Limits of Sequences

8/2/2019 Limits of Sequences

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4.3. LIMIT OF A SEQUENCE: THEOREMS 115

4.3 Limit of a Sequence: Theorems

These theorems fall in two categories. The rst category deals with ways tocombine sequences. Like numbers, sequences can be added, multiplied, divided,... Theorems from this category deal with the ways sequences can be combinedand how the limit of the result can be obtained. If a sequence can be writtenas the combination of several "simpler" sequences, the idea is that it should beeasier to nd the limit of the "simpler" sequences. These theorems allow us towrite a limit in terms of easier limits. however, we still have limits to evaluate.The second category of theorems deal with specic sequences and techniquesapplied to them. Usually, computing the limit of a sequence involves usingtheorems from both categories.

4.3.1 Limit Properties

We begin with a few technical theorems. They do not play an important role incomputing limits, but they play a role in proving certain results about limits.

Theorem 310 Let x be a number such that8 > 0, jxj < , then x = 0.Proof. See problems at the end of the section.

Theorem 311 If a sequence converges, then its limit is unique.Proof. We assume that an ! L1 and an ! L2 and show that L1 = L2. Given > 0 choose N1 such that n N1 =) jan L1j <

2. Similarly, choose N2

such that n N2 =) jan L2j < 2

. Let N = max (N1; N2). If n N, then

jL1

L2

j=

jL1

an + an

L2

j jan L1j + jan L2j 0 such that janj M for all n.Proof. ChooseN such thatn N =) jan Lj < 1. By the triangle inequality,we have

janj jLj jan Lj< 1

Thus, if n N, janj < 1 + jLj. LetM1 = max(ja1j ; ja2j ; :::; jaNj). Let M =max(M1; 1 + jLj). Then, clearly, janj < M

Theorem 313 If a sequence converges, then 8 > 09N : m; n N =)jan amj <


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116 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES

Proof. Given > 0, we can choose N such that n; m N =) jan Lj < 2

and jam Lj 0be given. Choose N such that nk N =) jank Lj < . Now, if k N, thennk N therefore

jbk Lj = jank Lj<

This theorem is often used to show that a given sequence diverges. To do so,it is enough to nd two subsequences which do not converge to the same limit.Alternatively, once can nd a subsequence which diverges.

Example 316 Study the convergence of cos nThe subsequence cos2n converges to 1, while the subsequence cos(2n + 1) converges to 1. Thus, cos2n must diverge.

In the next two sections, we look at theorems which give us more tools tocompute limits.


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4.3.2 Limit Laws

The theorems below are useful when nding the limit of a sequence. Finding thelimit using the denition is a long process which we will try to avoid wheneverpossible. Since all limits are taken as n ! 1, in the theorems below, we willwrite lim an for lim

n!1an.

Theorem 317 Let(an) and(bn) be two sequences such thatan ! a andbn ! bwith a and b real numbers. Then, the following results hold:

1. lim(an bn) = (lim an) (lim bn) = a b.

2. lim(anbn) = (lim an)(lim bn) = ab.

3. if lim bn = b 6= 0 then lim

an

bn

= lim a

n

lim bn= a

b.

4. lim janj = jlim anj = jaj.

5. if an 0 then lim an 0.

6. if an bn then lim an lim bn.

7. if lim an = a 0 then lim pan =p

lim an =p

a.

Proof. We prove some of these items. The remaining ones will be assigned as

problems at the end of the section.

1. We provelim(an + bn) = (lim an)+(lim bn). The proof of lim(an bn) =(lim an) (lim bn) is left as an exercise. We need to prove that8 > 0,9N : n N =) jan + bn (a + b)j < . Let > 0 be given, choose N1such that n N1 =) jan aj <

2. Choose N2 such that n N2 =)

jbn bj < 2

. Let N = max (N1; N2). If n N, then

jan + bn (a + b)j = j(an a) + (bn b)j jan aj + jbn bj by the triangle inequality 0, 9N : n N =) janbn abj < . Sincean converges, it is bounded, let M be the bound i.e. janj < M. ChooseN1 such that n N1 =) jan aj <

2 (jbj + 1) . Choose N2 such that


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n N2 =) jbn bj < 2 (M + 1)

. Let N = max(N1; N2). If n N then

janbn abj = janbn anb + anb abj= jan (bn b) + b (an a)j janj jbn bj + jbj jan aj< M

2 (M + 1)+ jbj

2 (jbj + 1) 0, 9N : n N =)

jjan

j ja

jj< . Let > 0 be given, choose N such that n

N =

)jan aj < (since an ! a) . Ifn N, then we have:jjanj jajj < jan aj by the triangle inequality

<

5. We prove it by contradiction. Assume that an ! a < 0. Choose N suchthat n N ) jan aj < 1

2a. Then,

an a < 12

a

) an < 12

a

) an < 0which is a contradiction.

