Notes on Introductory Real Analysis

MT294: Real Analysis - Notes 2014–15

Yiftach Barnea

December 12, 2014

2

Contents

1 Sequences 51.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Sequences of Functions 92.1 Pointwise Convergence and Uniform

Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Series 133.1 Convergence of Series . . . . . . . . . . . . . . . . . . . . . . . 133.2 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . 173.3 Power Series and Radius of Convergence . . . . . . . . . . . . 223.4 Rearrangement . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5 Conditional Convergence . . . . . . . . . . . . . . . . . . . . . 28

4 Limits of Functions 334.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Continuity 355.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2 Applications of Continuity . . . . . . . . . . . . . . . . . . . . 365.3 Monotone Functions and Continuity . . . . . . . . . . . . . . . 385.4 Limits of Sequences of Continuous

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

6 Differentiation 416.1 Basic Properties of Differentiation . . . . . . . . . . . . . . . . 416.2 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3

4 CONTENTS

6.3 Extreme Points and Mean ValueTheorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6.4 L’Hospital Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 506.5 Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 516.6 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 53

7 Riemann Integral 557.1 Riemann Integral and Its Basic

Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.2 The Fundamental Theorem of Calculus . . . . . . . . . . . . . 687.3 Riemann Integral and Continuous

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 697.4 Riemann Integral and Uniform

Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 737.5 Integration and Differentiation of Power Series . . . . . . . . . 747.6 The Exponential and Logarithmic

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Chapter 1

Sequences

1.1 Revision

We recall the following definitions and results from MT194:

Definition. Let (an) be a sequence in R we say the (an) converges to a ∈ Rif for all ϵ > 0 there exists N = N(ϵ) ∈ N such that for all n ≥ N we havethat |an − a| < ϵ. In this case a is called the limit of the sequence andwe write lim

n→∞an = a. If the sequence does not have a limit, we say that it

divergent.

Comment: Sometime limn→∞

an is used even when the limit is not known to

exist. In such case, we mean it conditionally, that is, a = limn→∞

an if the limit

exists.

Theorem 1.1 (Basic Properties of Convergent Sequences). Let (an) be asequence in R, then the following hold:

1. If (an) is convergent, then the limit of (an) is unique;

2. If (an) is convergent, then (an) is bounded, i.e., there exists M ∈ Rsuch that for all n ∈ N we have that |an| ≤M ;

3. Given a tail of (an), i.e., a sequence (am+n) for some constant m ∈N, then (an) convergent if and only if its tail is convergent. If bothsequences are convergent, then they converge to the same limit;

5

6 CHAPTER 1. SEQUENCES

4. If limn→∞

an = a, then every subsequence of (an) also converges to a;

5. If limn→∞

an = a and b and c are in R such that b ≤ an ≤ c for all n, then

b ≤ a ≤ c;

6. If (an) and (bn) are convergent and an ≤ bn for all n, then we have thatlimn→∞

an ≤ limn→∞

bn.

Theorem 1.2 (Arithmetic of Limits). Let (an) and (bn) be convergent se-quences in R with lim

n→∞an = a and lim

n→∞bn = b, then the following hold:

1. limn→∞

(an + bn) = a+ b;

2. limn→∞

anbn = ab;

3. If bn = 0 for all n and if b = 0, then

limn→∞

anbn

=a

b.

Theorem 1.3 (The Sandwich Theorem). Let (an), (bn) and (cn) be sequencesin R. Suppose that x = lim

n→∞an = lim

n→∞cn and that an ≤ bn ≤ cn for all n.

Then x = limn→∞

bn.

Definition. Let (an) be a sequence in R.

• We say that (an) is increasing if an ≤ an+1 for all n. Furthermore,if an < an+1 for all n, we say that (an) is strictly increasing.

• We say that (an) is decreasing if an ≥ an+1 for all n. Furthermore,if an > an+1 for all n, we say that (an) is strictly decreasing.

• We say that (an) is monotone if it is either increasing or decreasing.

Theorem 1.4 (Monotone Convergence Theorem). A monotone sequence isconvergent if and only if it is bounded.

Corollary 1.5. Let (an) be a monotone sequence in R and let (ani) be a

subsequence of it. Then (an) is convergent if and only if (ani) is convergent.

Furthermore, if they converge, then they converge to the same limit.

1.2. CAUCHY SEQUENCES 7

Proof. Homework.

Euler’s Number: Let an = (1+ 1n)n. Recall that (an) is a bounded increas-

ing sequence and therefore has a limit. We call that limit Euler’s Numberand we denote it by e = lim

n→∞(1 + 1

n)n.

Theorem 1.6. Every sequence has a monotone subsequence.

Theorem 1.7 (Bolzano-Weierstrass). A bounded sequence has a convergentsubsequence.

1.2 Cauchy Sequences

Definition. Let (an) be a sequence in R. We say that (an) is a Cauchysequence or that it satisfies the Cauchy criterion if for every ϵ > 0 thereexists N = N(ϵ) ∈ N such that for all m,n ≥ N we have that |an − am| < ϵ.

Example: The sequence(1n

)is a Cauchy sequence. Given ϵ > 0 take N

such that ϵ2> 1

N. If m,n ≥ N , then∣∣∣∣ 1n − 1

m

∣∣∣∣ ≤ ∣∣∣∣ 1n∣∣∣∣+ ∣∣∣∣ 1m

∣∣∣∣ ≤ 2

N< ϵ.

Lemma 1.8. Let (an) be a convergent sequence in R. Then (an) is a Cauchysequence.

Proof. Let a = limn→∞

an. Then for all ϵ > 0 there exists N = N(ϵ) such that

for all n ≥ N we have that |an − a| < ϵ. Given ϵ > 0, we take N = N(ϵ2

),

then if n,m ≥ N , we have that

|an − am| = |an − a+ a− am| ≤ |an − a|+ |a− am| <ϵ

2+ϵ

2= ϵ.

Lemma 1.9. Let (an) be a Cauchy sequence in R. Then (an) is bounded.

Proof. Take N such that if n,m ≥ N , then |an − am| < 1. In particular, forn ≥ N we have that |an−aN | < 1, i.e., aN − 1 < an < aN +1. So for n ≥ N ,−|aN | − 1 ≤ aN − 1 < an < aN + 1 ≤ |aN | + 1. Let K = max

k<N{|ak|} and

let M = max(K, |aN | + 1), then |an| ≤ M because if n < N we have that|an| ≤ K and if n ≥ N we have that |an| < |aN |+ 1.

8 CHAPTER 1. SEQUENCES

Theorem 1.10 (Cauchy Convergence Criterion). Let (an) be a sequence inR. Then (an) is convergent if and only if it is a Cauchy sequence.

Proof. If (an) is convergent, then by Lemma 1.8 it is a Cauchy sequence.Suppose that (an) is a Cauchy sequence. From the previous lemma (an) isbounded and therefore from Bolzano-Weierstrass Theorem it has a conver-gent subsequence (ani

), say that a = limi→∞

ani.

Given ϵ > 0 we take I = I(ϵ2

)∈ N such that for all i ≥ I we have that

|ani− a| < ϵ

2. We also take K = K

(ϵ2

)∈ N such that for all n,m ≥ K we

have that |an − am| < ϵ2. Set N = max(nI , K) and fix J such that nJ ≥ N .

Since nJ ≥ N ≥ nI and as (ni) is an increasing sequence J ≥ I. We deducethat |anJ

− a| < ϵ2.

Suppose that n > N . Both n, nJ ≥ N ≥ K, thus, |an − anJ| < ϵ

2. Hence,

|an − a| = |an − anJ+ anJ

− a| ≤ |an − anJ|+ |anJ

− a| < ϵ

2+ϵ

2= ϵ.

We conclude that a = limn→∞

an.

Example: Let an =n∑

k=1

1k. We claim that (an) is not a Cauchy sequence and

therefore does not converges. Indeed, take ϵ = 12. Suppose it is a Cauchy

sequence then there exists N such that for all m,n ≥ N , |an−am| < 12. Take

m = N and n = 2N , then

|an − am| =

∣∣∣∣∣2N∑k=1

1

k−

N∑k=1

1

k

∣∣∣∣∣ =2N∑

k=N+1

1

k≥

2N∑k=N+1

1

2N= N

1

2N=

1

2.

This is a contradiction, so (an) is not a Cauchy sequence and therefore itdoes not converge.

Chapter 2

Sequences of Functions

2.1 Pointwise Convergence and Uniform

Convergence

Definition. Let A ⊆ R and fn : A → R for each n ∈ N. We call (fn) asequence of functions on A to R.

We note that if a ∈ A, then (fn(a)) forms a sequence (of numbers).

Definition. Let A ⊆ R and let (fn) be a sequence of functions on A. We saythat (fn) converges pointwise or just converges to f on A, if for eachx ∈ A we have that (fn(x)) converges to f(x). In other words, for all x ∈ Aand for all ϵ > 0, there exists N = N(x, ϵ) ∈ N such that for all n ≥ N wehave that |fn(x)− f(x)| < ϵ.

Definition. Let A ⊆ R and let (fn) be a sequence of functions on A. Wesay that (fn) converges uniformly to f on A, if for all ϵ > 0, there existsN = N(ϵ) ∈ N such that for all x ∈ A and for all n ≥ N we have that|fn(x)− f(x)| < ϵ.

Notice that the difference between these two definitions is that in the firstone N depends on x while in the second one the same N works for all x.Clearly if (fn) converges uniformly to f on A, then it converges pointwise tof on A.

9

10 CHAPTER 2. SEQUENCES OF FUNCTIONS

Examples:

1. fn(x) =1nsin(x) converge uniformly to f(x) = 0 on R. Indeed, given

ϵ > 0 we take N > 1ϵ. Then for all n ≥ N and for all x ∈ R we have

that

|fn(x)− f(x)| =∣∣∣∣ 1n sin(x)

∣∣∣∣ ≤ 1

n<

1

N< ϵ.

2. Define

fn(x) =

{1− nx, 0 ≤ x < 1

n;

0, 1n≤ x ≤ 1.

Then we claim that (fn) converges pointwise to

f(x) =

{1, x = 00, 0 < x ≤ 1,

on [0, 1] but not uniformly.

As fn(0) = 1 for all n we have that (fn(0)) converges to 1 = f(1). For0 < x ≤ 1 we have that fn(x) = 0 for all n ≥ 1

x. So (fn(x)) converges

to 0 = f(x). Thus, fn converges pointwise to f on [0, 1].

On the other hand, suppose (fn) converges uniformly to f on [0, 1].Then there is N such that for all n ≥ N and for all x ∈ [0, 1] we havethat |fn(x)− f(x)| < 1

2. Take 0 < x < 1

2N. Then

|fN(x)− f(x)| = 1−Nx > 1−N1

2N=

1

2.

This is a contradiction so (fn) does not converge uniformly to f

Theorem 2.1 (Cauchy Criterion for Uniform Convergence). Let A ⊆ R andlet (fn) be a sequence of functions on A. Then (fn) converges uniformly onA if and only if for all ϵ > 0 there exists N = N(ϵ) ∈ N such that for allx ∈ A and for all m,n ≥ N we have that |fn(x)− fm(x)| < ϵ.

