IntroRealAnal-ch03

Chapter 3

Sequences

3.1 Basic Properties

Definition 3.1.1. A sequence is a function a : N ! R.

Instead of using the standard function notation of a(n) for sequences, it isusually more convenient to write the argument of the function as a subscript,an

.

Example 3.1.1. Let the sequence an

= 1 � 1/n. An easy calculation showsa1 = 0, a2 = 1/2, a3 = 2/3, etc.

Example 3.1.2. Let the sequence bn

= 2n. It’s easy to see b1 = 2, b2 = 4, b3 = 8,etc.

Example 3.1.3. Some sequences are not defined by an explicit formula, but aredefined recursively. This is an inductive method of definition in which successiveterms of the sequence are defined by using other terms of the sequence. The mostfamous of these is the Fibonacci sequence. To define the Fibonacci sequence,fn

, let f1 = 0, f2 = 1 and for n > 2, let fn

= fn�2 + f

n�1. The first few termsare 0, 1, 1, 2, 3, 5, 8, . . . . There actually is a simple formula that directly givesfn

, but we leave its derivation as an exercise.

It’s often inconvenient for the domain of a sequence to be N, as required byDefinition 3.1.1. For example, the sequence beginning 1, 2, 4, . . . can be written20, 21, 22, . . . and the sequence function written as 2n�1 with domain N, or 2n

with domain !. The second is simpler, and is usually chosen as the betterrepresentation. As long as there is a simple substitution to write the sequencefunction in the form of Definition 3.1.1, there’s no reason to adhere to the letterof the law. In general, the domain of a sequence can be any set of the form{n 2 Z : n � N} for some N 2 Z.

Definition 3.1.2. A sequence an

is bounded if {an

: n 2 N} is a bounded set.This definition is extended in the obvious way to bounded above and boundedbelow.

3-1

3-2 CHAPTER 3. SEQUENCES

The sequence of Example 3.1.1 is bounded, but the sequence of Example3.1.2 is not, although it is bounded below.


converges to L 2 R if for all " > 0 there existsan N 2 N such that whenever n � N , then |a

n

�L| < ". If a sequence does notconverge, then it is said to diverge.

When an

converges to L, we write limn!1 a

n

= L, or often, more simply,an

! L.

Example 3.1.4. Let an

be as in Example 3.1.1. We claim an

! 1. To see this,let " > 0 and choose N 2 N such that 1/N < ". Then, if n � N

|an

� 1| = |(1� 1/n)� 1| = 1/n 1/N < ",

so an

! 1.

Example 3.1.5. The sequence bn

of Example 3.1.2 diverges. To see this, supposenot. Then there is an L 2 R such that b

n

! L. If " = 1, there must be anN 2 N such that |b

n

� L| < " whenever n � N . Choose n � N . |L � 2n| < 1implies L < 2n + 1. But, then

bn+1 � L = 2n+1 � L > 2n+1 � (2n + 1) = 2n � 1 � 1 = ".

This violates the condition on N . We conclude that for every L 2 R thereexists an " > 0 such that for no N 2 N is it true that whenever n � N , then|b

n

� L| < ". Therefore, bn

diverges.


diverges to 1 if for every B > 0 there is anN 2 N such that n � N implies a

n

> B. The sequence an

is said to diverge to�1 if �a

n

diverges to 1.When a

n

diverges to 1, we write limn!1 a

n

= 1, or often, more simply,an

! 1.

A common mistake is to forget that an

! 1 actually means the sequencediverges in a particular way. Don’t be fooled by the suggestive notation intotreating 1 as a number!

Example 3.1.6. It is easy to prove that the sequence of Example 3.1.2 divergesto 1.

Theorem 3.1.1. If an

! L, then L is unique.