6. We apply the results found in parts 1 and 5 to the sequencean bn.7. See problems

Remark 318 Parts 1, 2 and 3 of the above theorem hold even when a and bare extended real numbers as long as the right hand side in each part is dened.You will recall the following rules when working with extended real numbers:

1. 1 + 1 = 1 1 = (1) (1) = 12. 11 = (1) 1 = 1 (1) = 13. If x is any real number, then

(a) 1 + x = x + 1 = 1


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(b) 1 + x = x 1 = 1

(c)

x

1 =x

1 = 0(d)

x

0=

1 if x > 01 if x < 0

(e) 1 x = x 1 = 1 if x > 0

1 if x < 0

(f) (1) x = x (1) = 1 if x > 0

1 if x < 0

4. However, the following are still indeterminate forms. Their behavior isunpredictable. Finding what they are equal to requires more advanced tech-niques such as lHpitals rule.

(a) 1 + 1 and 1 1(b) 0 1 and 1 0(c)

11 and

0

0

Remark 319 When using theorems from this category, it is important to re-member previous results since these theorems allow us to write a limit in termsof other limits, we hopefully know. The more limits we know, the better o weare.

Example 320 If c 6= 0, nd lim cn

. We know from an example in the previoussection that lim 1

n= 0. Therefore

limc

n= c lim

1

n= c 0= 0

Example 321 Find lim 1n2

. In the previous section, we computed this limitusing the denition. We can also do it as follows.

lim1

n2= lim

1

n 1

n

=

lim

1

n

lim

1

n

= 0 0= 0

Remark 322 From the above example, we can see that ifp is a natural number,lim 1

np= 0.


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Example 323 Find lim n2+3n

2n2+1.

This problem involves using a standard technique you should remember. We

show all the steps, then we will draw a general conclusion. We begin by factoringthe term of highest degree from both the numerator and denominator.

limn2 + 3n

2n2 + 1= lim

n2

1 + 3n

n2

2 + 1n2

= lim1 + 3

n

2 + 1n2

Now,

lim

1 +

3

n

= lim(1) + 3 lim

1

n

= 1

and

lim

2 +

1

n2

= lim (2) + lim

1

n2

= 2

Since both the limit of the numerator and denominator exist and the limit of thedenominator is not 0, we can write

limn2 + 3n

2n2 + 1=

lim

1 + 3n

lim

2 + 1

n2

=

1

2

Remark 324 The same technique can be applied to every fraction for whichthe numerator and denominator are polynomials in n. We see that the limitof such a fraction will be the same as the limit of the quotient of the terms ofhighest degree. Let us look at some examples:

Example 325 lim3n2 + 2n 10

2n + 5= lim

3n2

2n= lim

2

3n =

2

3lim n = 1

Example 326 lim5n3 2n + 1

2n4 + 5n2 2 = lim5n3

2n4= lim

5

2n=

5

2lim

1

n= 0

Example 327 lim

2n3

n2 + 2n + 1

n3 + 10n2 5 = lim2n3

n3 = lim2 = 2

4.3.3 More Theorems on Limits

In example 304 , we used an approximation to simplify the problem a little bit.In this particular example, the approximation was not really necessary, it was


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more to illustrate a point. Sometimes, if the problem is more complicated, itmay be necessary to use such an approximation in order to be able to nd the

condition n has to satisfy. In other words, when we try to satisfy jxn Lj < ,we usually simplify jxn Lj to some expression involving n. Let E1 (n) denotethis expression. This gives us the inequality E1 (n) < which we have to solvefor n. If it is too hard, we then try to nd a second expression we will call E2 (n)such that E1 (n) < E2 (n) < . E2 (n) should be such that solving the inequalityE2 (n) < is feasible and easier. In order to achieve this, several tricks are used.We recall some useful results, as well as some theorems below.

Theorem 328 (Bernoullis inequality) If x 1, and n is a natural num-ber, then (1 + x)

n 1 + nx.

Theorem 329 (binomial theorem) (a + b)n = an+nan1b+n (n 1)

2

an2b2+

n (n 1) (n 2)3!

an3b3 + ::: + nabn1 + bn.

Corollary 330 (1 + x)n = 1 + nx +n (n 1)

2x2 +

n (n 1) (n 2)3!

x3 + ::: +

nxn1 + xn.

In particular, when x 0, then (1 + x)n is greater than any part of the righthand side. For example, we obtain Bernoullis inequality: (1 + x)

n 1 + nx.We could also write (1 + x)

n n (n 1)2

x2 or (1 + x)n n (n 1) (n 2)

3!x3.