Proof. Suppose that (fn) converges uniformly on A to f . Given ϵ > 0 takeN = N( ϵ

2) ∈ N such that for all x ∈ A and for all n ≥ N we have that

|fn(x)− f(x)| < ϵ2. Then for all x ∈ A and for all m,n ≥ N we have that

|fn(x)− fm(x)| = |fn(x)− f(x) + f(x)− fm(x)| ≤

2.1. POINTWISE CONVERGENCE AND UNIFORMCONVERGENCE11

|fn(x)− f(x)|+ |f(x)− fm(x)| <ϵ

2+ϵ

2= ϵ

as required.On the other hand, suppose that for all ϵ > 0 there exists N = N(ϵ) ∈ N

such that for all x ∈ A and for all m,n ≥ N we have that |fn(x)−fm(x)| < ϵ.Clearly, fixing x we have that (fn(x)) is a Cauchy sequence and therefore fromCauchy Convergence Criterion for sequences (fn(x)) is convergent. We definef : A → R by f(x) = lim

n→∞fn(x). Given ϵ > 0 we take N = N( ϵ

2) ∈ N such

that for all x ∈ A and for all m,n ≥ N we have that |fn(x) − fm(x)| < ϵ2.

Since f(x) = limn→∞

fn(x) we can find K = K(x, ϵ2) ∈ N such that for all

m ≥ K we have |fm(x)− f(x)| < ϵ2. Take n ≥ N and m ≥ max(N,K), then

|fn(x)− f(x)| = |fn(x)− fm(x) + fm(x)− f(x)| ≤

|fn(x)− fm(x)|+ |fm(x)− f(x)| < ϵ

2+ϵ

2= ϵ

as required.

Theorem 2.2 (Arithmetic of Uniform Convergence). Let A ⊆ R and let (fn)and (gn) be sequences of functions on A converging uniformly on A to f andg respectively. Then

1. (fn + gn) converges uniformly on A to f + g and

2. if f and g are bounded on A, then (fngn) converges uniformly on A tofg.

Proof. 1. Given ϵ > 0. Let N ∈ N be such that for all n ≥ N and for allx ∈ A we have that |fn(x)− f(x)| < ϵ

2and |gn(x)− g(x)| < ϵ

2. Then

|(fn + gn)(x)− (f + g)(x)| = |fn(x)− f(x) + gn(x)− g(x)|

≤ |fn(x)− f(x)|+ |gn(x)− g(x)| < ϵ.

Thus, (fn + gn) converges uniformly on A to f + g.

2. As f and g are bounded on A we can find M such that for all x ∈ A|f(x)| < M and |g(x)| < M . Since (fn) and (gn) converge uniformly tof and g on A respectively we can find N such that for all n ≥ N and forall x ∈ A we have that |fn(x)− f(x)| < M and |gn(x)− g(x)| < M . In

12 CHAPTER 2. SEQUENCES OF FUNCTIONS

particular, for all n ≥ N and for all x ∈ A we have that |fn(x)| < 2Mand |gn(x)| < 2M .

Given ϵ > 0 we find K such that for all n ≥ K and for all x ∈ A wehave that |fn(x) − f(x)| < ϵ

4Mand |gn(x) − g(x)| < ϵ

4M. Then for all

n ≥ max(N,K) and for all x ∈ A we have that

|(fngn)(x)− (fg)(x)| = |(fngn)(x)− (fgn)(x) + (fgn)(x)− (fg)(x)|

≤ |(fn(x)− f(x))gn(x)|+ |f(x)(gn(x)− g(x))| ≤

|fn(x)− f(x)||gn(x)|+ |f(x)||gn(x)− g(x)| < ϵ

4M2M +M

ϵ

4M< ϵ.

Chapter 3

Series

3.1 Convergence of Series

Definition. Given a sequence (ai) in R we define the series generated by

(ai) to be the sequence sn =n∑

i=1

ai and we write∑ai = (sn). We call sn the

partial sums of the series∑ai.

If (sn) is convergent, we say that the series is convergent and we write∞∑i=1

ai = limn→∞

sn. Otherwise, we say that the series is divergent.

Comments:

• Sometime the index is omitted and we just write∑i

ai or even∑ai for

∞∑i=1

ai. I will try to avoid this as then∑ai has a double meaning one

as (sn) and the other one as limn→∞

sn. (However, it is usually very easy

from the context to see which one it is meant to be.)

• Sometime∞∑i=1

ai is used even when the limit is not known to exist. In

such case, we mean it conditionally, that is,∞∑i=1

ai = limn→∞

sn if the limit

exists.

13

14 CHAPTER 3. SERIES

Examples:

1. Take x ∈ R. Then for x = 1k∑

n=0

xn = 1−xk+1

1−x. Thus, for |x| < 1 the

series is convergent and∞∑n=0

xn = 11−x

. Otherwise, it is divergent.

2.m∑

n=1

1n(n+1)

=m∑

n=1

(1n− 1

n+1

)= 1− 1

m+1−→m→∞

1, thus,∞∑n=1

1n(n+1)

= 1.

3. We have seen that the harmonic series∑

1nis divergent.

Theorem 3.1. Let (an) be a sequence in R. If the series∑an is convergent,

then limn→∞

an = 0.

Proof. If∑an is convergent, then sm =

m∑n=1

an is a Cauchy sequence. There-

fore, given ϵ > 0, there exists N such that for i, j ≥ N we have that |si−sj| <ϵ. For all n ≥ N + 1 we have that n, n− 1 ≥ N , so |an| = |sn − sn−1| < ϵ asrequired.

Example: Notice that the Harmonic series is an example that shows thatyou cannot reverse the statement of the theorem, so lim

n→∞1n= 0, but

∑1nis

divergent.

Notation: Sometime the sequence (ai) in R starts with i = 1 but i = k,where k is some integer (like in our firat example above), not necessarilypositive or we might just like to consider the tail of the sequence starting

with k. We then could look at the sumsn∑

i=k

ai and if these sums converge we

write∞∑i=k

ai = limn→∞

n∑i=k

ai. In such case, of course, we need to be even more

careful about omitting the index. However, things are not too bad as wehave the following easy observation:

Lemma 3.2. Let (ai) be a sequence in R and k an integer. Then∞∑i=1

ai exists

if and only if∞∑i=k

ai exists.

3.1. CONVERGENCE OF SERIES 15

Proof. Let us assume k > 1, the case k < 1 is similar. We write sn =n∑

i=1

ai

for n ≥ 1 and tn =n∑

i=k

ai for n ≥ k. Set A =k−1∑i=1

ai. Then for n ≥ k we have

that sn = A+ tn. If∞∑i=k

ai = limn→∞

tn exists, then

∞∑i=1

ai = limn→∞

sn = limn→∞

(A+ tn) = limn→∞

A+ limn→∞

tn = A+∞∑i=k

ai

exists. On the other hand, if∞∑i=1

ai = limn→∞

sn exists, then

∞∑i=k

ai = limn→∞

tn = limn→∞

(sn − A) = limn→∞

sn − limn→∞

A =∞∑i=1

ai − A

exists.(Notice that we consider (sn) sometimes with n ≥ k and sometimes with

n ≥ 1, but this is fine as the limit of a tail of a sequence exists if and only ifthe limit of the sequence exists and in such case they are equal.)

Theorem 3.3. Let (an) be a sequence in R. Suppose an ≥ 0 for all n. Thenthe series

∑an is convergent if and only if the sequence of partial sums is

bounded.

Proof. If an ≥ 0 for all n, write the partial sums sm =m∑

n=1

an, then (sm)

is clearly an increasing sequence. The result follows immediately from theMonotone Convergence Theorem (Theorem 1.4).

Examples:

1. Here is another proof that the harmonic series∑

1nis divergent. Indeed,

s2n = 1+1

2+

(1

3+

1

4

)+

(1

5+ · · ·+ 1

8

)+· · ·+

(1

2n−1 + 1+ · · ·+ 1

2n

)≥

1 +1

2+

2

4+ · · ·+ 2n−1

2n= 1 +

n

2,

is unbounded subsequence and therefore does not converge. It followsfrom Theorem 1.1 part 4 that the Harmonic sequence does not converge.


2.∑

1n2 is convergent. Clearly, (sn) is an increasing sequence. Now,

s2m−1 =

1+

(1

22+

1

32

)+

(1

42+ · · ·+ 1

72

)+· · ·+

(1

(2m−1)2+ · · ·+ 1

(2m − 1)2

)≤

1 +2

22+ · · ·+ 2m−1

(2m−1)2= 1 +

1

2+ · · ·+ 1

2m−1≤ 2.

So (s2m−1) is bounded by 2 and therefore convergent by the Mono-tone Convergence Theorem. It follows from Corollary 1.5 that (sn) isconvergent.

Theorem 3.4 (Cauchy Criterion for Series). Let (ai) be a sequence in R.Then the series

∑ai is convergent if and only if for every ϵ > 0 there exists

N = N(ϵ) ∈ N such that for all n ≥ m ≥ N we have that

∣∣∣∣ n∑i=m+1

ai

∣∣∣∣ < ϵ.

Proof. This is just a reformulation of Cauchy Convergence Criterion that asequence is convergent if and only if it is a Cauchy sequence for (sk), where

sk =k∑

i=1

ai are the partial sums because sn − sm =n∑

i=m+1

ai.

Example: We used Cauchy criterion to show that the harmonic series is notconvergent.

Theorem 3.5 (Comparison Test). Let (ai) and (bi) be sequences in R. Sup-pose that there exists K ∈ N such that for all i ≥ K we have that 0 ≤ ai ≤ bi.Then:

1. If∑bi is convergent, then

∑ai is convergent;

2. If∑ai is divergent, then

∑bi is divergent.

Proof. 1. Suppose that∑bi is convergent, then it satisfies Cauchy Cri-

terion for Series. Thus, given ϵ > 0, there exists N = N(ϵ) ∈ N such

that for all n ≥ m ≥ N we have that

∣∣∣∣ n∑i=m+1

bi

∣∣∣∣ < ϵ. Then for all

n ≥ m ≥ max(K,N) we have that that∣∣∣∣∣n∑

i=m+1

ai

∣∣∣∣∣ =n∑

i=m+1

ai ≤n∑

i=m+1

bi =

∣∣∣∣∣n∑

i=m+1

bi

∣∣∣∣∣ < ϵ.

So∑ai satisfies Cauchy Criterion for Series and therefore converges.

3.2. ABSOLUTE CONVERGENCE 17

2. Suppose that∑ai is divergent. If

∑bi is convergent, then from the

previous part∑ai is convergent. This is a contradiction, so

∑bi is

divergent

Examples:

1. For 0 < p < 1 the series∑

1np is divergent by comparing it to the

Harmonic series as 1np ≥ 1

n.

2. The series∑

1n2 is convergent since 1

n2 ≤ 2n(n+1)

and because we have

seen that∞∑n=1

2n(n+1)

= 2∞∑n=1

1n(n+1)

= 2 so∑

2n(n+1)

is convergent.

3.2 Absolute Convergence

Definition. Given a sequence (an) in R we say that the series∑an is ab-

solutely convergent if∑

|an| converges. We say that it is conditionallyconvergent if it is convergent but not absolutely convergent.

Examples:

1. The series∞∑n=1

(−1)n+1

n2 is absolutely convergent as∞∑n=1

1n2 is convergent.

2.∞∑n=1

(−1)n+1

nis not absolutely convergent as

∞∑n=1

1nis not convergent. Now,

s2k =2k∑n=1

(−1)n+1

n=

(1− 1

2

)+

(1

3− 1

4

)+ · · ·+

(1

2k − 1− 1

2k

)=

1

2+

1

12+ · · ·+ 1

(2k − 1)(2k)=

k∑n=1

1

(2n− 1)(2n)≥ 1

2.