Proof. Suppose an

! L1 and an

! L2. Let " > 0. According to Definition3.1.2, there exist N1, N2 2 N such that n � N1 implies |a

n

� L1| < "/2 andn � N2 implies |a

n

� L2| < "/2. Set N = max{N1, N2}. If n � N , then

|L1 � L2| = |L1 � an

+ an

� L2| |L1 � an

|+ |an

� L2| < "/2 + "/2 = ".

Since " is an arbitrary positive number an application of Problem 2.7 showsL1 = L2.

July 24, 2012 http://math.louisville.edu/⇠lee/ira

3.1. BASIC PROPERTIES 3-3

Theorem 3.1.2. an

! L i↵ for all " > 0, the set {n : an

/2 (L � ", L + ")} isfinite.

Proof. ()) Let " > 0. According to Definition 3.1.2, there is anN 2 N such that{a

n

: n � N} ⇢ (L�", L+"). Then {n : an

/2 (L�", L+")} ⇢ {1, 2, . . . , N�1},which is finite.

(() Let " > 0. By assumption {n : an

/2 (L � ", L + ")} is finite, so letN = max{n : a

n

/2 (L� ", L+ ")}+ 1. If n � N , then an

2 (L� ", L+ "). ByDefinition 3.1.2, a

n

! L.

Corollary 3.1.3. If an

converges, then an

is bounded.

Proof. Suppose an

! L. According to Theorem 3.1.2 there are a finite numberof terms of the sequence lying outside (L � 1, L + 1). Since any finite set isbounded, the conclusion is obvious.

The converse of this theorem is not true. For example, an

= (�1)n isbounded, but does not converge. The main use of Corollary 3.1.3 is as a quickfirst check to see whether a sequence might converge. It’s usually pretty easyto determine whether a sequence is bounded. If it isn’t, it must diverge.

The following theorem lets us analyze some complicated sequences by break-ing them down into combinations of simpler sequences.

Theorem 3.1.4. Let an

and bn

be sequences such that an

! A and bn

! B.Then

(a) an

+ bn

! A+B,

(b) an

bn

! AB, and

(c) an

/bn

! A/B as long as bn

6= 0 for all n 2 N and B 6= 0.

Proof. (a) Let " > 0. There are N1, N2 2 N such that n � N1 implies |an

�A| < "/2 and n � N2 implies |b

n

� B| < "/2. Define N = max{N1, N2}.If n � N , then

|(an

+ bn

)� (A+B)| |an

�A|+ |bn

�B| < "/2 + "/2 = ".

Therefore an

+ bn

! A+B.

(b) Let " > 0 and ↵ > 0 be an upper bound for |an

|. Choose N1, N2 2 N suchthat n � N1 =) |a

n

� A| < "/2(|B| + 1) and n � N2 =) |bn

� B| <"/2↵. If n � N = max{N1, N2}, then

|an

bn

�AB| = |an

bn

� an

B + an

B �AB| |a

n

bn

� an

B|+ |an

B �AB|= |a

n

||bn

�B|+ ||B||an

�A|

< ↵"

2↵+ |B| "

2(|B|+ 1)

< "/2 + "/2 = ".



(c) First, notice that it su�ces to show that 1/bn

! 1/B, because part (b) ofthis theorem can be used to achieve the full result.

Let " > 0. Choose N 2 N so that the following two conditions are satisfied:n � N =) |b

n

| > |B|/2 and |bn

�B| < B2"/2. Then, when n � N ,

��1

bn

� 1

B

�� =��B � b

n

bn

B

�� <��B2"/2

(B/2)B

�� = ".

Therefore 1/bn

! 1/B.

It is often easier to prove that a sequence converges by comparing it with aknown sequence than it is to analyze it directly. For example, a sequence suchas a

n

= sin2 n/n3 can easily be seen to converge to 0 because it is dominatedby 1/n3. The following theorem makes this idea more precise. It’s called theSandwich Theorem here, but is also called the Squeeze, Pinching, Pliers orComparison Theorem in di↵erent texts.