And so on. This is useful to get approximations on quantities like 3n. Werewrite it as

3n = (1 + 2)n

Theorem 331 (squeeze theorem) If an ! L, cn ! L and an bn cn,then bn ! LProof. We need to prove that 8 > 09N : n N =) jbn Lj < . Let > 0be given. Choose N1 such that n N1 =) jan Lj < or < an L < .Similarly, choose N2 such that n N2 =) < cn L < . Let N =max(N1; N2). If n N then

an bn cn () an L bn L cn L() < an L bn L cn L <

() < bn

L <

() jbn Lj <

Theorem 332 If 0 < a < 1 then an ! 0


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Proof. Let x =1

a 1. Then, x > 0 and a = 1

1 + x. For n 1,

an =1

(1 + x)n

11 + nx

by Bernoullis inequality

1

x

So, given > 0, N; =1

xwill work.

We look at a few more examples, to see how all these results come into play.

Example 333 Find lim2n2 + 1

n2 + 1using the denition

Of course, we can nd this limit by using the theorems on limits. Here, we doit using the denition, as asked. We think the limit is 2. We want to show that

for every > 0, there exists N such that n N =)

2n2 + 1

n2 + 1 2

< . First,

we simplify the absolute value.2n2 + 1

n2 + 1 2

= 1n2 + 1

=

1

n2 + 1

1

, then we will have

2n2

+ 1n2 + 1

2 < . So, N = 1 will work.

Example 334 Find lim1

n2 + 2n 4We think the limit is 0. We need to prove that for every > 0, there exists N


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such that n N =)

1

n2 + 2n

4

< . We begin by noticing that

n2 + 2n 4 = n2 + 2 (n 2)> n2 if n > 2

Therefore, if n > 2,

1n2 + 2n 4 = 1n2 + 2n 4

0, there exists N such that n N =)4n

2n> M. First, we notice that

4n

2n=

(1 + 3)n

2n

>

n (n 1) 322

2nby using the binomial theorem

>9 (n 1)

4

We can see that

9 (n 1)4

> M () n 1 > 4M9

() n > 4M + 99

so, N =4M + 9

9will work.

Remark 340 We used the binomial theorem to deduce that(1 + 3)n n (n 1) 32

2.

Often, when using this theorem, students do not know which term to keep, theterm in x, or x2, or x3,... The answer is that it depends on what we are trying

to achieve. Here, we wanted to show that4n

2n> M for any M, in other words,


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4n

2ncan be made as big as we want simply by taking n large enough. Since we

could not solve for n, we replaced 4n

2n by a smaller term. That smaller term

should have similar properties, in other words, the smaller term should also getarbitrarily large. If we take the smaller term too small, it will not work. For

example, it is true that4n

2n>

1

2n. However,

1

2ncannot be made as large as

we want, so this does not help us. If instead of using(1 + 3)n n (n 1) 3

2

2(terms of second degree), we had used (1 + 3)

n 1 + 3n, then we would haveobtained

1 + 3n

2n> M () 2nM < 1 + 3n

() 2nM 3n < 1() n (2M 3) < 1() n < 1

2M 3So, this does not work. This tells us that 1+3n

2n> M only for a few values of n.

This is why we used terms of higher degree.

Example 341 Find limp

n + 2 pn

limp

n + 2 pn = limp

n + 2 pn pn + 2 + pnpn + 2 +

pn

= lim2p

n + 2 +p

n

= 2lim

pn + 2 +

pn

= 0

Remark 342 In many proofs or problems, dierent versions of the triangleinequality are often used. As a reminder, here are the dierent versions of thetriangle inequality students should remember.

jjaj jbj j ja bj jaj + jbjand

jjaj jbj j ja + bj jaj + jbj

Finally, we give a theorem which generalizes some of the examples we didabove.

Theorem 343 All limits are taken as n ! 1.

1. If p > 0 then lim1

np= 0.


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2. If p > 0 then lim np

p = 1.

3. limn

pn = 1.4. If p > 1 and 2 R then lim n

pn= 0.

5. If jpj < 1 then limpn = 0

6. 8p 2 R, lim pn

n!= 0.

Proof. We prove each part separately.

1. We can use the denition. Let > 0 be given. We show there exists

N > 0 : n N =)

1

np 0

< . We begin with what we want to achieve.

1np 0 < () 1np <

() np > 1

() n > 1pp

So, we see that if N is the smallest ineteger larger than 1pp , the result will

follow.

2. See problems

3. Letxn =

np

n 1. Proving the result amounts to proving that lim

xn = 0.From xn = npn 1, we can write

np

n = xn + 1

n = (xn + 1)n

Using the binomial theorem, we see that if n 2

n n (n 1)2

x2n

Thus

x2n 2

n 1

It follows that for all n 2 we have0 xn

r2

n 1

Since limq

2

n1 = 0, it follows using the squeeze theorem that lim xn = 0.