Using the Comparison Test with 1(2n−1)(2n)

≤ 1n2 we see that (s2k) is

convergent. Given ϵ > 0 we take N such that 1N< ϵ

4and using Cauchy

criterion for all 2i ≥ 2j ≥ N we have that |s2i − s2j| < ϵ2. Take n ≥

m ≥ N . Then if n = 2i and m = 2j are both even, then |sn − sm| <ϵ2. If n = 2i + 1 and m = 2j + 1 are both odd, then |sn − sm| =


|s2i+1 − s2j+1| =∣∣∣ 12i+1

+ s2i − s2j+2 +1

2j+2

∣∣∣ ≤ ∣∣ 12i+1

∣∣ + |s2i − s2j+2| +∣∣∣ 12j+2

∣∣∣ < ϵ4+ ϵ

2+ ϵ

4= ϵ. A similar argument works when n is even

and m is odd and vice versa, thus, it follows from Cauchy Criterion for

Series that∑ (−1)n+1

nis conditionally convergent.

Notice that the limit must be at least 12and in particular non-zero.

Theorem 3.6. Let (ai) be a sequence in R. If the series∑ai is absolutely

convergent, then it is convergent.

Proof. Suppose∑ai is absolutely convergent. Then

∑|ai| is convergent.

From Cauchy Criterion for Series we have that for all ϵ > 0 there exists N =

N(ϵ) ∈ N such that for n ≥ m ≥ N we have thatn∑

i=m+1

|ai| =∣∣∣∣ n∑i=m+1

|ai|∣∣∣∣ < ϵ.

Then

∣∣∣∣ n∑i=m+1

ai

∣∣∣∣ ≤ n∑i=m+1

|ai| < ϵ. Thus,∑ai satisfies Cauchy Criterion for

Series and therefore is convergent.

Theorem 3.7 (The Ratio Test). Let (an) be a sequence of nonzero elementsin R.

1. If there is K ∈ N and r < 1 such that for all n ≥ K we have that∣∣∣an+1

an

∣∣∣ ≤ r, then∑an is absolutely convergent;

2. If there is K ∈ N such that for all n ≥ K we have that∣∣∣an+1

an

∣∣∣ ≥ 1, then∑an is divergent.

In particular, if limn→∞

∣∣∣an+1

an

∣∣∣ = x, then∑an is absolutely convergent if x < 1

and is divergent if x > 1.

Proof. 1. We would like to use the Comparison Test. Clearly the series∑ aKrKrn is convergent as it is essentially a geometric series. Now, for all

n ≥ K we have that∣∣∣an+1

an

∣∣∣ ≤ r, then by induction on m for all m ≥ 1

we have that∣∣∣∣aK+m

aK

∣∣∣∣ = ∣∣∣∣aK+1

aK

∣∣∣∣ ∣∣∣∣aK+2

aK+1

∣∣∣∣ · · · ∣∣∣∣ aK+m

aK+m−1

∣∣∣∣ ≤ rm.


So for all n ≥ K we have that

|an| ≤∣∣aKrn−K

∣∣ = ∣∣∣aKrK

rn∣∣∣ .

From the Comparison Test the series is absolutely convergent.

If limn→∞

∣∣∣an+1

an

∣∣∣ = x < 1, then pick x < r < 1. From the definition of a

limit there is K such that for all n ≥ K we have that∣∣∣an+1

an

∣∣∣ ≤ r, so

the result follows from what we just proved.

2. If there is K ∈ N such that for all n ≥ K we have that∣∣∣an+1

an

∣∣∣ ≥ 1, then

for all n ≥ K we have that |an| ≥ |aK | so the sequence (an) does notconverge to 0 and therefore from Theorem 3.1

∑an is divergent.

If limn→∞

∣∣∣an+1

an

∣∣∣ = x > 1, then pick 1 < r < x. From the definition of a

limit there is K such that for all n ≥ K we have that∣∣∣an+1

an

∣∣∣ ≥ r, so

the result follows from what we just proved.

Examples:

1.∑

n5

3nconverges as

(n+1)5

3n+1

n5

3n

=

(1 + 1

n

)53

−→n→∞

1

3< 1.

2. Fix 0 = x ∈ R. Then∑

xn

n!is convergent. Indeed,

limn→∞

∣∣∣∣∣xn+1

(n+1)!

xn

n!

∣∣∣∣∣ = limn→∞

∣∣∣∣ x

n+ 1

∣∣∣∣ = 0 < 1.

We write E(x) =∞∑n=0

xn

n!.

Theorem 3.8 (Arithmetic of Series). Let (ai) and (bi) be sequences in R.Suppose

∑ai and

∑bi are convergent series. Then

1.∑

(ai + bi) is a convergent series and∞∑i=0

(ai + bi) =∞∑i=0

ai +∞∑i=0

bi.


2. If c ∈ R, then∑cai is convergent and

∞∑i=0

cai = c∞∑i=0

ai.

3. Write cn =n∑

i=0

aibn−i. If∑ai is absolutely convergent, then

∑cn is

convergent and∞∑n=0

cn =

(∞∑i=0

ai

)(∞∑i=0

bi

).

Proof. 1.

∞∑i=0

ai+∞∑i=0

bi = limn→∞

n∑i=0

ai+ limn→∞

n∑i=0

bi = limn→∞

n∑i=0

(ai+bi) =∞∑i=0

(ai+bi).

2.

c

∞∑i=0

ai = c limn→∞

n∑i=0

ai = limn→∞

c

n∑i=0

ai = limn→∞

n∑i=0

cai =∞∑i=0

cai.

3. Write A =∞∑i=0

ai, B =∞∑i=0

bi, An =n∑

i=0

ai, Bn =n∑

i=0

bi, Cn =n∑

i=0

ci and

βn = Bn −B. Then

Cn = a0b0 + (a0b1 + a1b0) + · · ·+ (a0bn + a1bn−1 + · · ·+ anb0) =

a0Bn+a1Bn−1+· · ·+anB0 = a0(B+βn)+a1(B+βn−1)+· · ·+an(B+β0) =

AnB + (a0βn + a1βn−1 + · · ·+ anβ0).

Put γn = a0βn + a1βn−1 + · · ·+ anβ0. Since limn→∞

AnB = AB it suffices

to show that limn→∞

γn = 0.

Let α =∞∑i=0

|ai|. If α = 0, then ai = 0 for all i, so γn = 0 for all

n and there is nothing much to prove. Otherwise, given ϵ > 0 sincelimn→∞

Bn = B we can choose N such that for all n ≥ N we have that

|βn| < ϵ2α. Then

|γn| = |a0βn + a1βn−1 + · · ·+ anβ0| =

|β0an + β1an−1 + · · ·+ βn−1a1 + βna0| ≤


|β0an + · · ·+ βN−1an−N+1|+ |βNan−N + · · ·+ βna0| ≤

|β0an + · · ·+ βN−1an−N+1|+ |βN ||an−N |+ · · ·+ |βn||a0| ≤

|β0an + · · ·+ βN−1an−N+1|+ϵ

2α(|an−N |+ · · ·+ |a0|) ≤

|β0an|+ · · ·+ |βN−1an−N+1|+αϵ

2α= |β0||an|+ · · ·+ |βN−1||an−N+1|+

ϵ

2.

Take M > |β0|, |β1|, . . . , |βN−1|. Since∑ai is convergent from Theo-

rem 3.1 we have that limn→∞

an = 0. Thus, we can find K such that for

n ≥ K we have that |an| < ϵ2NM

. Then for all n ≥ N +K and for all0 ≤ i ≤ N − 1 we have that n− i ≥ K. Hence, for all n ≥ N +K wehave that

|γn| < |β0|ϵ

2NM+ · · ·+ |βN−1|

ϵ

2NM+ϵ

2<

ϵ

2N+ · · ·+ ϵ

2N+ϵ

2=ϵ

2+ϵ

2= ϵ.

Thus, limn→∞

γn = 0 and therefore,∞∑n=0

cn = limn→∞

Cn = limn→∞

AnB +

limn→∞

γn = AB + 0 = AB as required.

Example: Let us see that if we do not require∑an to be absolutely con-

vergent we can get into troubles. Take a0 = b0 = 0 and an = bn = (−1)n+1√n

for

n ≥ 1. We have seen that∑ (−1)n+1

√n

is convergent. Then

cn =n∑

i=0

aibn−i =n∑

i=1

(−1)n+2√i(n− i)

= (−1)n+2

n∑i=1

1√i(n− i)

.

Since(n2

)2 ≥ i(n− i) we have that

|cn| =n∑

i=1

1√i(n− i)

≥n∑

i=1

1√(n2

)2 =n∑

i=1

1(n2

) =n(n2

) = 2.

In particular, (cn) does not converge to 0 and hence∑cn does not converge.


3.3 Power Series and Radius of Convergence

Definition. Let (an)∞n=0 be a sequence in R. Let c ∈ R, We call

∞∑n=0

an(x−c)n

a power series around c. If c = 0, we just call∞∑n=0

anxn a power series.

From now on for simplicity we will mainly deal with power series ratherthan power series around a point. Trivially by changing the variable by aconstant every power series around a point can be shifted into a power series.

Theorem 3.9 (Radius of Convergence). Let (an)∞n=0 be a sequence in R.

Then there exists R ≥ 0 (possibly R = 0 or R = ∞) such that for all x

with |x| < R we have that the power series∞∑n=0

anxn absolutely converges and

for all x with |x| > R the power series∞∑n=0

anxn diverges. Furthermore, if

limn→∞

∣∣∣ anan+1

∣∣∣ exists, then R = limn→∞

∣∣∣ anan+1

∣∣∣.Proof. Suppose

∞∑n=0

anyn converges and |x| < |y|. We claim that

∞∑n=0

anxn is

absolutely convergent. Since∞∑n=0

anyn converges we have that lim

n→∞|anyn| = 0.

In particular, there exists K such that |anyn| < 1 for all n ≥ K. Now,

|anxn| = |anyn|∣∣∣xy ∣∣∣n so for all n ≥ K we have that |anxn| <

∣∣∣xy ∣∣∣n. But∣∣∣xy ∣∣∣ < 1, hence,∞∑n=0

∣∣∣xy ∣∣∣n is convergent. Thus, from the Comparison Test

∞∑n=0

anxn is absolutely convergent.

Let

R = sup

{|y|

∣∣∣∣∣∞∑n=0

anyn converges

}.

Clearly if |x| > R, then∞∑n=0

anxn diverges. On the other hand, if |x| < R,

there exists y with |x| ≤ |y| < R such that∞∑n=0

anyn converges. From the

previous paragraph∞∑n=0

anxn absolutely converges.

3.3. POWER SERIES AND RADIUS OF CONVERGENCE 23

Suppose S = limn→∞

∣∣∣ anan+1

∣∣∣ exists. Thenlimn→∞

∣∣∣∣an+1xn+1

anxn

∣∣∣∣ = limn→∞

∣∣∣∣an+1x

an

∣∣∣∣ = |x|S.

From the Ratio Test we have that∞∑n=0

anxn converges if |x| < S and diverges

if |x| > S, thus, S = R.

We comment that if x = ±R, then we cannot say in general whether∞∑n=0

anxn is convergent or divergent (homework: find examples).

Definition. Let (an)∞n=0 be a sequence in R. Then R as in the theorem above

is called the radius of convergence of the power series∞∑n=0

anxn.