Theorem 3.1.5 (Sandwich Theorem). Suppose an

, bn

and cn

are sequencessuch that a

n

bn

cn

for all n 2 N.

(a) If an

! L and cn

! L, then bn

! L.

(b) If bn

! 1, then cn

! 1.

(c) If bn

! �1, then an

! �1.

Proof. (a) Let " > 0. There is an N 2 N large enough so that when n � N ,then L � " < a

n

and cn

< L + ". These inequalities imply L � " < an

bn

cn

< L+ ". Therefore, cn

! L.

(b) Let B > 0 and choose N 2 N so that n � N =) bn

> B. Thencn

� bn

> B whenever n � N . This shows cn

! 1.

(c) This is essentially the same as part (b).

3.2 Monotone Sequences

One of the problems with using the definition of convergence to prove that agiven sequence converges is that the limit of the sequence must be known inorder to verify that the sequence converges. This gives rise at best to a “chickenand egg” problem and in the worst case there is no nice representation of thelimit to use. The next few sections are ultimately concerned with removing thisdeficiency from Definition 3.1.2, but some interesting side-issues are exploredalong the way.

Not surprisingly, we begin with the simplest case.


3.2. MONOTONE SEQUENCES 3-5


is increasing, if an+1 � a

n

for all n 2 N. Itis strictly increasing if a

n+1 > an

for all n 2 N.A sequence a

n

is decreasing, if an+1 a

n

for all n 2 N. It is strictlydecreasing if a

n+1 < an

for all n 2 N.If a

n

is any of the four types listed above, then it is said to be a monotonesequence.

Notice the and � in the definitions of increasing and decreasing sequences,respectively. Many calculus texts use strict inequalities because they seem tobetter match the intuitive idea of what an increasing or decreasing sequenceshould do. For us, the non-strict inequalities are more convenient.

Theorem 3.2.1. A bounded monotone sequence converges.

Proof. Suppose an

is a bounded increasing sequence, L = lub {an

: n 2 N} and" > 0. Clearly, a

n

L for all n 2 N. According to Theorem 2.3.3, there existsan N 2 N such that a

N

> L � ". Then L � an

� aN

> L � " for all n � N .This shows a

n

! L.

If an

is decreasing, let bn

= �an

and apply the preceding argument.

The key idea of this proof is the existence of the least upper bound of thesequence when viewed as a set. This means the Completeness Axiom impliesTheorem 3.2.1. In fact, it isn’t hard to show Theorem 3.2.1 also implies theCompleteness Axiom, showing they are equivalent statements. Because of this,Theorem 3.2.1 is often used as the Completeness Axiom on R instead of theleast upper bound property we used in Axiom 8.

Example 3.2.1. The sequence en

=�1 + 1

n

�n

converges.

Looking at the first few terms of this sequence, e1 = 2, e2 = 2.25, e3 ⇡ 2.37,e4 ⇡ 2.44, it seems to be increasing. To show this is indeed the case, fix n 2 Nand use the binomial theorem to expand the product as

en

=nX

k=0

✓n

k

◆1

nk

(3.1)

and

en+1 =

n+1X

k=0

✓n+ 1

k

◆1

(n+ 1)k. (3.2)

For 1 k n, the kth term of (3.1) is



✓n

k

◆1

nk

=n(n� 1)(n� 2) · · · (n� (k + 1))

k!nk

=1

k!

✓1� 1

n

◆✓1� 2

n

◆· · ·

✓1� k + 1

n

◆

<1

k!

✓1� 1

n+ 1

◆✓1� 2

n+ 1

◆· · ·

✓1� k + 1

n+ 1

◆

=(n+ 1)n(n� 1)(n� 2) · · · (n+ 1� (k + 1))

k!(n+ 1)k

=

✓n+ 1

k

◆1

(n+ 1)k,

which is the kth term of (3.2). Since (3.2) also has one more positive term inthe sum, it follows that e

n

< en+1, and the sequence e

n

is increasing.Since 1/k! 1/2k�1 for k 2 N, equation (3.1) implies

✓n

k

◆1

nk

<1

k! 1

2k�1, k 2 N.

