Let K ∈ N (usually Iwould have K = 1.) Then integrals of the form

∫∞K f(x)dx are called improper integrals and

in more detail we have the following.For any K < r < s we have

0 ≤ F (r) =

∫ r

Kf(x)dx ≤

∫ r

Kf(x)dxF (r) +

∫ s

rf(x)dxF (r) =

∫ s

Kf(x)dx = F (s),

and so by the Monotone Convergence Theorem either

1. The sequence(∫ nK f(x)dx

)n≥K is bounded and hence convergent, in which case we write

its limit as∫∞K f(x)dx and say that “

∫∞K f(x)dx converges”.

2. Alternatively the sequence(∫ nK f(x)dx

)n≥K is unbounded and hence tends to infinity. In

this case we say “∫∞K f(x)dx diverges” or indeed that

∫∞K f(x)dx =∞.

59

So we paraphrase Theorem 9.2.1 as saying that∑∞n=1 f(n) converges ⇐⇒

∫∞1 f(x) converges.

(Finally, note that talking about improper integrals usually involves a somewhat different kind oflimit, namely the limit of the F (r) as the real number r tends to infinity. Also, if the conditions(i)–(iii) do not hold, then you have to be much more careful with the definitions and propertiesof improper integrals.)

Exercise 9.2.8. If f is a function to which the Integral Test applies and if it tells us that we

have convergence, then we certainly cannot conclude that∞∑n=1

f(n) =∫∞1 f(x)dx.

Examine the proof of the Integral Test to show that

∞∑n=2

f(n) ≤∫ ∞1

f(x)dx ≤∞∑n=1

f(n)

whenever either side converges. Show, in fact, that we always have strict inequality here.

Example 9.2.9. Consider the series∑∞

n=2

1

n lnn. To investigate convergence we take f(x) =

1

x lnxin the Integral Test and consider the integral

∫ n2

1

x lnxdx. In order to evaluate it we make

the substitution u = lnx. Then du = dxx and the new range of integration is from u = ln 2 to

u = lnn. Thus ∫ n

2

1

x lnxdx =

∫ lnn

ln 2

du

u= [lnu]lnnln 2 = ln(lnn)− ln(ln 2).

Now (exercise) ln(lnn) → ∞ as n → ∞ and so the sequence

(∫ n2

1

x lnxdx

)n∈N

diverges.

Hence, by the Integral Test, the series∑∞

n=2

1

n lnndiverges.

Exercise 9.2.10. Let p ∈ R. Prove that the series

∞∑n=2

1

n(lnn)p

converges if and only if p > 1. What about the series

∞∑n=3

1

n lnn (ln lnn)p?

(We take the lower limit of summation to be 3 here because ln ln 2 is negative and so (ln ln 2)p

might not be defined.)

Final Remark: In the integral test you do have to be careful to check that conditions (i - iii)hold. The trouble is that without them it is easy to get silly counterexamples. For example,suppose we take the function y(x) = 1 + cos((2n+ 1)π); this function has been chosen so thatf(n) = 0 for all n and f(x) ≥ 0 for all x ≥ 0. So, here

∑f(n) = 0 + · · · + 0 = 0 is certainly

convergent, whereas the integral∫∞x=1 f(x)dx =∞.

Exercises 9.2.11. No doubt when you first saw the techniques of integration it took a whileto get the intuition about which technique to use for which example. The same goes for ourtechniques for testing the convergence of infinite series

∑an (and sequences). The only way

to develop this intuition is to do lots of examples. Here are some examples, with partial hintsabout how to approach them:

60

Remember the geometric series and the series∑np. Can you easily compare the given

sequence with these?If there is a dominant term of the form n! or cn the Ratio test will probably work. But

something like∑ n!

(n+ 1)!or∑ n!

(n+ 1)!simplify to

∑ 1

n+ 1respectively

∑ 1

(n+ 2)(n+ 1)so nothing is perfect. If limn→∞ an 6= 0 then the “Test for Divergence” (meaning 9.1.1) willwork. If the function can easily be integrated, do so. If the terms have alternating positive andnegative terms, see the next chapter.

So, what about:

(1)

∞∑n=1

n− 1

2n+ 1. Answer: Diverges by 9.1.1 as limn→∞ an 6= 0.

(2)∞∑n=1

√n3 + 1

3n3 + 4n2 + 2. Answer: Converges by comparison with

∑bn for bn =

√n3

n3= n−3/2

which converges as 3/2 > 1. Here you have to put a little work into the comparison test, but,

since we want to get convergence we should show that an =

√n3 + 1

3n3 + 4n2 + 2≤ λbn for some λ.

So certainly

∞∑n=1

√n3 + 1

3n3 + 4n2 + 2≤∞∑n=1

√n3 + 1

3n3=

1

3

∞∑n=1

√n−3 + n−6 ≤ 2

3

∞∑n=1

√n−3.

And, finally this converges by CT and 9.2.3.