Theorem 3.10 (Arithmetic of Power Series). Let (an)∞n=0 and (bn)

∞n=0 be

sequences in R. Let R1 be the radius of convergence of∞∑n=0

anxn and R2 be

the radius of convergence of∞∑n=0

bnxn.

1. If |x| < R1 and |x| < R2, then

∞∑n=0

anxn +

∞∑n=0

bnxn =

∞∑n=0

(an + bn)xn.

2. If |x| < R1 and |y| < R2, then(∞∑n=0

anxn

)(∞∑n=0

bnyn

)=

∞∑n=0

(n∑

i=0

aixibn−iy

n−i

).

Proof. This follows immediately from Theorem 3.8 and the fact that powerseries are absolutely convergent for x with |x| less than the radius of conver-gent.


Example: For x ∈ R let E(x) =∞∑n=0

xn

n!. We have see that E(x) is absolutely

convergent for any x. Then

E(x)E(y) =

(∞∑n=0

xn

n!

)(∞∑n=0

yn

n!

)=

∞∑n=0

(n∑

i=0

xi

i!

yn−i

(n− i)!

)=

∞∑n=0

(n∑

i=0

n!

n!

xi

i!

yn−i

(n− i)!

)=

∞∑n=0

n∑i=0

1

n!

(n!

i!(n− i)!xiyn−i

)=

∞∑n=0

1

n!

(n∑

i=0

(n

i

)xiyn−i

)=

∞∑n=0

(x+ y)n

n!= E(x+ y).

Theorem 3.11 (Uniform Convergence of Power Series). Let (ai)∞i=0 be a

sequence in R and let R be the radius of convergence of∞∑i=0

aixi. Set the

functions fn(x) =n∑

i=0

aixi. Let J = [a, b] ⊆ (−R,R) be a closed and bounded

interval in R. Then the sequence (fn) converges uniformly to f(x) =∞∑i=0

aixi

on J .

Proof. Take |a|, |b| < r < R. By Theorem 3.9∞∑i=0

|airi| is convergent. Thus,

given ϵ > 0 there exists N such for all n ≥ m ≥ N we have thatn∑

i=m+1

|airi| <

ϵ. Hence, for all x ∈ [a, b] we have that for all n ≥ m ≥ N

|fn(x)− fm(x)| =

∣∣∣∣∣n∑

i=m+1

aixi

∣∣∣∣∣ ≤n∑

i=m+1

|aixi| ≤n∑

i=m+1

|airi| < ϵ.

Theorem 2.1 implies that (fn) converges uniformly on J to a function g(x).

It is clear that (fn) converges pointwise to f(x) =∞∑i=0

aixi on J , but if it

converges uniformly to g(x) it also converges pointwise to it. Since the limit

is unique g(x) = f(x) =∞∑i=0

aixi.

3.4. REARRANGEMENT 25

3.4 Rearrangement

When we add a finite number of real numbers changing the order of thesummation does not change the sum. What happens with an infinite sum?

Example: Let A =∞∑n=1

(−1)n+1

n. As mention above A = 0. Now, let us

change the order of our summation:

∞∑k=1

(1

2k − 1− 1

2(2k − 1)− 1

4k

)=

1− 1

2− 1

4+

1

3− 1

6− 1

8+

1

5− 1

10− 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

2A = A.

Definition. Let (an) and (bn) be a couple of sequences in R. We say that(bn) is a rearrangement of (an) if there exists a bijection f : N → N suchthat bn = af(n) for all n ∈ N.

Example: Let an = n2 and f : N → N defined by f(2k − 1) = 2k andf(2k) = 2k − 1. Then b2k−1 = af(2k−1) = a2k = 4k2 and b2k = af(2k) =a2k−1 = (2k − 1)2, i.e., (an) is 1, 4, 9, 16, . . . and (bn) is 4, 1, 16, 9, . . ..

Theorem 3.12 (Rearrangement Theorem for ACS). Let (ai) be a sequencein R and let (bi) be a rearrangement of it. If the series

∑ai is absolutely

convergent, then∑bi is convergent and

∞∑i=1

ai =∞∑i=1

bi.

Proof. Since∑ai is absolutely convergent it is convergent, write A =

∞∑i=1

ai.

As∑ai is absolutely convergent

∑|ai| is convergent and therefore satisfies

Cauchy’s Criterion. We deduce that given ϵ > 0 we can find N = N( ϵ2) ∈ N

such that for n > m ≥ N we have both

∣∣∣∣ m∑i=1

ai − A

∣∣∣∣ < ϵ2and

n∑i=m+1

|ai| < ϵ2.

Suppose that f : N → N is a bijection such that bn = af(n) for all n.As f is a bijection we can define kn to be the integer such that n = f(kn)


for all n, i.e., kn = f−1(n). So bkn = af(kn) = an for all n. Let K =max {k1, k2, . . . , kN}. Then for all n ≥ K we have that each bk1 = a1, bk2 =

a2, . . . , bkN = aN appears inn∑

i=1

bi as a summand. Take some M ≥ f(i) for

all 1 ≤ i ≤ n, then∣∣∣∣∣n∑

i=1

bi − A

∣∣∣∣∣ =∣∣∣∣∣

n∑i=1

af(i) − A

∣∣∣∣∣ ≤∣∣∣∣∣

N∑i=1

ai − A

∣∣∣∣∣+M∑

i=N+1

|ai| <ϵ

2+ϵ

2= ϵ.

It is natural to ask what happens if∑ai is conditionally convergent series.

Theorem 3.13 (Rearrangement Theorem for CCS). Let (ai) be a sequencein R. If the series

∑ai is conditionally convergent, then for any B ∈ R,

there exists a rearrangement (bi) of (ai) such that∞∑i=1

bi = B.

Proof. Let i1 < i2 < i3 < . . . be the indices such that aik ≥ 0 and j1 < j2 <

j3 < . . . be the indices such that ajk < 0. We first notice that∞∑k=1

|ajk | =

−∞∑k=1

ajk . Now, if∑aik and

∑ajk both converge, then both are bounded,

say∞∑k=1

aik =

∣∣∣∣ ∞∑k=1

aik

∣∣∣∣ ≤ M and −∞∑k=1

ajk =

∣∣∣∣ ∞∑k=1

ajk

∣∣∣∣ ≤ M . We conclude

that the partial sums of∑

|ai| are bounded by 2M . From Theorem 3.3 wededuce that

∑ai is absolutely convergent. This is a contradiction, thus, at

least one of them does not converge and as its elements have the same signits partial sums are not bounded. It is now easy to verify that if only oneof them is bounded, then

∑ai is also unbounded (homework). But that is

a contradiction to the fact that it is convergent. We deduce that both areunbounded.

We will now rearrange our sequence. Let r1 ≥ 1 to be the smallest integersuch that

r1∑k=1

aik > B,

such r1 exists since∑aik is unbounded (from above). Set b1 = ai1 , b2 =

ai2 , . . . , br1 = air1 . Let s1 ≥ 1 to be the smallest integer such that

r1∑k=1

aik +

s1∑k=1

ajk < B,

3.4. REARRANGEMENT 27

such s1 exists since∑ajk is unbounded (from below). Set br1+1 = aj1 , br1+2 =

aj2 , . . . , br1+s1 = ajs1 . We continue by induction. Supposed we have pickedr1, s1, . . . , rm, sm such that for all 1 ≤ p ≤ m we have that rp is the smallestinteger such that

r1+r2+···+rp∑k=1

aik +

s1+s2+···+sp−1∑k=1

ajk > B

and sp is the smallest integer such that

r1+r2+···+rp∑k=1

aik +

s1+s2+···+sp∑k=1

ajk < B.

We also assume that we have arranged bi for 1 ≤ i ≤ r1 + r2 + · · · + rm +s1 + s2 + · · · + sm in a similar way too the case m = 1. Then we pick rm+1

to be the smallest integer such that

r1+r2+···+rm+1∑k=1

aik +

s1+s2+···+sm∑k=1

ajk > B.

Set br1+s1+r2+s2···+rm+sm+1 = air1+r2+···+rm+1 , br1+s1+r2+s2···+rm+sm+2 = air1+r2+···+rm+2 ,. . ., br1+s1+r2+s2···+rm+sm+rm+1 = air1+r2+···+rm+rm+1

. We also pick sm+1 to bethe smallest integer such that

r1+r2+···+rm+1∑k=1

aik +

s1+s2+···+sm+1∑k=1

ajk < B.

Set br1+s1+r2+s2···+rm+sm+rm+1+1 = ajs1+s2+···+sm+1 , br1+s1+r2+s2···+rm+sm+rm+1+2 =ajs1+s2+···+sm+2 , . . ., br1+s1+r2+s2···+rm+sm+rm+1+sm+1 = ajs1+s2+···+sm+sm+1

.

We now need to show that∞∑i=1

bi = B. Given n, then either r1 + s1 + r2 +

s2 + · · ·+ rm−1 + sm−1 < n ≤ r1 + s1 + r2 + s2 + · · ·+ rm−1 + sm−1 + rm forsome m or r1 + s1 + r2 + s2 + · · ·+ rm < n ≤ r1 + s1 + r2 + s2 · · ·+ rm + smfor some m. In the first case we have that

r1+s1+r2+s2+···+rm−1+sm−1∑i=1

bi <

n∑i=1

bi ≤r1+s1+r2+s2···+rm−1+sm−1+rm∑

i=1

bi.


So

ajsm−1<

r1+s1+r2+s2+···+rm−1+sm−1∑i=1

bi −B <n∑

i=1

bi −B

≤r1+s1+r2+s2···+rm−1+sm−1+rm∑

i=1

bi −B ≤ airm .

In the second case we have that

r1+s1+r2+s2+···+rm−1+sm−1+rm∑i=1

bi >

n∑i=1

bi ≥r1+s1+r2+s2···++rm+sm∑

i=1

bi.

So

airm >

r1+s1+r2+s2+···+rm−1+sm−1+rm∑i=1

bi −B >n∑

i=1

bi −B

≥r1+s1+r2+s2···++rm+sm∑

i=1

bi −B ≥ ajsm .

As∑ai is convergent, we have from Theorem 3.1 that lim

i→∞ai = 0. It is not

difficult to see that when i tends to infinity so are jsm−1 ,irm , and jsm . Thus,

from the Sandwich Theorem∞∑i=1

bi = B.

Example: Try to rearrange∞∑n=1

(−1)n+1

nso it will convert to 0. (Homework:

find the first 20 elements in the rearrangement.)

3.5 Conditional Convergence

Theorem 3.14 (Alternating Series Test). Let (an) be a decreasing sequencein R of strictly positive numbers. Suppose lim

n→∞an = 0. Then the series∑

(−1)n+1an is convergent.

Proof. Let sk =k∑

n=1

(−1)n+1an. Then

s2k =2k∑n=1

(−1)n+1an = (a1 − a2) + (a3 − a4) + · · ·+ (a2k−1 − a2k).

3.5. CONDITIONAL CONVERGENCE 29

As (an) is decreasing sequence in R we have that a2n−1 − a2n ≥ 0 for all n,therefore, (s2k) is an increasing sequence. Also,

s2k =2k∑n=1

(−1)n+1an = a1 − (a2 − a3)− · · · − (a2k−2 − a2k−1)− a2k.

Again, a2n − a2n+1 ≥ 0 and so is a2k, thus, s2k ≤ a1. From the MonotoneConvergence Theorem (s2k) converges, say to A. Since s2k − s2k−1 = a2ktends to zero it follows that (sk) converges to A (why?).

Example:∑ (−1)n+1

√n

is convergent. Notice that it is not absolutely conver-gent by comparing it to the harmonic series.

