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Overview (MA2730,2812,2815) lecture 17
Lecture slides for MA2730 Analysis I
Simon Shawpeople.brunel.ac.uk/~icsrsss
College of Engineering, Design and Physical Sciencesbicom & Materials and Manufacturing Research InstituteBrunel University
November 10, 2015
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Contents of the teaching and assessment blocks
MA2730: Analysis I
Analysis — taming infinity
Maclaurin and Taylor series.
Sequences.
Improper Integrals.
Series.
Convergence.
LATEX2ε assignment in December.
Question(s) in January class test.
Question(s) in end of year exam.
Web Page:http://people.brunel.ac.uk/~icsrsss/teaching/ma2730
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Lecture 17
MA2730: topics for Lecture 17
Lecture 17
Conditional and absolute convergence
The Leibniz test for alternating series
Rearrangements of conditionally convergent series: Jedi maths
Examples and Exercises
Reference: The Handbook, Chapter 5, Section 5.3.Homework: Finish Exercise Sheet 4aSeminar: Q5
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Time Management — Tip 3
Put First Things FirstStephen Covey, The Seven Habits of Highly Effective People
IMPORTANT NOT IMPORTANT
URGENT
Next LecturesTomorrow’s testBuy birthday present
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NOT URGENT
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Playstation
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Lecture 17
Reference: Stewart, calculus, sixth edition, Chapter 12.5.
Recall that in Lecture 13 we ‘concluded’ that ln(2) = 2 ln(2).
It’s false, of course: we did it by taking x = 1 in theMaclaurin series for ln(1 + x) to get,
ln(2) = 1− 1
2+
1
3− 1
4+
1
5− 1
6+
1
7− 1
8+ · · ·
and then re-ordering the way in which the terms added up.
Today we study the material in Subsection 5.3 of TheHandbook. We’ll then see what went wrong. . .
Specifically, we will begin with Alternating Series and thenmove on to understand conditional and absolute convergence,as well as some striking implications for working withalternating series.
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Lecture 17
We start with the concept of Alternating Series as described inSubsection 5.3.1 of The Handbook.
Definition 5.12 in The Handbook
The sequence {an} is alternating if the sequence {(−1)nan} is ofone sign, that is, every term is either non-positive or non-negative.Such sequences generate an alternating series
∑∞n=1 an.
Example
The sequence {an} for an =cos(πn)
n2generates the alternating
series,
∞∑
n=1
cos(πn)
n2= −1 +
1
22− 1
32+
1
42− 1
52+ · · ·
We see that (−1)nan is a sequence of positive terms.
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Lecture 17
Until now a few of our convergence tests for series cannot beapplied to alternating series because they required the terms of theseries to be of one sign.
But we can work with geometric series,∑
ark, for which r < 0.
The ratio and root tests also allowed for alternating series.
And we have seen this — our first test:
Our first test. Lemma 4.14: absolute value convergence
If∑∞
k=1 |ak| converges, then∑∞
k=1 ak converges.
We’ll return to this result later.
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Lecture 17
In a moment we shall state and prove the Leibniz test foralternating series.
Afterwards we’ll move to a discussion of the big issues that arisewhen dealing with alternating series.
Here’s the Leibniz test.
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Lecture 17
Leibniz Test, Theorem 5.13 in The Handbook
Let {an} be an alternating sequence such that {|an|} is strictlymonotonically decreasing with limn→∞ an = 0.
Then the series∑∞
n=1 an converges.
Moreover, the rate of convergence of the partial sum sn to thelimit sum s can be estimated by |s− sn| 6 |an+1| for n > 1.
Proof: boardwork
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Examples
∞∑
n=1
(−1)n+1
n
∞∑
n=1
(−1)n+1
n2
∞∑
n=1
(−1)n
∞∑
n=0
(−1)n
(2n)!
Boardwork and discussion
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The first of those examples was the Macalurin expansion of ln(2):
ln(2) =
∞∑
n=1
(−1)n+1
n= 1− 1
2+
1
3− 1
4+
1
5− 1
6+
1
7− 1
8+ · · ·
It is in fact perfectly well behaved. But if we change the negativesigns to positive we get,
∞∑
n=1
1
n= 1 +
1
2+
1
3+
1
4+
1
5+
1
6+
1
7+
1
8+ · · ·
This is the Harmonic series. We’ve seen from the integral test thatit diverges to ∞.
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So, we have an alternating harmonic series that converges eventhough the harmonic series diverges:
1− 1
2+
1
3− 1
4+
1
5− 1
6+
1
7− 1
8+ · · · =
∞∑
n=1
(−1)n+1
n= ln(2)
1 +1
2+
1
3+
1
4+
1
5+
1
6+
1
7+
1
8+ · · · =
∞∑
n=1
1
n= ∞
What’s going on?
First let’s recall our first test, from Lecture 15. . .
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Lecture 17
Our first test. Lemma 4.14: absolute value convergence
If∑∞
k=1 |ak| converges, then∑∞
k=1 ak converges.
In particular, this says that
If∞∑
n=1
1
nconverges, then
∞∑
n=1
(−1)n+1
nconverges.
But this is of no use because the harmonic series on the left doesnot converge.
This, obviously, cannot be true in general:
If∑∞
k=1 ak converges, then∑∞
k=1 |ak| converges.
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So, to summarize,
If∑∞
k=1 |ak| converges, then∑∞
k=1 ak converges.
If∑∞
k=1 ak converges, then∑∞
k=1 |ak| might converge. . .
. . . or it might not!
Examples:∑ 1n2 and
∑ (−1)n
n2 both converge.