(3)∞∑n=1

ne−n . Answer: This converges, either by the ratio test or the integral test. By

integrating by parts∫ n

x=1xe−xdx = −(x+ 1)e−x

∣∣∣∣nx=1

= −(n+ 1)e−n + 2e−1 → 2e−1 <∞

as n→∞.

(4)

∞∑n=1

2n

n!and

∞∑n=1

n!

2n. Answer: The Ratio Test will work to show the first converges and

the second diverges. (There is an easier test to use for the second one—do you see it?)

(5)∞∑n=1

1

2 + 3n. Answer: Since this is closely related to a geometric series we should use

the Comparison Test to compare it to∑ 1

3n and deduce that it converges.

61

Chapter 10

Series with Positive and NegativeTerms

10.1 Alternating Series

We know that for a series∞∑n=1

an to converge it is definitely not sufficient that the sequence

(an)n∈N of terms be null. (It is necessary, but not sufficient.) For example, 1n → 0 as n → ∞,

but∞∑n=1

1n does not converge. However, it turns out that if the terms an decrease in modulus

and alternate in sign:a1 > 0, a2 < 0, a3 > 0, . . .

then it is both necessary and sufficient for the convergence of∞∑n=1

an that (an)n∈N be null.

For example,∞∑n=1

(−1)n+1

ndoes converge as the following argument suggests:

We have

1− 1

2+

1

3− · · ·+ 1

2n+ 1− 1

2n+ 2+ · · ·

=

(1− 1

2

)+

(1

3− 1

4

)+ · · ·

(1

2n+ 1− 1

2n+ 2

)+ · · ·

=1

2+

1

12+ · · ·+ 1

(2n+ 1)(2n+ 2)

=∞∑n=0

1

(2n+ 1)(2n+ 2),

and this series converges by the Comparison Test (1

(2n+ 1)(2n+ 2)≤ 1

(n+ 1)2for all n).

(I say “suggests” here because the argument is not completely rigorous. Why not?) For anargument which gives the sum exactly, see the end of the chapter.

Theorem 10.1.1 (The Alternating Series Test). Let (an)n∈N be a decreasing sequence of positiveterms such that an → 0 as n→∞. Then the series

∞∑n=1

(−1)n+1an

converges.

62

Proof. Let

sn =n∑k=1

(−1)k+1ak

be a typical partial sum. Then

s2n = a1 − a2 + a3 − a4 + · · ·+ a2n−1 − a2n

= a1 − (a2 − a3)− (a4 − a5)− · · · − (a2n−2 − a2n−1)− a2n.Notice that all the terms in brackets here are non-negative since the sequence (an)n∈N is de-creasing. Also a2n > 0. It follows that for all n ∈ N, s2n ≤ a1.

Also s2(n+1) = s2n + a2n+1 − a2n+2. Since a2n+1 ≥ a2n+2, we obtain that s2(n+1) ≥ s2n forall n ∈ N. Hence (s2n)n≥1 is an increasing sequence which is bounded above (by a1). Thereforeit converges by the Monotone Convergence Theorem. Let its limit be `. So s2n → ` as n→∞.

Now consider the sequence (s2n+1)n≥1. Then s2n+1 = s2n + a2n+1, and both the series s2nand a2n+1 converge. So, by the Algebra of Limits Theorem

limn→∞

s2n+1 = limn→∞

s2n − limn→∞

a2n+1 = `− 0 = `.

It now follows easily in this situation that sn → ` as n → ∞, as well, which (by definition)

implies that∞∑n=1

an converges, as required.

In more detail, let ε > 0 be given. Choose N1 so that |s2n− `| < ε for all 2n ≥ N1 and N2 sothat |s2n+1−`| < ε for all 2n+1 ≥ N2. Then clearly |sm−`| < ε for all m ≥ N = max{N1, N2}.So, yes, sn → ` as n→∞ as required. 2

Remark: Sometimes it is useful to slightly modify the series in the Alternating series. Thesame argument will work (or we can use the version proved) to deduce either of the followingslightly modified versions of the theorem:

(A) Let (an)n∈N be a decreasing sequence of positive terms such that an → 0 as n → ∞.Then the series

∞∑n=1

(−1)nan

converges.(B) Let (an)n≥0 be a decreasing sequence of positive terms such that an → 0 as n → ∞.

Then the series∞∑n=0

(−1)nan

converges.Remark: It might be tempting to say that any alternating sum converges. But that is false: Letan be positive numbers such that an ≥ 0 for all n and (an) is decreasing. Then

∑n≥1(−1)nan

converges if and only if (an) is null.Proof. The direction ⇐ is The Alternating Series Test 10.1.1. The direction ⇒ is the “NullityTheorem 8.1.4.” 2

Example 10.1.2. Let p ∈ R. Then∞∑n=1

(−1)n+1

npconverges if and only if p > 0.

Proof: If p ≤ 0, then for all n,

∣∣∣∣(−1)n+1

np

∣∣∣∣ = n−p ≥ 1. So

((−1)n+1

np

)n∈N

is not a null sequence

and hence the series∞∑n=1

(−1)n+1

npcannot converge, by 8.1.4.