Lemma 3.15 (Abel’s Lemma). Let (ai) and (bi) be sequences in R. Write

sn =n∑

i=1

bi for the partial sums of∑bi with s0 = 0. If m > n, then

m∑i=n+1

aibi = (amsm − an+1sn) +m−1∑i=n+1

(ai − ai+1)si.

Proof.

(amsm − an+1sn) +m−1∑i=n+1

(ai − ai+1)si =

am(sm−sm−1)+am−1(sm−1−sm−2)+· · ·+an+2(sn+2−sn+1)+an+1(sn+1−sn) =

ambm + · · ·+ an+1bn+1 =m∑

i=n+1

aibi.

Theorem 3.16 (Dirichlet’s Test). Let (ai) and (bi) be sequences in R. If(ai) is decreasing and lim

i→∞ai = 0 and if (sn) the partial sums of

∑bi are

bounded, then∑aibi is convergent.

Proof. We would like to use Cauchy’s Criterion. Suppose |sn| ≤ S for all n.As (ai) is decreasing and converges to 0 we have that ai ≥ 0 for all i and


since (ai) is decreasing ai − ai+1 ≥ 0 also for all i. Using Abel’s Lemma wehave that ∣∣∣∣∣

m∑i=n+1

aibi

∣∣∣∣∣ =∣∣∣∣∣(amsm − an+1sn) +

m−1∑i=n+1

(ai − ai+1)si

∣∣∣∣∣ ≤|amsm|+ |an+1sn|+

m−1∑i=n+1

|(ai − ai+1)si| ≤

(am + an+1)S +m−1∑i=n+1

(ai − ai+1)S =

(am + an+1)S + (an+1 − am)S = 2an+1S.

As limi→∞

ai = 0 we deduce that∑aibi satisfies Cauchy’s Criterion and thus

converges.

Example: We recall that 2 sin(α) cos(β) = sin(α + β)− sin(β − α). Hence,

2 sin(x2

)(cos(x) + cos(2x) + · · ·+ cos(nx)) =(

sin

(3x

2

)− sin

(x2

))+ · · ·+

(sin

((2n+ 1)x

2

)− sin

((2n− 1)x

2

))=

sin

((2n+ 1)x

2

)− sin

(x2

).

Then for x = 2πk we have that

| cos(x) + cos(2x) + · · ·+ cos(nx)| ≤

∣∣∣∣∣∣sin(

(2n+1)x2

)− sin

(x2

)2 sin

(x2

)∣∣∣∣∣∣ ≤

∣∣∣∣∣ 1

sin(x2

)∣∣∣∣∣ .We deduce that the partial sums of

∑cos(nx) are bounded. Take (an) to

be any decreasing sequence with limn→∞

an = 0. Then∑an cos(nx) converges,

e.g.,∑ cos(nx)

nconverges.

Theorem 3.17 (Abel’s Test). Let (ai) and (bi) be sequences in R. If (ai)is a convergent monotone sequence and the series

∑bi is convergent, then∑

aibi is also convergent.

3.5. CONDITIONAL CONVERGENCE 31

Proof. Without loss of generalization we assume that (ai) is a decreasingsequence. Suppose lim

i→∞ai = a. Then (ai−a) is decreasing and lim

i→∞ai−a = 0.

Suppose∞∑i=1

bi = b and let (sn) be the partial sums of∑bi. Then obviously

(sn) is bounded. Hence, from Dirichlet’s Test we have that∑

(ai − a)bi is

convergent. Say∞∑i=1

(ai − a)bi = c. We deduce that

∞∑i=1

aibi =∞∑i=1

((ai − a)bi + abi) =∞∑i=1

(ai − a)bi +∞∑i=1

abi =

c+ a∞∑i=1

bi = c+ ab.

Example: Fix x = 2πk. Take an =(1 + 1

n

)nand bn = cos(nx)

n. We know that

(an) is an increasing sequence that converges to e and that∑bi is convergent.

Then from Abel’s Test∑ (1+ 1

n)n cos(nx)

n=∑ (n+1)n cos(nx)

nn+1 is convergent.


Chapter 4

Limits of Functions

4.1 Revision

We recall the following:

Definition. Let A ⊆ R we call c ∈ R a limit point or a cluster point ofA if there exists a sequence (xi) in A such that xi = c for all i and lim

i→∞xi = c.

Note: c may or may not be in A.

Definition. Let A ⊆ R, f : A → R and let c ∈ R be a limit point of A.Then we say that f(x) converges to ξ as x tends to c or tends to ξ asx tends to c and write lim

x→cf(x) = ξ if for every sequence (xn) in A such

that xn = c for all n and limn→∞

xn = c we have that limn→∞

f(xn) = ξ.

Comment: As with the previous definitions, sometime limx→c

f(x) is used even

when the limit is not known to exist. In such case, we mean it conditionally,that is, ξ = lim

x→cf(x) if the limit exists.

Theorem 4.1 (Uniqueness of Limit). Let A ⊆ R, f : A → R and let c ∈ Rbe a limit point of A. Then if lim

x→cf(x) exists, then it is unique.

Theorem 4.2 (Arithmetic Properties of Limits). Let A ⊆ R, f, g : A → Rand let c be a limit point of A. Suppose that lim

x→cf(x) = F and lim

x→cg(x) = G

Then

33

34 CHAPTER 4. LIMITS OF FUNCTIONS

1. limx→c

(f + g)(x) = F +G;

2. limx→c

(fg)(x) = FG;

3. if g(x) = 0 for all x ∈ A and G = 0, then limx→c

(fg

)(x) = F

G.

Theorem 4.3 (Inequality of Limits). Let A ⊆ R, f, g : A → R and let c bea limit point of A. Suppose that f(x) ≤ g(x) for all x ∈ A and that lim

x→cf(x)

and limx→c

g(x) exist. Then limx→c

f(x) ≤ limx→c

g(x).

Theorem 4.4 (The Sandwich Theorem for Functions). Let A ⊆ R,f, g, h : A → R and let c be a limit point of A. Suppose that lim

x→cg(x) =

ξ = limx→c

h(x) and that g(x) ≤ f(x) ≤ h(x) for all x ∈ A. Then limx→c

f(x) = ξ.

Theorem 4.5 (ϵ-δ Theorem of Limits). Let A ⊆ R, f : A→ R and let c be alimit point of A. Then lim

x→cf(x) = ξ if and only if for every ϵ > 0 there exists

δ = δ(c, ϵ) > 0 such that if x ∈ A, x = c and |x− c| < δ, then |f(x)− ξ| < ϵ.

Chapter 5

Continuity

5.1 Revision

We recall the following:

Definition. Let A ⊆ R, f : A → R and a ∈ A. We say that f is con-tinuous at a if lim

x→af(x) exists and lim

x→af(x) = f(a). We say that f is

continuous on A if f is continuous at c for all c ∈ A.

Theorem 5.1 (Arithmetic of Continuous Functions). Let A ⊆ R and leta ∈ A. Suppose that f, g : A→ R are continuous at a ∈ A. Then

1. f + g is continuous at a;

2. fg is continuous at a;

3. if g(x) = 0 for all x ∈ A, then fgis continuous at a.

Theorem 5.2 (Composition of Continuous Functions). Let A,B ⊆ R, letf : A→ R such that f(x) ∈ B for all x ∈ A, let g : B → R and let a ∈ A. Iff is continuous at a and g is continuous at b = f(a), then g ◦f is continuousat a.

Theorem 5.3 (Extreme Points of Continuous Functions). Let I = [a, b] ⊂R be a closed and bounded interval. Suppose f : I → R is a continuousfunction on I. Then f is bounded on I and there exist c, d ∈ I such that

f(c) = sup{f(x)

∣∣∣x ∈ I}

and f(d) = inf{f(x)

∣∣∣x ∈ I}.

35

36 CHAPTER 5. CONTINUITY

Theorem 5.4 (Intermediate Value Theorem). Let I = [a, b] ⊂ R be a closedand bounded interval. Suppose f : I → R is a continuous function on I. Letξ ∈ R be between f(a) and f(b). Then there exists c ∈ I such that f(c) = ξ.

Theorem 5.5 (ϵ-δ Theorem of Continuity). Let A ⊆ R and f : A → R.Then f is continuous at a ∈ A if and only if for every ϵ > 0 there existsδ = δ(a, ϵ) > 0 such that if x ∈ A and |x− a| < δ, then |f(x)− f(a)| < ϵ.

5.2 Applications of Continuity

Definition. Let A ⊆ R and f : A → R. We say that f is uniformlycontinuous on A if for every ϵ > 0 there exists δ = δ(ϵ) > 0 such that ifx, y ∈ A and |y − x| < δ, then |f(y)− f(x)| < ϵ.

Examples:

1. f(x) = x is uniformly continuous on R: we can take δ = ϵ;

2. f(x) = x2 is not uniformly continuous on R: Take ϵ = 1. Suppose f isuniformly continuous on R. Then there is δ > 0 such that if x, y ∈ Rand |y − x| < δ, then |f(y) − f(x)| < 1. Take x > 1

δand y = x + δ

2.

Then |y − x| = δ2< δ. Nevertheless,

|f(y)− f(x)| =

∣∣∣∣∣(x+

δ

2

)2

− x2

∣∣∣∣∣ = |xδ + δ2/4| > |xδ| > 1

δδ = 1 = ϵ.

This is a contradiction, thus, f is not uniformly continuous on R.

Theorem 5.6. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R be a continuous function on I. Then f is uniformly continuouson I.

Proof. Suppose f is not uniformly continuous. Then there exists ϵ > 0 suchthat for all δ > 0 there exist x, y ∈ I (which depend on δ) with |y − x| < δbut |f(y)− f(x)| ≥ ϵ.

For each n ∈ N we choose xn, yn ∈ I such that |yn − xn| < 1n, but

|f(yn) − f(xn)| ≥ ϵ. As I is bounded we have that (xn) is bounded. FromBolzano-Weierstrass Theorem (xn) has a convergent subsequence, say (xni

),

5.2. APPLICATIONS OF CONTINUITY 37

denote x = limi→∞

xni. Since a ≤ xn ≤ b for all n we have that a ≤ x ≤ b, so

x ∈ I.As f is a continuous function by Theorem 5.5 there exists δ > 0 such

that if |t− x| < δ and t ∈ I, then |f(t)− f(x)| < ϵ2. Take i big enough such

that both 1ni< δ

2and |xni

− x| < δ2hold. Since |xni

− x| < δ2we have that

|f(xni)− f(x)| < ϵ

2. Because 1

ni< δ

2we have that

|yni− x| ≤ |yni

− xni|+ |xni

− x| < 1

ni

+δ

2<δ

2+δ

2= δ

and hence, |f(yni)− f(x)| < ϵ

2. We deduce that

ϵ ≤ |f(yni)− f(xni

)| ≤ |f(yni)− f(x)|+ |f(x)− f(xni

)| < ϵ

2+ϵ

2= ϵ.

This is a contradiction and therefore there is no such ϵ. In other words,for every ϵ > 0 there exists δ = δ(ϵ) > 0 such that if x, y ∈ I and |y−x| < δ,then |f(y)− f(x)| < ϵ.

Example: Take I = [0, 1] and f = x2. Given ϵ > 0, take δ = ϵ2. If |y−x| < δ,

then

|f(y)− f(x)| = |y2 − x2| = |(y + x)(y − x)| ≤ |y + x||y − x| < 2δ = ϵ.