Substituting this into (3.1) yields

en

=nX

k=0

✓n

k

◆1

nk

< 1 + 1 +1

2+

1

4+ · · ·+ 1

2n�1

= 1 +1� 1

2n

1� 12

< 3,

so en

is bounded.Since e

n

is increasing and bounded, Theorem 3.2.1 implies en

converges.Of course, you probably remember from your calculus course that e

n

! e ⇡2.71828.

Theorem 3.2.2. An unbounded monotone sequence diverges to 1 or �1,depending on whether it is increasing or decreasing, respectively.

Proof. Suppose an

is increasing and unbounded. If B > 0, the fact that an

is unbounded yields an N 2 N such that aN

> B. Since an

is increasing,an

� aN

> B for all n � N . This shows an

! 1.The proof when the sequence decreases is similar.

3.3 Subsequences and the Bolzano-WeierstrassTheorem

Definition 3.3.1. Let an

be a sequence and � : N ! N be a function such thatm < n implies �(m) < �(n); i.e., � is strictly increasing sequence of natural


3.3. SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM3-7

numbers. Then bn

= a � �(n) = a�(n) is a subsequence of a

n

.

The idea here is that the subsequence bn

is a new sequence formed from anold sequence a

n

by possibly leaving terms out of an

. In other words, we see thatall the terms of b

n

must also appear in an

, and they must appear in the sameorder.

Example 3.3.1. If an

= sin(n⇡/2), then some possible subsequences are

bn

= a4n+1 =) bn

= 1,

cn

= a2n =) cn

= 0,

anddn

= an

2 =) dn

= (1 + (�1)n+1)/2.

Theorem 3.3.1. an

! L i↵ every subsequence of an

converges to L.

Proof. ()) Suppose � : N ! N is strictly increasing, as in the precedingdefinition. Clearly, �(1) � 1. Suppose �(n) � n for some n 2 N. Then�(n+ 1) > �(n) � n ) �(n+ 1) � n+ 1. This simple induction argument hasestablished �(n) � n for all n 2 N.

Now, suppose an

! L and bn

= a�(n) is a subsequence of a

n

. If " > 0, thereis an N 2 N such that n � N implies a

n

2 (L � ", L + "). From the precedingparagraph, it follows that when n � N , then b

n

= a�(n) = a

m

for some m � n.So, b

n

2 (L� ", L+ ") and bn

! L.(() Since a

n

is a subsequence of itself, it is obvious that an

! L.

The main use of Theorem 3.3.1 is not to show that sequences converge, but,rather to show they diverge. It gives two strategies for doing this: find twosubsequences converging to di↵erent limits, or find a divergent subsequence. InExample 3.3.1, the subsequences b

n

and cn

demonstrate the first strategy, whiledn

demonstrates the second.Even if the original sequence diverges, it is possible there are convergent

subsequences. For example, consider the divergent sequence an

= (�1)n. Inthis case, a

n

diverges, but the two subsequences a2n = 1 and a2n+1 = �1 areconstant sequences, so they converge.

Theorem 3.3.2. Every sequence has a monotone subsequence.

Proof. Let an

be a sequence and T = {n 2 N : m > n =) am

� an

}. Thereare two cases to consider, depending on whether T is finite.

First, assume T is infinite. Define �(1) = minT and assuming �(n) isdefined, set �(n+ 1) = minT \ {�(1),�(2), . . . ,�(n)}. This inductively definesa strictly increasing function � : N ! N. The definition of T guarantees a

�(n)

is an increasing subsequence of an

.Now, assume T is finite. Let �(1) = maxT + 1. If �(n) has been chosen for

some n > maxT , then the definition of T implies there is an m > �(n) such thatam

a�(n). Set �(n + 1) = m. This inductively defines the strictly increasing

function � : N ! N such that a�(n) is a decreasing subsequence of a

n

.