63

On the other hand, if p > 0, then

(1

np

)n∈N

is a decreasing null sequence of positive terms.

Hence the series∞∑n=1

(−1)n+1

npconverges by the Alternating Series Test.

We end this section with a precise formula for the alternating sum∞∑n=1

(−1)n+1

n. The last

step in this argument requires the definition of an integral as a limit of sums of areas. If youhave not seen this before, do not worry, since it is not something required for this course. Butthe computation (pointed out by Carolyn Dean) is fun and interesting. This computation isnot something you need to remember for this course.

Example 10.1.3. The series∞∑n=1

(−1)n+1

nconverges to ln(2).

Proof. This follows form a more careful analysis of the partial sums

s2N =

2N∑n=1

an = 1− 1

2+

1

3+ · · ·+ 1

2N − 1− 1

2N.

Claim: s2N =1

N + 1+

1

N + 2+ · · ·+ 1

2N.

Proof of the Claim: It is clear that it holds for N = 1 since s2 = 1− 12 = 1

2 . So suppose thatit holds for some N . Then, by definition

s2(N+1) = 1− 1

2+

1

3+ · · ·+ 1

2N + 1− 1

2N + 2

= s2N +1

2N + 1− 1

2N + 2

=

(1

N + 1+

1

N + 2+ · · ·+ 1

2N

)+

1

2N + 1− 1

2

1

N + 1by the inductive hypothesis

=

(1

N + 2+ · · ·+ 1

2N

)+

1

2N + 1+

1

2

1

N + 1by a simple manipulation.

Thus, by induction, the Claim is proven.This proves that for all n we have

s2n =1

n

[1

1 + 1n

+1

1 + 2n

+ · · ·+ 1

1 + nn

].

Finally, this is the Riemann Sum for the integral

∫ 1

x=0

1

1 + xdx where one is dividing the region

[0, 1] into n subdivisions. (You might not have seen this before, but it’s a simple enough idea,and will be done properly next year.) So, by definition

limn→∞

s2n =

∫ 1

x=0

1

1 + xdx = ln(x)

∣∣∣∣21

= ln(2).

64

10.2 Absolute Convergence

Suppose that we are given a series∞∑n=1

an where some of the an are positive and some negative.

Then (apart from the Alternating Series Test) there are not so many good rules for deciding

whether∞∑n=1

an converges or not. For example something like

1 +1

2− 1

3+

1

4+

1

5− 1

6· · ·

(where every third term is negative) is not something covered by one of our rules. The one other

rule we will have is that the series∞∑n=1

an does converge provided the series∞∑n=1|an| converges.

Before stating the result we make a definition.

Definition 10.2.1. Let∞∑n=1

an be any series. We say that∞∑n=1

an is absolutely convergent

if the series∞∑n=1|an| is convergent. If

∞∑n=1

an is convergent, but not absolutely convergent, then

we say it is conditionally convergent.

For example, the series∞∑n=1

(−1)n+1

n2is absolutely convergent because

∞∑n=1

∣∣∣∣(−1)n+1

n2

∣∣∣∣ =

∞∑n=1

1

n2which is convergent. However, the series

∞∑n=1

(−1)n+1

nis conditionally convergent because

it converges (see Example 10.1.2) but∞∑n=1

∣∣∣∣(−1)n+1

n

∣∣∣∣ =∞∑n=1

1

nwhich diverges.

The definition above rather presupposes that absolute convergence is stronger than conver-gence and we next give a proof of this fact.

Theorem 10.2.2. If the series∞∑n=1

an is absolutely convergent, then it is convergent.

Proof. We are given that∞∑n=1|an| converges. Let

pn =

{an if an ≥ 0,

0 if an ≤ 0.

Then∞∑n=1

pn is a series of positive terms and for all n ∈ N, pn ≤ |an|. Hence∞∑n=1

pn converges

by the Comparison Test. Similarly, if

qn =

{|an| if an < 0,

0 if an ≥ 0

then∞∑n=1

qn converges.

Hence, by the Algebra of Infinite Sums Theorem 8.1.5,∞∑n=1

(pn − qn) is convergent. But for

all n ∈ N, pn − qn = an and we are done. 2

65

Example 10.2.3.∞∑n=1

1

n2sin(nπ

4

)converges absolutely and hence converges.

Proof: The sine terms are 0,1√2, 1,

1√2, 0 − 1√

2, , · · · , so the alternating series test does not

help. However, as sin(x) ≤ 1, we do know that 0 ≤∣∣∣∣ 1

n2sin(nπ

4

)∣∣∣∣ ≤ 1

n2for all n. Hence

∞∑n=1

∣∣∣∣ 1

n2sin(nπ

4

)∣∣∣∣ converges by the ratio test. .

The same sort of argument means that we can modify some of our earlier tests to work forseries with positive and negative terms. For example:

Theorem 10.2.4. (The Modified Ratio Test) Suppose that an for all n ∈ N are any real

numbers and assume that

∣∣∣∣an+1

an

∣∣∣∣→ l as n→∞.