Theorem 5.7. Let I ⊆ R be an interval and f : I → R be a continuous

function on I. Then J = f(I) ={f(i)

∣∣∣i ∈ I}

is an interval. Furthermore,

if I = [a, b] is closed and bounded, then J is closed and bounded.

Proof. Let c = inf J and d = sup J , where we write c = −∞ and d = ∞ ifJ is unbounded from below or from above respectively. Clearly J ⊆ [c, d] wewill show that (c, d) ⊆ J . Since there are at most four subsets between (c, d)and [c, d], namely (c, d), [c, d), (c, d] and [c, d], and because all of these foursubsets are intervals we will obtain that J must be an interval.

Suppose that s ∈ (c, d). Then by definitions of infimum and supremum we

have r, t ∈ J such that c ≤ r ≤ s and s ≤ t ≤ d. As r, t ∈ J ={f(i)

∣∣∣i ∈ I}

we have x, z ∈ I such that f(x) = r and f(z) = t. It follows from theIntermediate Value Theorem that s = f(y) for some y ∈ I, i.e., s ∈ J . Weconclude that (c, d) ⊆ J .

Now, if I = [a, b] is closed and bounded, then by the Extreme Points ofContinuous Functions Theorem we have that c = −∞, d = ∞ and c, d ∈ J ,thus, J = [c, d].


Example: Take I = [−2, 3] and f(x) = x2. Then f(I) = [0, 9].

5.3 Monotone Functions and Continuity

Theorem 5.8. Let I = [a, b] ⊂ R be a closed and bounded interval. Supposef : I → R is a strictly monotone function and write J = f(I). Then thereexists g : J → I such that g is the inverse of f and g is a strictly monotonecontinuous function.

Proof. By definition f : I → J is surjective. Without loss of generality f isstrictly increasing. Suppose x1, x2 ∈ I and say x1 < x2, then f(x1) < f(x2)and thus, f is injective. Hence, f is bijective and it has an inverse g : J → I.

Take y1, y2 ∈ J with y1 < y2. As f is surjective there are x1, x2 ∈ Isuch that f(xi) = yi. If x1 > x2, then y1 = f(x1) > f(x2) = y2, this is acontradiction to our assumption, so x1 < x2. We deduce that

g(y1) = g(f(x1)) = x1 < x2 = g(f(x2)) = y2

and we conclude that g is strictly increasing.Let y ∈ J . We would like to show that g is continuous at y. Suppose it

is not, then by Theorem 5.5 there exists ϵ > 0 such that for all δ > 0 thereexists yδ ∈ J with |yδ − y| < δ, but |g(yδ) − g(y)| ≥ ϵ. Let x = g(y) andxδ = g(yδ). So either xδ − x ≥ ϵ or x− xδ ≥ ϵ. In the first case we have thatx < x+ ϵ

2< xδ and as x, xδ ∈ I and I is an interval we have that x+ ϵ

2∈ I.

Because f is strictly increasing we have that

f(x) < f(x+

ϵ

2

)< f(xδ),

so

yδ − y = f(xδ)− f(x) > f(x+

ϵ

2

)− f(x) > 0.

In the second case we have that x > x− ϵ2> xδ and as x, xδ ∈ I and I is

an interval we have that x− ϵ2∈ I. Because f is strictly increasing we have

that

f(x) > f(x− ϵ

2

)> f(xδ),

so

y − yδ = f(x)− f(xδ) > f(x)− f(x− ϵ

2

)> 0.

5.4. LIMITS OF SEQUENCES OF CONTINUOUS FUNCTIONS 39

Let µ = min(f(x+ ϵ

2

)− f(x), f(x)− f

(x− ϵ

2

)). Notice that µ is inde-

pendent of our choice of δ. As f(x+ ϵ

2

)−f(x) > 0 and f(x)−f

(x− ϵ

2

)> 0

we have that µ > 0. Since |yδ − y| ≥ f(x+ ϵ

2

)− f(x) or |yδ − y| ≥

f(x)− f(x− ϵ

2

)we have that

δ > |yδ − y| ≥ µ.

As µ is independent of our choice of δ we derive a contradiction by takingδ < µ. We conclude that g is continuous.

Note: We did not require that f is continuous. However, if f is not con-tinuous, J may not be an interval. For the application we have in mind,computing the derivative of the inverse, we will need J to be an interval, butin that case we do not need to worry as indeed f will be continuous.

Example: Take I = [2, 3] and f(x) = x2. Then J = [4, 9] and g(y) =√y.

5.4 Limits of Sequences of Continuous

Functions

Theorem 5.9 (Uniform Convergence of Continuous Functions). Let A ⊆ Rand let (fn) be a sequence of functions on A converging uniformly to f onA. If for all n ∈ N we have that fn is continuous at c ∈ A, then f is alsocontinuous at c. Moreover, if for all n ∈ N we have that fn is uniformlycontinuous on A, then so is f .

Proof. Suppose that fn is continuous at c ∈ A for all n. Given ϵ > 0we pick N ∈ N such that for all n ≥ N and for all x ∈ A we have that|fn(x)− f(x)| < ϵ

3. As fN is continuous at c we can find δ > 0 such that for

all x ∈ A with |x − c| < δ we have that |fN(x) − fN(c)| < ϵ3. Thus, for all

x ∈ A with |x− c| < δ we have that

|f(x)− f(c)| = |f(x)− fN(x) + fN(x)− fN(c) + fN(c)− f(c)| ≤

|f(x)− fN(x)|+ |fN(x)− fN(c)|+ |fN(c)− f(c)| < ϵ

3+ϵ

3+ϵ

3= ϵ.

We conclude that f is continuous at c.Suppose that fn is uniformly continuous on A for all n. Given ϵ > 0

we pick N ∈ N such that for all n ≥ N and for all x ∈ A we have that


|fn(x) − f(x)| < ϵ3. As fN is uniformly continuous on A we can find δ > 0

such that for all x, y ∈ A with |x− y| < δ we have that |fN(x)− fN(y)| < ϵ3.

Thus, for all x, y ∈ A with |x− y| < δ we have that

|f(x)− f(y)| = |f(x)− fN(x) + fN(x)− fN(y) + fN(y)− f(y)| ≤

|f(x)− fN(x)|+ |fN(x)− fN(y)|+ |fN(y)− f(y)| < ϵ

3+ϵ

3+ϵ

3= ϵ.

We conclude that f is uniformly continuous on A.

Examples:

1. fn(x) =1nsin(x) which are continuous and converge uniformly to f(x) =

0 which is continuous.

2.

fn(x) =

{1− nx, 0 ≤ x < 1

n;

0, 1n≤ x ≤ 1

which are continuous and converge pointwise to

f(x) =

{1, x = 00, 0 < x ≤ 1

which is not continuous, thus, (fn) does not converge uniformly to f .

Corollary 5.10 (Continuity of Power Series). Let (ai)∞i=0 be a sequence in R

and let R be the radius of convergence of∞∑i=0

aixi. Then

∞∑i=0

aixi is continuous

function on the interval (−R,R).

Proof. Let c ∈ (−R,R). Take |c| < r < R. Let fn(x) =n∑

i=0

aixi. Now, (fn) is

a sequence of continuous functions uniformly converging to∞∑i=0

aixi on [−r, r].

From Theorem 5.9∞∑i=0

aixi is continuous on [−r, r], in particular at c.

Chapter 6

Differentiation

You have seen most of the results in this sections in MT171, so we will notgive many examples. We will mainly focus on the proofs.

6.1 Basic Properties of Differentiation

Definition. Let I ⊆ R be an interval containing a point c and let f : I → R.We say that f is differentiable at c if

limx→c

f(x)− f(c)

x− c

exists. In such case, we write

f ′(c) =df

dx

∣∣∣∣c

= limx→c

f(x)− f(c)

x− c

and we call it the derivative of f at c. (Recall that in the definition of thelimit above we required x to be in I.) If f is differentiable at c for all c ∈ I,we say that f is differentiable on I and we write f ′(x) for the functionx 7→ f ′(x) defined on I.

Example: Let f(x) = a. Then

f ′(c) = limx→c

f(x)− f(c)

x− c= lim

x→c

a− a

x− c= lim

x→c

0

x− c= 0.

41

42 CHAPTER 6. DIFFERENTIATION

Example: We will show that (sin)′(x) = cos(x). We first recall that sin(x)−sin(c) = 2 sin

(x−c2

)cos(x+c2

). So

sin(x)− sin(c)

x− c=

2 sin(x−c2

)cos(x+c2

)x− c

=2 sin

(x−c2

)2x−c

2

cos

(x+ c

2

)=

sin(x−c2

)x−c2

cos

(x+ c

2

).

Then since cos(x+c2

)is a continuous function we have that

(sin)′(c) = limx→c

sin(x)− sin(c)

x− c= lim

x→c

sin(x−c2

)x−c2

cos

(x+ c

2

)=

limx→c

sin(x−c2

)x−c2

limx→c

cos

(x+ c

2

)=

limh→0

sin(h)

hcos(c).

So we are reduced to showing that limh→0

sin(h)h

= 1. We look at a picture

of a circle with radius 1 and we draw an angel of h. Then the area of thewhole circle is π. So the area of the section with angle h is h

2ππ = h

2. That

section contains a right angle triangle with sides of lengths cos(h) and sin(h)

and therefore an area sin(h) cos(h)2

. Also it is contained in a right angle trianglewith sides of lengths 1 and tan(h) and therefore an area tan(h)/2. We deducethat

sin(h) cos(h)

2≤ h

2≤ tan(h)

2=

sin(h)

2 cos(h).

Hence,

cos(h) ≤ sin(h)

h≤ 1

cos(h).

As limh→0

cos(h) = cos(0) = 1 the Sandwich Theorem implies that limh→0

sin(h)h

= 1

as required.

Theorem 6.1. Let I ⊆ R be an interval, let f : I → R and let c ∈ I. If fis differentiable at c, then it is continuous at c.

6.1. BASIC PROPERTIES OF DIFFERENTIATION 43

Proof. Let k = f ′(c). Given ϵ > 0 it follows from Theorem 4.5 that there

exists δ0 > 0 such that∣∣∣f(x)−f(c)

x−c− k∣∣∣ < ϵ for all x ∈ I with |x− c| < δ0. In

particular, for all x ∈ I with |x−c| < δ0 we have that∣∣∣f(x)−f(c)

x−c

∣∣∣ < |k|+ϵ, i.e.,|f(x)−f(c)| < (|k|+ ϵ)|x− c|. Pick δ = min(δ0,

ϵ|k|+ϵ

), then for all x ∈ I with

|x− c| < δ ≤ δ0 we have that |f(x)− f(c)| ≤ (|k|+ ϵ)|x− c| < (|k|+ ϵ)δ ≤ ϵ.From Theorem 5.5 we conclude that f is continuous at c.

The other direction does not work, there are functions which are contin-uous at c but not differentiable at c.

Examples:

1. f(x) = |x| is continuous at 0, but not differentiable at 0 because if

x > 0, f(x)−f(0)x−0

= 1 and if x < 0, f(x)−f(0)x−0

= −1.

2. For I = [0,∞) we have that f(x) = x23 is continuous at 0 but f(x)−f(0)

x−0=

x−13 which does not converge when x tends to 0 so f is not differentiable

at 0.