If the sequence in Theorem 3.3.2 is bounded, then the corresponding mono-tone subsequence is also bounded. Recalling Theorem 3.2.1, we see the following.

Theorem 3.3.3 (Bolzano-Weierstrass). Every bounded sequence has a conver-gent subsequence.

3.4 Lower and Upper Limits of a Sequence

There are an uncountable number of strictly increasing functions � : N ! N, soevery sequence a

n

has an uncountable number of subsequences. If an

converges,then Theorem 3.3.1 shows all of these subsequences converge to the same limit.It’s also apparent that when a

n

! 1 or an

! �1, then all its subsequencesdiverge in the same way. When a

n

does not converge, the situation is a bit moredi�cult.

Example 3.4.1. Let Q = {qn

: n 2 N} and ↵ 2 R. Since every interval containsan infinite number of rational numbers, it is possible to choose �(1) = min{k :|q

k

� ↵| < 1}. In general, assuming �(n) has been chosen, choose �(n + 1) =min{k > �(n) : |q

k

� ↵| < 1/n}. Such a choice is always possible becauseQ \ (↵ � 1/n,↵ + 1/n) \ {q

k

: k �(n)} is infinite. This induction yields asubsequence q

�(n) of qn converging to ↵.

If an

is a sequence and bn

is a convergent subsequence of an

with bn

! L,then L is called a limit point of a

n

. A convergent sequence has only one limitpoint, but a divergent sequence may have many limit points, as shown above.

To make some sense out of this, suppose an

is a bounded sequence, andTn

= {ak

: k � n}. Define

`n

= glbTn

and µn

= lubTn

.

Because Tn

� Tn+1, it follows that for all n 2 N,

`1 `n

`n+1 µ

n+1 µn

µ1.

Theorem 3.2.1 implies both `n

and µn

converge. If `n

! ` and µn

! µ, clearly` µ. These numbers, ` and µ are lower and upper bounds for the limits of allconvergent subsequences of a

n

. These ideas are made precise by the followingdefinitions and theorems.

Definition 3.4.1. Let an

be a sequence. If an

is bounded below, then the lowerlimit of a

n

islim inf a

n

= limn!1

glb {ak

: k � n}.

If an

is bounded above, then the upper limit of an

is

lim sup an

= limn!1

lub {ak

: k � n}.

When an

is unbounded, the lower and upper limits are set to appropriate infinitevalues, while recalling the usual warnings about 1 not being a number.


3.5. THE NESTED INTERVAL THEOREM 3-9

The next theorem summarizes the observations presented above.


is a bounded sequence, then lim inf an

lim sup an

andlim inf a

n

= lim sup an

i↵ an

converges.

Suppose that an

is bounded above and both µn

and µ are as in the discussionpreceding the definition. Choose �(1) so a

�(1) > µ1�1. If �(n) has been chosenfor some n 2 N, then choose �(n+ 1) > �(n) to satisfy

µ�(n) � a

�(n+1) > lub {ak

: k � �(n)}� 1/n = u�(n) � 1/n.

This inductively defines a subsequence a�(n) ! µ, where the convergence is

guaranteed by the Sandwich Theorem. A similar argument can be made forlim inf a

n

. The following theorem has been proved.


is a sequence, then an

has subsequences bn

and cn

suchthat b

n

! lim sup an

and cn

! lim inf an

. Moreover, if dn

is any convergentsubsequence of a

n

, then

lim inf an

limn!1

dn

lim sup an

.

If L is the set of all limit points of a sequence an

, then the theorem showsL ⇢ [lim inf a

n

, lim sup an

].