(i) If l < 1, then∞∑n=1

an is absolutely convergent and hence convergent.

(ii) If l > 1, then∞∑n=1

an is divergent.

(iii) If l = 1, then we still cannot conclude whether∞∑n=1

an is convergent or divergent.

Proof. (i) In this case the Ratio Test 9.1.7 says that∑

n≥1 |an| converges; i.e. that∞∑n=1

an is

absolutely convergent.(ii) In this case we know from the proof of 9.1.7 that |an| → ∞ as n → ∞. So, certainly

an 6→ 0 as n→∞. Thus, by Theorem 8.1.4,∞∑n=1

an is divergent. 2

Example 10.2.5.∞∑n=1

xn (and indeed∞∑n=1

(−1)nxn) converges if |x| < 1 and diverges if |x| > 1.

In this case, one can also check that it diverges when |x| = 1.

Final Comments. In Example 9.2.11, we saw how one might approach testing series forconvergence. Of course, there, the sequences all had positive terms. So how should one approachgeneral series? In a sense the rules are easier, as there are really only three possibilities for aseries

∑∞n=1 an:

1. Is limn→∞ = 0? If not then the series diverges by the n-th term test Theorem 8.1.4.

2. Does the Alternating Series Test 10.1.1 apply? If so, use it!

3. Otherwise you had better hope that the Absolute Convergence Test (Theorem 10.2.2)applies, in which case you are back to the ideas of Chapter 9. As we will see in the nextchapter, one of the most important cases where this case applies is when one can use theModified Ratio Test 10.2.4.

For examples of all these cases, see the next Exercise Sheet.

66

Chapter 11

Power Series

We now consider (the simplest case of) series where the nth term depends on a real variable xand we ask for which values of x does the series converge.

Definition 11.0.6. A series of the form∞∑n=1

anxn (also written

∑n≥1 anx

n) is called a power

series (in the variable x).

For example, as we have mentioned before, we can write ex =∞∑n=1

1

n!xn as a power series.

Similarly, the geometric series∞∑n=1

xn is a power series (with an = 1 for all n). As we saw

in Example 10.2.5, this converges absolutely for |x| < 1 and diverges otherwise. For anotherexample, consider

Example 11.0.7.∞∑n=1

1

nxn.

Let’s use the (modified) Ratio Test to study this example. Since it is customary to write

the series as∞∑n=1

anxn as we have done, when using the tests we should set (say) cn =

1

nxn, and

apply that rule to∞∑n=1

cn. So,

∣∣∣∣cn+1

cn

∣∣∣∣ =

∣∣∣∣ xn+1

n+ 1· nxn

∣∣∣∣ = |x| · n

n+ 1→ |x| as n→∞.

Thus, if |x| < 1 the series converges absolutely, by the Modified Ratio Test 10.2.4, while itdiverges if |x| > 1. When |x| = 1 we get two more cases:

If x = 1 the series is∞∑n=1

1

nwhich we know diverges.

If x = −1 the series is∞∑n=1

(−1)n

nwhich we know converges (by the Alternating Test 10.1.1

and the Remark just after that theorem).

The pattern we see in these last two examples is actually a general phenomenon: using

the ratio test we find that for any series there exists some number R ≥ 0 such that∞∑n=1

anxn

converges absolutely if |x| < R, it diverges if |x| > R and may or may not converge when|x| = R.

(In the last two examples R = 1 but we will see other examples where it can be any positivenumber. We will also see examples where R = 0 and R =∞)

67

11.1 The Radius of Convergence of a Power Series

The main result concerning convergence of power series is the following.

Theorem 11.1.1. Let∞∑n=1

anxn be a power series. Then there are three possibilities concerning

convergence.

Case (i):∞∑n=1

anxn converges only for x = 0.

Case (ii):∞∑n=1

anxn converges for all x ∈ R.

Case(iii): There exists a unique positive real number R such that

(a)∞∑n=1

anxn converges absolutely for all x with |x| < R, and

(b)∞∑n=1

anxn diverges for all x with |x| > R.

Before we go through the proof we require a lemma.

Lemma 11.1.2. If the power series∞∑n=1

anxn converges when x = r (where r is a nonzero real

number) and u is a real number satisfying 0 < |u| < |r|, then it converges absolutely for x = u.

Proof. We are given that∞∑n=1

anrn converges. In particular anr

n → 0 as n → ∞ (see 8.1.4)

and hence (by Theorem 2.3.9) there is some M > 0 such that |anrn| ≤M for all n ∈ N.

Let t =|u||r|

. Then 0 < t < 1 and

|anun| = |anrn|tn ≤Mtn

for all n ∈ N.

But∞∑n=1

Mtn is convergent (since the geometric series∞∑n=1

tn is convergent as 0 < t < 1) and

hence∞∑n=1|anun| is convergent by the Comparison Test. Thus

∞∑n=1

anun is absolutely convergent

as required. 2

Proof. [of 11.1.1]Assume that neither case (i) nor case (ii) in the statement of Theorem 11.1.1 holds. Then

there exists some b > 0 such that∞∑n=1

anbn is convergent, and some c > 0 such that

∞∑n=1

ancn is

divergent.