Theorem 6.2 (Arithmetic of Differentiable Functions). Let I ⊆ R be aninterval containing a point c and let f, g : I → R. Suppose that f and g aredifferentiable at c. Then

1. f + g is differentiable at c and (f + g)′(c) = f ′(c) + g′(c);

2. fg is differentiable at c and (fg)′(c) = f ′(c)g(c) + f(c)g′(c);

3. if g(x) = 0 for all x ∈ I, then fgis differentiable at c and(

f

g

)′

(c) =f ′(c)g(c)− f(c)g′(c)

g(c)2.

Proof. 1.

(f+g)′(c) = limx→c

(f + g)(x)− (f + g)(c)

x− c= lim

x→c

(f(x)− f(c)

x− c+g(x)− g(c)

x− c

)=

limx→c

f(x)− f(c)

x− c+ lim

x→c

g(x)− g(c)

x− c= f ′(c) + g′(c);


2. Using the continuity of g at c we obtain

(fg)′(c) = limx→c

(fg)(x)− (fg)(c)

x− c= lim

x→c

f(x)g(x)− f(c)g(c)

x− c=

limx→c

f(x)g(x)− f(c)g(x) + f(c)g(x)− f(c)g(c)

x− c=

limx→c

(f(x)− f(c))g(x)

x− c+ lim

x→c

f(c)(g(x)− g(c))

x− c=

limx→c

f(x)− f(c)

x− climx→c

g(x) + f(c) limx→c

g(x)− g(c)

x− c= f ′(c)g(c) + f(c)g′(c);

3. Using the continuity of g at c we obtain(f

g

)′

(c) = limx→c

fg(x)− f

g(c)

x− c= lim

x→c

f(x)g(x)

− f(c)g(c)

x− c=

limx→c

f(x)g(c)− f(c)g(x)

g(x)g(c)(x− c)= lim

x→c

f(x)g(c)− f(c)g(c) + f(c)g(c)− f(c)g(x)

g(x)g(c)(x− c)=

limx→c

f(x)− f(c)

x− climx→c

g(c)

g(x)g(c)+ lim

x→c

−f(c)g(x)g(c)

limx→c

g(x)− g(c)

x− c=

f ′(c)g(c)− f(c)g′(c)

g2(c).

6.2 The Chain Rule

Theorem 6.3 (Caratheodory’s Theorem). Let I ⊆ R be an interval contain-ing a point c and let f : I → R. Then f is differentiable at c if and only ifthere exists a function φ(x) : I → R satisfying f(x)−f(c) = φ(x)(x− c) andφ is continuous at c. In such case, φ(c) = f ′(c).

Proof. Suppose there exists a function φ(x) that staisfies f(x) − f(c) =φ(x)(x− c) and φ is continuous at c. Then

f ′(c) = limx→c

f(x)− f(c)

x− c= lim

x→cφ(x) = φ(c).

6.2. THE CHAIN RULE 45

On the other hand, suppose that f is differentiable at c. Then define

φ(x) =

{f(x)−f(c)

x−cx = c

f ′(c) x = c

Obviously, f(x)− f(c) = φ(x)(x− c). Now,

limx→c

φ(x) = limx→c

f(x)− f(c)

x− c= f ′(c) = φ(c).

Thus, φ is continuous at c.

Theorem 6.4 (The Derivative of the Inverse of a Function). Let I ⊆ R bean interval containing a point a and let f : I → R be a strictly monotoneand continuous function on I. Let J = f(I) and let g : J → I be the strictlymonotone and continuous function inverse to f . If f is differentiable at aand f ′(a) = 0, then g is differentiable at b = f(a) and

g′(b) =1

f ′(a)=

1

f ′(g(b)).

Proof. Firstly, by Theorem 5.8 f indeed has an inverse g that is strictlymonotone and continuous. We are given that f ′(g(b)) = f ′(a) = 0. Wewould like to apply Caratheodory’s Theorem and show that

ψ(y) =

{g(y)−g(b)

y−by = b

1f ′(g(b))

y = b

is continuous at b.Let φ : I → R defined by

φ(x) =

{f(x)−f(a)

x−ax = a

f ′(a) x = a.

Since f is strictly monotone we have that if x = a, then f(x) = f(a), soφ(x) = 0 and φ(a) = f ′(a) = 0. By Caratheodory’s Theorem φ is continuousat a. For y ∈ J write x = g(y), then y − b = f(x)− f(a) = φ(x)(x− a). Asφ(x) = 0 for all x ∈ I we can divide by φ(x) and obtain

g(y)− g(b) = x− a =1

φ(x)(y − b) =

1

φ(g(y))(y − b).


We notice that since f is continuous at a and g is continuous at b we havethat x tends to a if and only if y tends to b. Since φ(a) = f ′(a) = 0 andcontinuous at a we get that

limy→b

ψ(y) = limy→b

g(y)− g(b)

y − b= lim

y→b

1

φ(g(y))=

limx→a

1

φ(x)=

1

f ′(a)=

1

f ′(g(b))= ψ(b).

Thus, ψ : J → R is continuous at b so by Caratheodory’s Theorem g isdifferentiable at b and g′(b) = 1

f ′(g(b)).

Example: Let I = [−π2, π2] and f(x) = sin x. Then sin is strictly monotone

on I and continuous. Let J = f(I) = [−1, 1], then f has an inverse on J ,f−1 = arcsin. Also, (sin)′ = cos so (sin)′(x) = cos x = 0 for all x = ±π

2. So

arcsin is differential for all y = ±1.

(arcsin)′(y) =1

f ′(arcsin(y))=

1

cos(arcsin(y))=

1√1− sin(arcsin(y))2

=1√

1− y2

(the sign of the square root is positive because arcsin is an increasing functionon J).

Theorem 6.5 (Chain Rule). Let I, J ⊆ R be intervals, let a ∈ I and letf : I → R and g : J → R. Suppose that for all x ∈ I we have that f(x) ∈ J .If f is differentiable at a and g is differentiable at b = f(a), then g ◦ f isdifferentiable at a and (g ◦ f)′(a) = g′(f(a))f ′(a).

Proof. Let φ : I → R defined by

φ(x) =

{f(x)−f(a)

x−ax = a

f ′(a) x = a

and Let ψ : J → R defined by

ψ(y) =

{ g(y)−g(b)y−b

y = b

g′(b) y = b

6.3. EXTREME POINTS AND MEAN VALUE THEOREMS 47

Then from Caratheodory’s Theorem φ is continuous at a and ψ is continuousat b. Now, for all x ∈ I

g◦f(x)−g◦f(a) = g(f(x))−g(f(a)) = g(f(x))−g(b) = ψ(f(x))(f(x)−b) =

ψ(f(x))(f(x)− f(a)) = ψ(f(x))φ(x)(x− a) = ((ψ ◦ f)φ)(x)(x− a).

Define τ : I → R by

τ(x) = ((ψ ◦ f)φ)(x) ={ψ(f(x))φ(x) x = ag′(f(a))f ′(a) x = a

We have just shown that g ◦ f(x) − g ◦ f(a) = τ(x)(x − a). Since f iscontinuous at a, ψ is continuous at b = f(a) and φ is continuous at a itfollows from theorems 5.1 and 5.2 that τ is continuous at a. Caratheodory’sTheorem implies the result.

Example: Let f(x) = sin(x) g(x) = x2 and I = J = R. Then f ′(x) = cos(x)and g′(x) = 2x. So (

sin2(x))′= 2 sin(x) cos(x) = sin(2x).

6.3 Extreme Points and Mean Value

Theorems

Theorem 6.6 (Interior Extremum Theorem). Let I = [a, b] ⊂ R be a closedand bounded interval with a = b , let a < c < b and let f : I → R. If c is alocal extremum point of f and f is differentiable at c, then f ′(c) = 0.

Proof. Without loss of generalization c is a local maximum. That is thereexist a ≤ e < c < f ≤ b such that f(c) ≥ f(x) for all x ∈ [e, f ]. Take (xn) tobe a sequence in [e, c) converging to c. Then

f ′(c) = limn→∞

f(xn)− f(c)

xn − c.

As f(c) ≥ f(xn) and c > xn for all n we have that f(xn)−f(c)xn−c

≥ 0 for all nand thus f ′(c) ≥ 0. Now, take (xn) to be a sequence in (c, f ] converging toc. Then

f ′(c) = limn→∞

f(xn)− f(c)

xn − c.


As f(c) ≥ f(xn) and c < xn for all n we have that f(xn)−f(c)xn−c

≤ 0 for all nand thus f ′(c) ≤ 0. We conclude that f ′(c) = 0.

Theorem 6.7 (Rolle’s Theorem). Let I = [a, b] ⊂ R be a closed and boundedinterval with a = b and let f : I → R. Suppose f is continuous on I anddifferentiable at any x ∈ (a, b). If f(a) = f(b), then there exists c ∈ (a, b),such that f ′(c) = 0.

Proof. There are two possibilities either f is a constant function and theargument is obvious as f ′(x) = 0 for all x ∈ I or there exists x ∈ (a, b) withf(x) = f(a) = f(b). Since f is continuous on I from Theorem 5.3 there existc, d ∈ I such that f(c) is minimum for f(I) and f(d) is maximum for f(I).By the previous argument either f(c) = f(a) or f(d) = f(a) without lossof generality say f(c) = f(a). So a < c < b from the Interior ExtremumTheorem f ′(c) = 0.

Example: Let P (x) be a polynomial of degree n. Then P ′(x) is a polynomialof degree n − 1. Suppose P has n distinct real roots α1 < α2 < · · · < αn.Then from Rolle’s Theorem P ′(x) has a root between αi and αi+1 for all i.Thus, it has exactly n− 1 distinct roots.

Let us generalize Rolle’s Theorem:

Theorem 6.8 (Mean Value Theorem). Let I = [a, b] ⊂ R be a closed andbounded interval with a = b and let f : I → R. Suppose f is continuous onI and differentiable at any x ∈ (a, b). Then there exists c ∈ (a, b) such that

f(b)− f(a) = f ′(c)(b− a).

Proof. Let

g(x) = f(x)− f(a)− f(b)− f(a)

b− a(x− a).

Clearly, g(x) is continuous on I, differentiable at any x ∈ (a, b) and g(a) =g(b) = 0. So from Rolle’s Theorem there exists c ∈ (a, b) such that

0 = g′(c) = f ′(c)− f(b)− f(a)

b− a.

So

f ′(c) =f(b)− f(a)

b− a

and therefore f(b)− f(a) = f ′(c)(b− a).

6.3. EXTREME POINTS AND MEAN VALUE THEOREMS 49

Corollary 6.9. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R. Suppose f is continuous on I and differentiable at any x ∈ (a, b)and f ′(x) = 0. Then f is constant on I.

Proof. Take x < y ∈ I. Then from the Mean Value Theorem there existsz ∈ (x, y) with f(y) − f(x) = f ′(z)(y − x) = 0. Thus, f(x) = f(y) asrequired.

Corollary 6.10. Let I = [a, b] ⊂ R be a closed and bounded interval and letf, g : I → R. Suppose f and g are continuous on I, differentiable at any x ∈(a, b) and f ′(x) = g′(x). Then there exists a constat C such that f = g + Con I.

Proof. Look at h(x) = f(x)− g(x) and apply the previous corollary.

Corollary 6.11. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R. Suppose f is continuous on I and differentiable at all x ∈ (a, b).

1. f is increasing if and only if f ′(c) ≥ 0 for all c ∈ (a, b);

2. f is decreasing if and only if f ′(c) ≤ 0 for all c ∈ (a, b).