3.5 The Nested Interval Theorem

Definition 3.5.1. A collection of sets {Sn

: n 2 N} is said to be nested, ifSn+1 ⇢ S

n

for all n 2 N.

Theorem 3.5.1 (Nested Interval Theorem). If In

= [an

, bn

] is a nested collec-tion of closed intervals such that lim

n!1(bn

� an

) = 0, then there is an x 2 Rsuch that

Tn2N I

n

= {x}.

Proof. Since the intervals are nested, it’s clear that an

is an increasing sequencebounded above by b1 and b

n

is a decreasing sequence bounded below by a1.Applying Theorem 3.2.1 twice, we find there are ↵,� 2 R such that a

n

! ↵and b

n

! �.We claim ↵ = �. To see this, let " > 0 and use the “shrinking” condition

on the intervals to pick N 2 N so that bN

� aN

< ". The nestedness of theintervals implies a

N

an

< bn

bN

for all n � N . Therefore

aN

lub {an

: n � N} = ↵ bN

and aN

glb {bn

: n � N} = � bN

.

This shows |↵ � �| |bN

� aN

| < ". Since " > 0 was chosen arbitrarily, weconclude ↵ = �.

Let x = ↵ = �. It remains to show thatT

n2N In

= {x}.First, we show that x 2

Tn2N I

n

. To do this, fix N 2 N. Since an

increasesto x, it’s clear that x � a

N

. Similarly, x bN

. Therefore x 2 [aN

, bN

]. BecauseN was chosen arbitrarily, it follows that x 2

Tn2N I

n

.



Next, suppose there are x, y 2T

n2N In

and let " > 0. Choose N 2 N suchthat b

N

� aN

< ". Then {x, y} ⇢T

n2N In

⇢ [aN

, bN

] implies |x� y| < ". Since" was chosen arbitrarily, we see x = y. Therefore

Tn2N I

n

= {x}.

Example 3.5.1. If In

= (0, 1/n] for all n 2 N, then the collection {In

: n 2 N}is nested, but

Tn2N I

n

= ;. This shows the assumption that the intervals beclosed in the Nested Interval Theorem is necessary.

Example 3.5.2. If In

= [n,1) then the collection {In

: n 2 N} is nested, butTn2N I

n

= ;. This shows that the assumption that the lengths of the intervalsbe bounded is necessary. (It will be shown in Corollary 5.1.6 that when theirlengths don’t go to 0, then the intersection is nonempty, but the uniqueness ofx is lost.)

3.6 Cauchy Sequences

Often the biggest problem with showing that a sequence converges using thetechniques we have seen so far is that we must know ahead of time to what itconverges. This is the “chicken and egg” problem mentioned above. An escapefrom this dilemma is provided by Cauchy sequences.


is a Cauchy sequence if for all " > 0 there isan N 2 N such that n,m � N implies |a

n

� am

| < ".

Theorem 3.6.1. A sequence converges i↵ it is a Cauchy sequence.

Proof. ()) Suppose an

! L and " > 0. There is an N 2 N such that g � Nimplies |a

n

� L| < "/2. If m,n � N , then

|am

� an

| = |am

� L+ L� an

| |am

� L|+ |L� am

| < "/e+ "/2 = ".

This shows an

is a Cauchy sequence.(() Let a

n

be a Cauchy sequence. First, we claim that an

is bounded. Tosee this, let " = 1 and choose N 2 N such that n,m � N implies |a

n

� am

| < 1.In this case, a

N

� 1 < an

< aN

+1 for all n � N , so {an

: n � N} is a boundedset. The set {a

n

: n < N}, being finite, is also bounded. Since {an

: n 2 N} isthe union of these two bounded sets, it too must be bounded.

Because an

is a bounded sequence, Theorem 3.3.3 implies it has a convergentsubsequence b

n

= a�(n) ! L. Let " > 0 and choose N 2 N so that n,m � N

implies |an

� am

| < "/2 and |bn

� L| < "/2. If n � N , then �(n) � n and

|an

� L| = |an

� bn

+ bn

� L| |an

� bn

|+ |bn

� L| < "/2 + "/2 = ".