Now consider the set

S = {γ > 0 :∞∑n=1

|anxn| converges for all x such that |x| < γ}.

68

Now S is not empty because b ∈ S by Lemma 11.1.2. Also S is bounded above by c. (Since

if c′ ∈ S for some c′ > c then by definition of S,∞∑n=1|ancn| would converge, and hence so would

∞∑n=1

ancn, by 10.2.2 - a contradiction.)

Thus S is a non-empty set of real numbers which is bounded above. Hence by the Com-pleteness Property 2.4.6 it has a supremum, R say. We now show that this R has the desiredproperties.

(a) Suppose that x ∈ R and |x| < R. Then there is some y ∈ S with |x| < y (by Definition 2.4.1).

But then∞∑n=1|anxn| converges by the definition of S. That is, the series

∞∑n=1

anxn converges

absolutely.

(b) Suppose that x ∈ R and |x| > R. Choose any y with |x| > y > R. Suppose, for a contra-

diction, that∞∑n=1

anxn converges. Then by 11.1.2,

∞∑n=1|anun| converges for all u with |u| < |x|.

In particular, this series converges for all u with |u| < y, which implies that y ∈ S. But this isimpossible since y > R = sup(S).

There is one last thing to prove, namely that the R above is unique. It is left as an exerciseto show that there can be at most one real number R having properties (a) and (b). 2

Definition 11.1.3. If the real number R has the property that the series∞∑n=1

anxn converges

for all x with |x| < R, and diverges for all x with |x| > R, then R is called the radius of

convergence (RoC for short) of the power series∞∑n=1

anxn.

We also extend this definition by putting R = 0 if∞∑n=1

anxn only converges for x = 0, and

putting R =∞ if∞∑n=1

anxn converges for all x ∈ R.

By the theorem this definition does cover all possible cases and∞∑n=1

anxn converges absolutely

for |x| < R.

Finding the Radius of Convergence:To calculate the Radius of Convergence of a given series we always use the Ratio Test. In

the course of the calculations in the following examples we tacitly assume that x 6= 0. This isjustified since a power series always converges for x = 0 (and since we have been making theassumption that our series begin at n = 1, the sum is always 0 when x = 0.)

The Interval of Convergence. Even after one has found the radius of convergence R of a

power series∞∑n=1

anxn, you can also ask what happens when |x| = R.

Of course, there is nothing to check if R = 0 or R = ∞ so suppose that 0 < R < ∞. In

this case the Ratio Test will not help and you need to use some other rule to see if∞∑n=1

anxn

converges. The set {x :

∞∑n=1

anxn converges

}69

is called the interval of convergence. Certainly this contains (−R,R) and it may or may notcontain the end points ±R.

Example 11.1.4.∞∑n=1

xn - Here we already saw that the Radius of Convergence (RoC) is R = 1

and the interval of convergence is (−1, 1).

Example 11.1.5.∞∑n=1

1

nxn - Here we already saw that the RoC is R = 1 and the interval of

convergence is [−1, 1); that is it converges if x = −1 but diverges if x = +1.

Example 11.1.6. Consider the series∞∑n=1

nxn. To find the RoC we let cn = |nxn|. Then∞∑n=1

cn

is a series of positive terms and

cn+1

cn=

(n+ 1)|x|n+1

n|x|n=

(1 +

1

n

)· |x| → (1 + 0) · |x| = |x|, as n→∞.

So the (Modified) Ratio Test now tells us that if |x| < 1 then∞∑n=1

cn converges absolutely,

whereas if |x| > 1 then∞∑n=1

cn diverges. This shows that the RoC is 1. It is left for you to check

that it diverges if x = ±1.

Example 11.1.7. Consider the series∞∑n=1

(3n)!

(n!)3xn. We shall show that its radius of convergence

is1

27.

As above we let cn =

∣∣∣∣(3n)!

(n!)3xn∣∣∣∣ Then cn > 0 and

cn+1

cn=

(3(n+ 1))! · |x|n+1

((n+ 1)!)3· (n!)3

(3n)! · |x|n

=(3n)!(3n+ 1)(3n+ 2)(3n+ 3) · |x|n+1

(n!)3(n+ 1)3· (n!)3

(3n)! · |x|n

=(3n+ 1)(3n+ 2)(3n+ 3)

(n+ 1)(n+ 1)(n+ 1)· |x|

=(3 + 1

n)(3 + 2n)(3 + 3

n)

(1 + 1n)(1 + 1

n)(1 + 1n)· |x|

→ (3 + 0)(3 + 0)(3 + 0)

(1 + 0)(1 + 0)(1 + 0)· |x| (as n→∞)

= 27|x|.

So by the Ratio Test,∞∑n=1

cn converges for 27|x| < 1, i.e. for |x| < 1

27, and diverges for |x| > 1

27.

Therefore∞∑n=1

∣∣∣∣(3n)!