∑ (−1)n
n converges to ln 2 BUT∑ 1
n diverges to ∞.
What a mess! Let’s tidy our thoughts up a bit.
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We are starting to uncover an important concept. Here it is.
Definition 5.11
The series∑∞
n=1 an is said to be absolutely convergent if the series∑∞n=1 |an| converges.
The series∑∞
n=1 an is said to be conditionally convergent if∑∞n=1 an converges but
∑∞n=1 |an| diverges.
So, for example:
∑ (−1)n
n2is absolutely convergent.
∑ (−1)n
nis conditionally convergent.
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What have found?
Let {an} be an alternating sequence.
We now see that an alternating series may converge whereasits one-signed counterpart might not.
Such a series,∑
an, is conditionally convergent.
Example:∑ (−1)n
n converges to ln 2 BUT∑ 1
n diverges to ∞and so
∑ (−1)n
n is conditionally convergent.
When both∑
an and∑ |an| converge the alternating series
is called absolutely convergent
Example:∑ (−1)n
n2is absolutely convergent.
Any ideas as to what is going on here?Any intuition? Any guesses?
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Lecture 17
Some Intuition. . .
Let’s return to the harmonic series and its alternating counterpart:
1− 1
2+
1
3− 1
4+
1
5− 1
6+
1
7− 1
8+ · · · =
∞∑
n=1
(−1)n+1
n= ln(2)
1 +1
2+
1
3+
1
4+
1
5+
1
6+
1
7+
1
8+ · · · =
∞∑
n=1
1
n= ∞
The clue is in the proof technique used earlier for the Leibniz test.
In a divergent series of positive terms,∑ |an|, the terms do not
approach zero sufficiently fast for the sum to have a limit.
In an alternating series,∑
an, there is the possibility that thenegative terms offset the accumulation of the positive terms. Thesum may then have a limit, and the alternating series will converge.
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This raises an interesting question. . .
If the negative terms are cancelling out some of the accumulationof the positive terms then. . .
. . . if we move the terms around can we change this cancellationand get a different sum?
The answer is YES — and we’ve already seen an example of itback in Lecture 13.
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Lecture 17
Alternating series are W E I R D!
Recall this from Lecture 13:
ln(2) = 1− 1
2+
1
3−1
4+
1
5− 1
6+
1
7−1
8+
1
9− 1
10+
1
11− 1
12+ · · ·
=
(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− 1
6+ · · ·
)
=1
2ln(2) So ln(2) =
1
2ln(2).
Discussion. . . by moving terms around we get a different sum forthe series. Which equals sign is WRONG?
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Rearrangements: Jedi maths!
To answer this we need to think about what a rearrangementactually is.
Let φ : N → N be a bijection (that is: φ is one-to-one and onto).
This means that as we feed each n ∈ N into φ our output willeventually produce the whole of N without duplication.
Two examples:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, . . .} → {2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, . . .}{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, . . .} → {1, 3, 5, 7, 9, 2, 4, 6, 8, 10, 11, 13, . . .}
So, we can allow for any such rearrangement φ. . .
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Definition 5.14 in The Handbook
A series∑∞
m=1 bm is a re-arrangement of a series∑∞
n=1 an if thereexists a bijection φ : N → N such that bm = an for m = φ(n).
Rearrangements Theorem, Theorem 5.15
Every rearrangement of an absolutely convergent series isabsolutely convergent and all have the same sum.
Discussion: boardwork
Remark: Since ln(2) = −∑(−1)nn−1 is conditionally convergent
we cannot expect Theorem 5.15 to generalise to conditionallyconvergent series.
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Conditionally convergent series therefore have the potential todiverge — after some rearrangement.
This suggests that there must be some divergent behaviourembedded into each conditionally convergent series. This is thesubject of our next result.
Lemma 5.16
If∑∞
n=1 an is only conditionally convergent, then the series of itspositive terms, and of its negative terms, are both divergent.
Proof: we assume that none of the an are zero and set,
a+n =
{an if an > 00 if an < 0
and a−n =
{0 if an > 0an if an < 0
so that we can expect∞∑
n=1
an =
∞∑
n=1
a+n +
∞∑
n=1
a−n .
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We expect that∞∑
n=1
an =
∞∑
n=1
a+n +
∞∑
n=1
a−n .
There are now just three possibilities for∑
a+n and∑
a−n . . .
1) Both converge: P =∑
a+n and M =∑
a−n .
But this means that∑ |an| = P −M which is not possible
because∑
an is not absolutely convergent.
2) Only one converges: M =∑
a−n and ∞ =∑
a+n for example.
But now∑
an = ∞+M which is not possible because∑
an isgiven as conditionally convergent.
3) Both∑
a+n and∑
a−n diverge.
This completes the proof.
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Remarkably, a conditionally convergent series can be rearranged togive any value whatsoever, including ±∞
Riemann’s rearrangment theorem, Theorem 5.17
If a series∑∞
n=1 an converges conditionally, then for any numbers ∈ R there exists a re-arrangement, φ : N → N, such thats =
∑∞m=1 bm, where bm = an for m = φ(n).
Discussion: boardwork
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Summary
We can:
recognise an alternating series and find its absolute valuecounterpart.
apply and prove the Leibniz convergence test for alternatingseries.
test for absolute and conditional convergence.
apply the rearrangement theorem.
appreciate the result of Riemann’s rearrangement theorem.
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End of Lecture
Computational andαpplie∂ Mathematics
Put First Things FirstStephen Covey, The Seven Habits of Highly Effective People
Reference: The Handbook, Chapter 5, Section 5.3.Homework: Finish Exercise Sheet 4aSeminar: Q5
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