Therefore, an

! L.

According to this theorem, we can prove that a sequence converges withoutever knowing precisely to what it converges. An example of the usefulness ofthis idea is contained in the following definition and theorem.

1July 24, 2012

c�Lee Larson ([email protected])


3: Cauchy Sequences 3-11

Definition 3.6.2. A sequence xn

is contractive if there is a c 2 (0, 1) such that|x

k+1 � xk

| c|xk

� xk�1| for all k > 1. c is called the contraction constant.

Theorem 3.6.2. If a sequence is contractive, then it converges.

Proof. Let xk

be a contractive sequence with contraction constant c 2 (0, 1).We first claim that if n 2 N, then

|xn

� xn+1| cn�1|x1 � x2|. (3.3)

This is proved by induction. When n = 1, the statement is |x1 � x2| c0|x1 �x2| = |x1�x2|, which is trivially true. Suppose that |x

n

�xn+1| cn�1|x1�x2|

for some n 2 N. Then, from the definition of a contractive sequence and theinductive hypothesis,

|xn+1 � x

n+2| c|xn

� xn+1| c(cn�1|x1 � x2|) = cn|x1 � x2|.

This shows the claim is true in the case n + 1. Therefore, by induction, theclaim is true for all n 2 N.

To show xn

is a Cauchy sequence, let " > 0. Since cn ! 0, we can chooseN 2 N so that

cN�1

(1� c)<

"

|x1 � x2|. (3.4)

Let n > m � N . Then

|xn

� xm

| = |xn

� xn�1 + x

n�1 � xn�2 + x

n�2 � · · ·� xm+1 + x

m+1 � xm

| |x

n

� xn�1|+ |x

n�1 � xn�2|+ · · ·+ |x

m+1 � xm

|

Now, use (3.3) on each of these terms.

cn�2|x1 � x2|+ cn�3|x1 � x2|+ · · ·+ cm�1|x1 � x2|= |x1 � x2|(cn�2 + cn�3 + · · ·+ cm�1)

Apply the formula for a geometric sum.

= |x1 � x2|cm�1 1� cn�m

1� c

< |x1 � x2|cm�1

1� c

Use (3.4) to estimate the following.

|x1 � x2|cN�1

1� c

< |x1 � x2|"

|x1 � x2|= "

This shows xn

is a Cauchy sequence.



Example 3.6.1. Let �1 < r < 1 and define the sequence sn

=P

n

k=0 rk. If r = 0,

the convergence of sn

is trivial. So, suppose r 6= 0. In this case,

|sn+1 � s

n

||s

n

� sn�1|

=

��rn+1

rn

�� = |r| < 1

and sn

is contractive. Theorem 3.6.2 implies it converges.

Example 3.6.2. Suppose f(x) = 2 + 1/x, a1 = 2 and an+1 = f(a

n

) for n 2 N.It is evident that a

n

� 2 for all n. Some algebra gives

��an+1 � a

n

an

� an�1

�� =��f(f(a

n�1))� f(an�1)

f(an�1)� a

n�1

�� =1

1 + 2an�1

1

5.

This shows an

is a contractive sequence and, according to Theorem 3.6.2, an

!L for some L � 2. Since, a

n+1 = 2 + 1/an

, taking the limit as n ! 1 ofboth sides gives L = 2 + 1/L. A bit more algebra shows L = 1 +

p2. In fact,

a11 � 1 ⇡ 1.414213562 isp2 accurate to 10 decimal places.

One of the mistakes made by students new to Cauchy sequences is to thinkit’s enough to show that |a

n

� an+1| becomes small. In the following example,

|an

� an+1| ! 0 even though a

n

! 1.