(n!)3xn∣∣∣∣ converges for |x| < 1

27and diverges for |x| > 1

27. Thus, as in the

70

previous example (and, indeed, all examples like this), we conclude that the RoC of the series∞∑n=1

(3n)!

(n!)3xn is

1

27.

Example 11.1.8. Consider the series∑∞

n=0

xn

n!. (It is often more natural to start the series at

n = 0 rather than, as we have been doing, n = 1. This makes no difference at all to convergenceissues. In particular, it does not affect the radius of convergence.)

Here we let cn =

∣∣∣∣xnn!

∣∣∣∣ and, just as in Example 9.1.9, we see thatcn+1

cn→ 0 as n → ∞

no matter what x is. Since 0 < 1 it follows that the radius of convergence of this series (theexponential series) is ∞.

Example 11.1.9. Finally, consider the series∑

n≥0 n! xn. We have, for cn = |n! xn|, that

cn+1

cn=

(n+ 1)! xn+1

n! xn= (n+ 1)x.

Notice that here for any x 6= 0, (n+1)x→∞ > 1 as n→∞. Thus the Ratio Test says that theseries diverges for any x 6= 0. Thus the RoC of the series

∑n≥0 n! xn is 0: the series converges

for no value of x except x = 0.

71

Chapter 12

Further Results on Power Series -further reading

This chapter discusses some results that will be useful for next year even if they are getting alittle outside the scope of this year’s course. In particular, you are not required to know theseresults for the exams.

12.1 More General Taylor Series.

First, we have only been discussing power series of the form∑

n≥0 anxn. However when working

with Taylor series one often wants to work with expansions around values α other than zero, inwhich case one would get an expression like

∑n≥1 an(x−α)n. Using a substitution y = (x−α)

allows one to reduce to the case we have been considering and so essentially all the sametheorems will hold. For example, one gets the following variant of Theorem 11.1.1:

Theorem 12.1.1. Consider the series∞∑n=1

an(x − α)n, for some fixed number α. Then there

are three possibilities concerning convergence.

Case (i):∞∑n=1

an(x− α)n converges only for x = α.

Case (ii):∞∑n=1

an(x− α)n converges for all x ∈ R.

Case(iii): There exists a unique positive real number R such that

(a)∞∑n=1

an(x− α)n converges absolutely for all x with |x− α| < R, and

(b)∞∑n=1

an(x− α)n diverges for all x with |x− α| > R.

Proof. Set y = x − α. Then our series becomes∞∑n=1

anyn. So, now apply Theorem 11.1.1 to

that series and you will find the conclusion you get is exactly the present theorem. For example,

if∞∑n=1

anyn has radius of convergence R then

∞∑n=1

anyn converges if |y| < R, which is the same

72

as saying that∞∑n=1

an(x − α)n =∞∑n=1

anyn converges if |x − α| < R. Similarly it diverges if

|x− α| > R. 2

12.2 Rearranging Series

Let us begin with the following remarkable example. Given a series∑∞

n=1 an, then a rear-rangement of this series means a series of the form

∑∞n=1 bn which is got by rearranging the

terms of the first one. More formally, there exists a bijection φ : N → N such that bn = aφ(n)for each n.

Here is a famous example.

Example 12.2.1. Recall from Example 10.1.3 that

1− 1

2+

1

3− 1

4+ · · · =

∞∑n=1

(−1)n+1

n= ln(2).

Here is a rearrangement with sumln(2)

2. To do this we use the rearrangement where we always

have two negative terms in succession but only one positive one:

1− 1

2− 1

4+

1

3− 1

6− 1

8+

1

5− · · ·

=

(1− 1

2− 1

4

)+

(1

3− 1

6− 1

8

)+

(1

5− 1

10− 1

12

)· · ·

=

(1

2− 1

4

)+

(1

6− 1

8

)+

(1

10− 1

12

)· · ·

=1

2

(1− 1

2+

1

3− 1

4+

1

5· · ·)

=ln(2)

2.

In fact we can get any number we want in this way.

Proposition 12.2.2. Suppose that∑

n≥1 an is a conditionally convergent series, and let αbe any real number. Then there exists some rearrangement

∑n≥1 bn of this series for which∑

n≥1 bn = α.

Proof. (Outline) We start with some notation essentially coming from Question 3 on theExercise sheet for Week 11:

Notation 12.2.3. Given a series∑

n≥1 an, write

a+n =

{an if an ≥ 0

0 if an ≤ 0and a−n =

{0 if an ≥ 0

an if an ≤ 0

By that exercise sheet, the series a+n and a−n are divergent. Hence∑a+n =∞, since all the

terms are positive and, similarly,∑a−n = −∞.

Now, we construct the sequence {bn}: Since∑a+n =∞ we can therefore take the first few

positive numbers ak1 , ak2 . . . , akr so that X1 = ak1 + ak2 + · · · + akr > α. (More precisely wechoose the smallest possible r with this property.) Then, as

∑a−n = −∞, we can add to this

the first few negative numbers akr+1 , akr+2 . . . , aks so that

X2 = (ak1 + ak2 + · · ·+ akr) +(akr+1 + akr+2 + · · ·+ aks

)< α.