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4. Here again, we will use the binomial theorem. Since p > 1, we can writep = 1 + q with q > 0. Therefore, if k is a positive integer such that k > ,

we have

pn = (1 + q)n

>n (n 1) ::: (n k + 1)

k!qk

Now, if n > 2k, then k < 12

n. It follows thatn k + 1 > 12

n + 1 > 12

n. Itfollows that

n (n 1) ::: (n k + 1)k!

>nk

2kk!

and therefore

0 n

pn 2

kk!

qk 1

nk

Since

limn!1

2kk!

qk 1

nk=

2kk!

qklimn!1

1

nk

= 0

It follows from the squeeze theorem that limn

pn= 0

5. See problems (hint: write p = 1q

and use part 4 of the theorem with

= 0).

6. See problems.

4.3.4 Exercises

1. Prove that5n

n!! 0.

2. Prove thatn!

nn! 0.

3. Finish proving theorem 343.

4. Consider (xn) a sequence of real numbers such that limn!1

xn = x where

x > 0, prove there exists an integer N > 0 such that n N =) xn > 0.

5. Consider (xn) a sequence of real numbers such that xn 0 for any n. Ifx is a partial limit of (xn), prove that x 0. Prove the same result ifx = lim xn. Use this to show that if xn yn then lim xn lim yn.

6. Consider (xn) a sequence of real numbers and let x 2 R. Prove the twoconditions below are equivalent.


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(a) xn ! x as n ! 1.(b)

jxn

xj !

0 as n! 1

.

7. Consider (xn) a sequence of real numbers and let x 2 R. If lim xn = x,prove that lim jxnj = jxj.

8. Show by examples that ifjxnj ! jxj then xn does not necessarily converge.If it does converge, it does not necessarily converge to x.

9. Consider (xn) a sequence of real numbers and let x 2 R. Suppose thatlim xn = x. Dene yn = xn+p for some integer p. Prove that lim yn = x.

10. Given that an bn for every n and that lim an = 1, prove that lim bn =1.

11. Suppose that (an) and (bn) are sequences of real numbers such thatjan bnj < 1 for every n. Prove that if 1 is a partial limit of (an)then 1 is also a partial limit of (bn).

12. Two sequences are said to be eventually close if 8 > 0, 9N > 0 : n N =) jan bnj < .

(a) Prove that if two sequences (an) and (bn) are eventually close and ifa number x is the limit of the sequence (an) then x is also the limitof the sequence (bn).

(b) Prove that if two sequences (an) and (bn) are eventually close and ifa number x is a partial limit of the sequence (an) then x is also apartial limit of the sequence (bn).

13. Determine if the sequences below have limits. In each case, if a limit Lexists, given > 0, nd N such that n N =) jxn Lj < .

(a)n 1n + 1

(b)3n2 + 1

n2 + 1

(c)5n3 + n 4

2n3 + 3

(d)n3

n + 1

(e)4n2 + 1

n3 + n

(f)1

n2+

2

n2+ ::: +

n

n2

(g) (1)2n1

1 12n


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(h)

1 1

2

1 1

3

1 1

4

:::

1 1

n

(i)5n

2n

(j) Prove that if a > 1 then a2 > a >p

a > 1

(k) Prove thatn

2n! 0, n

2

2n! 0, n

3

2n! 0, n

4

2n! 0

14. Prove theorem 310

15. Finish proving theorem 317

16. Let (xn) and (yn) be two sequences. Assume further that lim xn = x andy is a partial limit of(yn). Prove that x + y is a partial limit for (xn + yn).

17. State and prove similar results for subtraction, multiplication and division.

18. Give an example of two divergent sequences (xn) and (yn) such that(xn + yn) is convergent.

19. Give an example of two sequences (xn) and (yn) such that xn ! 0, yn !1 and:

(a) xnyn ! 0(b) xnyn ! c where 0 < c < 1(c) xnyn ! 1(d) (xnyn) is bounded but has no limit.

20. Let (xn) be a sequence of real numbers such that xn ! 0. Prove thatx1 + x2 + ::: + xn

n7! 0

21. Let (xn) be a sequence of real numbers such that xn ! x for some realnumber x. Prove that

x1 + x2 + ::: + xnn

7! x

22. Prove that jxnj ! 0 () xn ! 0.23. Show by examples that if (xn) and (yn) are divergent sequences then

(xn + yn) is not necessarily divergent. Do the same for (xnyn)

24. Show that if (xn) ! 0 and (yn) is bounded then (xnyn) ! 0.25. Prove that if (xn) is a sequence in a set S R, then every nite partial

limit of (xn) must belong to S.


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26. Prove that ifS R and x 2 S then there exists a sequence (xn) in S suchthat xn

!x.

27. Let S R. Prove that the following conditions are equivalent:

(a) S is unbounded above.

(b) There exists a sequence (xn) in S such that xn ! 1.

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