Example 3.6.3. Suppose an

=P

n

k=1 1/k for n 2 N. There’s a trick for showingthe sequence a

n

diverges. First, note that an

is strictly increasing. For anyn 2 N, consider

a2n�1 =2n�1X

k=1

1

k=

n�1X

j=0

2j�1X

k=0

1

2j + k

>

n�1X

j=0

2j�1X

k=0

1

2j+1=

n�1X

j=0

1

2=

n

2! 1

Hence, the subsequence a2n�1 is unbounded and the sequence an

diverges. (Tosee how this works, write out the first few sums of the form a2n�1.)

The convergence of Cauchy sequences is equivalent to the CompletenessAxiom. In fact, this is the definition of completeness used most often because itallows completeness to be defined on a space using only metric properties andnot requiring order.

3.7 Exercises

3.1. Show that the sequence an

=3n+ 1

2n+ 3converges.

3.2. Let an

be a sequence such that a2n ! A and a2n � a2n�1 ! 0. Thenan

! A.


3.7. EXERCISES 3-13

3.3. If an

! L, then what can you say about

�n

=a1 + a2 + · · ·+ a

n

n?

Is there a divergent sequence an

such that �n

converges?

3.4. Determine the limit of an

= n

pn!. (Hint: If n is even, then n! > (n/2)n/2.)

3.5. A sequence an

converges to 0 i↵ |an

| converges to 0.

3.6. Define the sequence an

=pn for n 2 N. Show that |a

n+1 � an

| ! 0, butan

is not a Cauchy sequence.

3.7. Show that n

pn! ! 1. (Hint: When n is even n! >

�n

2

�n/2

.)

3.8. Show that n1/n ! 1. (Hint: xn � 1 = (x� 1)(xn�1 + xn�2 + · · ·+ 1).)

3.9. If an

is a sequence such that a

n

�1a

n

+1 ! 0, then does limn!1 a

n

exist?

3.10. Suppose a sequence is defined by a1 = 0, a1 = 1 and an+1 = 1

2 (an+an�1)

for n � 2. Prove an

converges, and determine its limit.

3.11. The golden ratio is the number � = (1 +p5)/2. Prove that

fn

=�n�1 � (1� �)n�1

p5

is a closed-form expression for the Fibonacci sequence of Example 3.1.3.

3.12. Prove that the sequence an

= n3/n! converges.

3.13. Let an

and bn

be sequences. Prove that both sequences an

and bn

convergei↵ both a

n

+ bn

and an

� bn

converge.

3.14. A sequence an

converges in the mean if an

= 1n

Pn

k=1 ak converges. Provethat if a

n

! L, then an

! L, but the converse is not true.

3.15. Find a sequence xn

such that for all n 2 N there is a subsequence of xn

converging to n.

3.16. Let an

be a sequence with range {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Prove thatP1n=1 an10

�n converges.

3.17. If an

is a sequence such that every subsequence of an

has a further sub-sequence converging to 0, then a

n

! 0.

3.18. If an

� 0 for all n 2 N and an

! L, thenpan

!pL.

3.19. If an

is a Cauchy sequence and bn

is a subsequence of an

such that bn

! L,then a

n

! L.



3.20. If an

! L 2 R and � : N ! N is a bijection, then bn

= a�(n) converges to

L. (The new sequence bn

is sometimes called a rearrangement of an

. Note thatbn

might not be a subsequence of an

.)

3.21. If xn

and yn

are sequences such that limn!1 x

n

= L 6= 0 and limn!1 x

n

yn

exists, then limn!1 y

n

exists.

3.22. If an

is a sequence, then there is a subsequence bn

of an

such thatlim

n!1 bn

= lim sup an

.

3.23. an

= 1/n is not contractive.

3.24. Approximate a solution of x3�5x+1 = 0 to within 10�4 using a Cauchysequence.


IntroRealAnal-ch03

Documents

Transcript of IntroRealAnal-ch03