73

(Again we take s minimal with this property.) Now keep going; adding positive terms aj to getbigger than α then immediately adding more negative ones so that the sum becomes less thanα, then immediately adding more positive terms et cetera.

Finally, we must check that the sum actually converges to α. This is the same idea as inthe proof of the Alternating Series test, but the notation is messy. More formally, we haveconstructed the numbers X1 > α, then X2 < α, and X3 > α, etc. Let a`u be the last numberadded to make Xu; thus in the last paragraph a`1 = akr while a`2 = aks . The key point tonotice is that, as each a`j is chosen minimally, |Xj − α| ≤ |a`j | in each case. But, by the nth

term test Theorem 8.1.4, limj→∞ |a`j | → 0 as j → ∞. Therefore, by the Sandwich Theorem3.1.4, |Xj − α| → 0 as j →∞. Which is exactly what we wanted to prove. 2

Strangely enough, for absolutely convergent series, taking rearrangements does not changethe sum.

Theorem 12.2.4. Let∑∞

n=1 an be an absolutely convergent series, say with∑∞

n=1 an = `. Thenfor any rearrangement

∑∞n=1 bn of

∑∞n=1 an we have

∑∞n=1 bn = `, as well.

Proof. Define the a+n , a−n and similarly b+n , b

−n as in Notation 12.2.3. By Question 3(a) of the

Exercise sheet for Week 11,∑∞

n=1 a+n is convergent, say with A =

∑∞n=1 a

+n . Also, obviously the

partial sums {u` =∑`

n=1 a+n } form an increasing sequence. Now, for any t, the terms b+1 , . . . , b

+t

appear in {a+1 , . . . , a+` } for some ` and hence the partial sums

vt =t∑

n=1

b+n ≤∑n=1

a+n ≤∞∑n=1

a+n .

Hence {vt}t≥1 is an ascending and bounded sequence. Thus, by the Monotone ConvergenceTheorem, it has a limit, say B. Notice also that B ≤ A by Lemma 4.2.5. In particular this alsoshows that

∑bn is absolutely convergent (if not then Question 3 of the Exercise sheet for Week

11 would imply that∑b+n = ∞). So we can reverse this argument and see that each u` ≤ B

and hence that A ≤ B.In other words A = B.Now repeat this argument for the a−n (with decreasing sequences of partial sums) to show

that∑

n≥1 a−n =

∑n≥1 b

−n . Now

∑n≥1 an =

∑n≥1 a

+n +

∑n≥1 a

−n (use Question 3(a) of the

Exercise sheet for Week 11, again), and similarly∑

n≥1 bn =∑

n≥1 b+n +

∑n≥1 b

−n . Therefore

we conclude that ∑n≥1

an =∑n≥1

a+n +∑n≥1

a−n =∑n≥1

b+n +∑n≥1

b−n =∑n≥1

bn.

2

One significant application of this result is that it tells us what happens when we multiplypower series. To set this up, recall the equation ex · ey = ex+y. In terms of power series thisshould mean that ∑

n≥0

xn

n!

∑n≥0

yn

n!

=

∑n≥0

(x+ y)n

n!

.

(“should” because when you construct exponentials formally in next year’s Real Analysis courseyou will proceed in the opposite direction: you define the exponential ex as the power series∑∞

n=0

xn

n!. This for example solves the problem of what exactly e is, but it does mean that rules

like exey = ex+y are no longer “obvious”.)

74

Anyway, this displayed equation is true and actually works much more generally. To provethis, we start with an analogous result for series. Here it is more natural to start summing ourpower series at n = 0 rather than n = 1 and so the next few results will be phrased in that way.

Theorem 12.2.5. Let∑∞

n=0 an and∑∞

n=0 bn be absolutely convergent series, say with∑∞

n=0 an =A and

∑∞n=0 bn = B.

Write∑

i,j≥0 aibj for the infinite sum consisting of the product, in any order, of every termof the first series multiplied by every term of the second series. Then

∑i,j≥0 aibj is absolutely

convergent (in the sense that the sum∑

i,j≥1 |aibj | is also convergent) and∑

i,j≥0 aibj = AB.

Remark: One use of this theorem is when we need to make sense of a summation with twoindices in an expression like

∑i,j≥0 aibj . The problem is that one does not want to take some-

thing like∑∞

j=0 (∑∞

i=0 aibj), since now one has an infinite sum of infinite sums—which causes awhole new set of problems! We get around this by making it into a single infinite sum

∑m≥0 cm.

But then one runs into the problem that there are many different ways of choosing the sum.The theorem says this does not matter since whatever way you do it you get the same answer.Proof. Recall from the Foundations of Pure Mathematics course that Z≥0 ×Z≥0 is countable,so pick your favourite bijection φ : Z≥0 → Z≥0 × Z≥0, and write it as φ(m) = (φ1(m), φ2(m)).Now set cm = aφ1(m)bφ2(m); thus the definition of

∑i,j≥0 aibj in the statement of the theorem

just means∑

m≥0 cm. Then, by considering the partial sums as in the previous theorem, we see

that the partial sums γt =∑t

n=0 |cm| form an increasing sequence of non-negative terms andcertainly

γt ≤ X =

( ∞∑n=0

|an|

)( ∞∑n=0

|bn|

)for t ∈ N.

Therefore, by the Monotone Convergence Theorem again, the sequence {γt} has a limit boundedabove by X. In other words

γ =

∞∑m=0

|cm| ≤ X =

( ∞∑n=0

|an|

)( ∞∑n=0

|bn|

).

But we can now apply Theorem 12.2.4 to conclude that, however we ordered the terms,∑i,j≥1 aibj must give the same number γ. Now, one way of constructing the doubly infinite

sum is for the (N + 1)2 term to be(∑N

n=0 an

)(∑Nn=0 bn

). Since

(N∑n=0

an

)(N∑n=0

bn

)→ AB as N →∞,

it follows that in fact our sum∑

i,j≥0 aibj equals AB, as well.Notice that since we only get one possible sum, Proposition 12.2.2 says the series must be

absolutely convergent. 2

As a special case of the theorem we get:

Corollary 12.2.6. (Cauchy Product) Let∑∞

n=0 an and∑∞

n=0 bn be absolutely convergent series,say with

∑∞n=0 an = A and

∑∞n=0 bn = B. Set cn =

∑nr=0 arbn−r for each n.

Then∑∞

n=0 cn converges absolutely with∑∞

n=0 cn = AB.

Proof. Once again, the given sequence of partial sums∑t

n=0 cn is a subsequence of somesequence of partial sums constructed from

∑i,j≥0 aibj . Therefore, by Theorem 6.1.3, it also

75

converges to AB. Once again, as we only get one possible sum, Proposition 12.2.2 says theseries must be absolutely convergent. 2

Finally we get the result on exponentials that we wanted:

Corollary 12.2.7. Let E(x) =∑∞

n=0

xn

n!. Then E(x) is absolutely convergent for all x. More-

over for any x, y ∈ R one has E(x)E(y) = E(x+ y).

Proof. Of course E(x) is absolutely convergent for all x, by Example 11.1.8. The second sen-tence follows from the Corollary 12.2.6 once one notices that (by using the Binomial Theorem)

(x+ y)n

n!=

1

n!

n∑r=0

n!

r!(n− r)!xryn−r =

n∑r=0

(xr

r!

)(yn−r

(n− r)!

).

2

As you might imagine, the Cauchy Product Theorem 12.2.6 and its variants can be used toprove lots of other product formulas.

12.3 Analytic Functions

This section gives an indication of where you will go with power series and related topics infuture courses.

Let R be a positive real number or ∞ and suppose that the series∑∞

n=0 anxn has radius of

convergence R. Then for each x in the interval (−R,R) the series∑∞

n=0 anxn converges. Let

us call its sum f(x). Then f : (−R,R)→ R and

f(x) =∞∑n=0

anxn.

Functions that can be obtained this way are called analytic (on (−R,R)).

One can now show that this analytic function f can be differentiated and that f ′(x) is givenby the result of differentiating the series term by term:

f ′(x) =∞∑n=0

nanxn−1.

A fundamental fact is that the differentiated series has the same radius of convergence as theoriginal series, namely R. So we may differentiate again:

f ′′(x) =

∞∑n=0

n(n− 1)anxn−2

and repeat this for any k ∈ N to obtain

f (k)(x) =

∞∑n=0

n(n− 1)(n− 2) · · · (n− k + 1)anxn−k.

Now the negative exponents of x do not actually appear (as their coefficients contain a 0,so vanish) and we usually rewrite this series (by changing the variable of summation from n ton+ k) as

f (k)(x) =

∞∑n=0

an+k(n+ k)(n+ k − 1)(n+ k − 2) · · · (n+ 1)xn.

76

Now if we put x = 0 in this expression then all the terms except the first vanish, and weobtain the following formula for the coefficients of the original series in terms of the function f :

f (k)(0) = k!ak

or

ak =f (k)(0)

k!.

So, to summarise, analytic functions behave very nicely with respect to differentiation (andintegration and almost all other operations) and the coefficients an can easily be determined.These functions therefore are very convenient to work with.

In the other direction suppose now that, rather than a power series, we are given a functionf : (−R,R) → R which can be differentiated as many times as we please. Does it follow thatit is an analytic function? Certainly we know what its power series must be, by the formulaabove, namely

∞∑n=0

f (n)(0)

n!xn.

This power series is called the Taylor series of the function f and one might guess that itconverges to f(x) for all x ∈ (−R,R). If you have calculated examples of Taylor series before,like for ex, sinx or ln(1 + x), then this has always been the case. And, indeed, it is true withany “reasonable” function.

However, one can easily write down functions for which the Taylor series does not converge(except at x = 0) and, rather more shockingly, there are also examples of functions f : R→ Rwhose Taylor series converges for all x (i.e. the radius of convergence is∞) but do not convergeto f(x) ...

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