Proof. Clearly 1 implies 2 by looking at−f . So it enough to prove 1. Supposef is increasing, then if x > c we have that f(x) ≥ f(c) so f(x)−f(c)

x−c≥ 0 and

if x < c we have that f(x) ≤ f(c) so again f(x)−f(c)x−c

≥ 0. Thus, f ′(c) is thelimit of non-negative numbers and therefore, non-negative.

On the other hand, suppose f ′(c) ≥ 0 for all c ∈ (a, b), then take y >x ∈ I. Then from the Mean Value Theorem there exists c ∈ (x, y) such thatf(y)− f(x) = f ′(c)(y − x) ≥ 0. We deduce that f is increasing.

Theorem 6.12 (Cauchy Mean Value Theorem). Let I = [a, b] ⊂ R be aclosed and bounded interval with a = b and let f, g : I → R. Suppose f and gare continuous on I and differentiable at all x ∈ (a, b) and assume g′(x) = 0for all x ∈ (a, b). Then there exists c ∈ (a, b) such that

f(b)− f(a)

g(b)− g(a)=f ′(c)

g′(c).

Proof. First we notice that since g′(x) = 0 for all x ∈ (a, b) form Rolle’sTheorem g(b) = g(a). Thus, we can indeed divide by g(b)− g(a). Let

h(x) =f(b)− f(a)

g(b)− g(a)(g(x)− g(a))− (f(x)− f(a)).


Then h(a) = 0 = h(b). From Rolle’s Theorem there exists c ∈ (a, b) suchthat

0 = h′(c) =f(b)− f(a)

g(b)− g(a)g′(c)− f ′(c).

We deduce thatf(b)− f(a)

g(b)− g(a)=f ′(c)

g′(c).

6.4 L’Hospital Rule

Cauchy Mean Value Theorem is not just abstract but it has an applicationfor computation:

Theorem 6.13 (L’Hospital Rule). Let I = [a, b] ⊂ R be a closed and boundedinterval and let f, g : I → R be continuous functions on I which are differ-entiable on (a, b). Let c ∈ I, suppose that f(c) = g(c) = 0 and that g(x) = 0and g′(x) = 0 for all c = x ∈ (a, b). If

limx→c

f ′(x)

g′(x)

exists, then we have that

limx→c

f(x)

g(x)= lim

x→c

f ′(x)

g′(x).

Proof. From Cauchy Mean Value Theorem we have that for any c = x ∈ Ithere exists tx between x and c such that

f(x)

g(x)=f(x)− f(c)

g(x)− g(c)=f ′(tx)

g′(tx).

Now when x converges to c so does tx as it is between x and c. Hence,

limx→c

f(x)

g(x)= lim

x→c

f ′(tx)

g′(tx)= lim

t→c

f ′(t)

g′(t)

as claimed.

6.5. TAYLOR’S THEOREM 51

Comments:

• The condition that limx→c

f ′(x)g′(x)

exists is necessary and we will see at home-

work an example where this limit does not exists but neverthelesslimx→c

f(x)g(x)

does exists.

• There are many variation on L’Hospital Rule which follows either fromvariation on the proof or direct application of the rule. We will notexplore all of them due to time limitations, but you should be able tofind them in many books.

Example: Take f(x) = sin(x), g(x) =√x and c = 0. First we notice that

f(0) = g(0) = 0. We also know that f and g are differentiable on I. Now,f ′(x) = cos(x) and g′(x) = 1

2√x. Since

limx→0

f ′(x)

g′(x)= lim

x→0

cos(x)1

2√x

= limx→0

2√x cos(x) = 0

from L’Hospital Rule we have that limx→0

sin(x)√x

= 0.

We can actually verify that the theorem works in this case by computingthe limit directly

limx→0

sin(x)√x

= limx→0

√x sin(x)

x=(limx→0

√x)(

limx→0

sin(x)

x

)= 0× 1 = 0.

6.5 Taylor’s Theorem

Another application of Rolle’s Theorem is to approximate a function by apolynomial using its derivatives. First we need to define the higher deriva-tives:

Definition. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R be differentiable on I. If c ∈ I and f ′(x) is differentiable at c,then we call (f ′)′(c) the second derivative of f at c and we write f ′′(c) orf (2)(c) for (f ′)′(c). Similarly, we can the define the third derivative of fat c and write f ′′′(c) or f (3)(c) for (f ′′)′(c). We also write f (0)(c) for f(c).In general, if f has an n − 1 derivative which is differentiable on I we cantalk about the n-derivative of f at c and write f (n)(c) for (f (n−1))′(c).


Theorem 6.14 (Taylor’s Theorem). Let n ∈ N, let I = [a, b] ⊂ R be a closedand bounded interval and let f : I → R be such that its i-derivatives existand are continuous for all 1 ≤ i ≤ n on I and that f (n+1) exists on (a, b).Fix x0 ∈ I and define P : I → R by

P (x) = f(x0) + f ′(x0)(x− x0) +f ′′(x0)

2!(x− x0)

2 + · · ·

+f (n)(x0)

n!(x− x0)

n =n∑

i=0

f (i)(x0)

i!(x− x0)

i.

Then for any x ∈ I there exists a point c = c(x) between x and x0 such that

f(x) = P (x) +f (n+1)(c)

(n+ 1)!(x− x0)

n+1.

Proof. We notice that P (x) is a polynomial of degree n and therefore P (i)(x) =0 for i > n. We also notice that P (i)(x0) = f (i)(x0) for 0 ≤ i ≤ n. Fix x ∈ Iand set M to be a number such that f(x)−P (x) =M(x− x0)

n+1 (if x = x0

we take M = f (n+1)(x0)(n+1)!

). We need to show that

M =f (n+1)(c)

(n+ 1)!

for some c between x and x0.

For t between x and x0 we define g(t) = f(t) − P (t) − M(t − x0)n+1.

Then for 0 ≤ i ≤ n we have that g(i)(x0) = 0. By the definition of M wehave that g(x) = 0. Therefore, from Rolle’s Theorem we have c0 betweenx and x0 such that g′(c0) = 0. Applying Rolle’s Theorem again we have c1between c0 and x0 (and therefore between x and x0) such that g(2)(c1) = 0.We continue by induction until we find c = cn between x and x0 such that0 = g(n+1)(c) = f (n+1)(c)− (n+ 1)!M . We conclude that

M =f (n+1)(c)

(n+ 1)!

as required.

6.6. TRIGONOMETRIC FUNCTIONS 53

6.6 Trigonometric Functions

Theorem 6.15 (Taylor’s Series of Trigonometric Functions). For all x ∈ Rsin(x) =

∞∑n=0

(−1)n

(2n+1)!x2n+1 and cos(x) =

∞∑n=0

(−1)n

(2n)!x2n.

Proof. Since (sin(x))′ = cos(x) and (cos(x))′ = − sin(x) it is easy to proveby induction on n that sin(2n)(0) = 0 and sin(2n+1)(0) = (−1)n. Denote

sn(x) =n∑

i=0

(−1)i

(2i+1)!x2i+1. Then from Taylor’s Theorem

sin(x)− sn(x) =(sin(c))(2n+2)

(2n+ 2)!x2n+2

for some c between 0 and x. Since∣∣(sin(c))(2n+2)

∣∣ = |sin(c)| ≤ 1 we obtainthat

| sin(x)− sn(x)| ≤|x|2n+2

(2n+ 2)!−→n→∞

0.

We deduce that∞∑n=0

(−1)n

(2n+1)!x2n+1 = lim

n→∞sn(x) = sin(x). Similar argument

shows that cos(x) =∞∑n=0

(−1)n

(2n)!x2n.


Chapter 7

Riemann Integral

7.1 Riemann Integral and Its Basic

Properties

Definition. Let I = [a, b] ⊂ R be a closed and bounded interval. A partitionof I is an ordered set P = {ai}ni=0, where a = a0 < a1 < · · · < an−1 < an =b. The mesh of P is δ(P) = max

i{ai − ai−1}. A partition Q is called a

refinement of P if P ⊆ Q.

Example: P ={

in

∣∣∣0 ≤ i ≤ n}

is a partition of [0, 1] with mesh δP = 1n.

Q ={

imn

∣∣∣0 ≤ i ≤ mn}is a refinement of P .

Lemma 7.1. Let I = [a, b] ⊂ R be a closed and bounded interval and letP and Q be partitions of I. Then there exists a partition of I which is arefinement of both P and Q.

Proof. Take P ∪Q.

Definition. Let I = [a, b] ⊂ R be a closed and bounded interval, let f : I → Rbe a bounded function and let P = {ai}ni=0 be a partition of I. We define thelower sum of P to be

σ(f,P) = s(f,P) =n∑

i=1

(ai − ai−1) inf{f(x)

∣∣∣x ∈ [ai−1, ai]}

55

56 CHAPTER 7. RIEMANN INTEGRAL

and the upper sum of P to be

Σ(f,P) = S(f,P) =n∑

i=1

(ai − ai−1) sup{f(x)

∣∣∣x ∈ [ai−1, ai]}.

Examples:

1. Let I = [x, y], let P = {ai}ni=0 and let f : I → R be a constant functionf(t) = c. Then

σ(f,P) =n∑

i=1

(ai − ai−1)c = cn∑

i=1

(ai − ai−1) = c(y − x).

Similarly, Σ(f,P) = c(y − x).

2. Let I = [0, 1], let P ={

in

∣∣∣0 ≤ i ≤ n}

and let f : I → R be f(x) = x.

Then

σ(f,P) =n∑

i=1

(i

n− i− 1

n

)i− 1

n=

n∑i=1

1

n

(i− 1)

n=

1

n2

n∑i=1

(i− 1) =n(n− 1)

2n2=n− 1

2n=

1

2− 1

2n.

Similarly,

Σ(f,P) =n∑

i=1

(i

n− i− 1

n

)i

n=

1

n2

n∑i=1

i =(n+ 1)n

2n2=n+ 1

2n=

1

2+

1

2n.

3. Here is a question from a recent exam:

Suppose that J = [0, 2], g(x) = (x− 1)2 and Q ={0, 1

2, 43, 2}.

(a) What is δQ, i.e., the mesh of Q?

δQ = 43− 1

2= 5

6.

(b) Compute σ(g,Q) and Σ(g,Q).

σ(g,Q) =

(1

2− 0

)(1

2− 1

)2

+

(4

3− 1

2

)(1− 1)2+

(2− 4

3

)(4

3− 1

)2

=

7.1. RIEMANN INTEGRAL AND ITS BASIC PROPERTIES 57

1

2× 1

4+

2

3× 1

9=

1

8+

2

27=

43

216.

Σ(g,Q) =

(1

2− 0

)(0−1)2+

(4

3− 1

2

)(1

2− 1

)2

+

(2− 4

3

)(2−1)2 =

1

2+

5

6× 1

4+

2

3=

12

24+

5

24×+

16

24=

33

24=

11

8.

Lemma 7.2. Let I = [a, b] ⊂ R be a closed and bounded interval, letf : I → R be a bounded function and let P and Q be partitions of I. Supposethat Q is a refinement of P. Then we have that σ(f,P) ≤ σ(f,Q) andΣ(f,P) ≥ Σ(f,Q).

Proof. Suppose P = {ai}ni=0 and Q = {bj}mj=0. As P ⊆ Q for each i we havethat ai ∈ Q so we can find ji such that ai = bji . We notice that

ai−1 = bji−1< bji−1+1 < bji−1+2 < · · · < bji−2 < bji−1 < bji = ai.

We write ki = inf{f(x)

∣∣∣x ∈ [ai−1, ai]}andmj = inf

{f(x)

∣∣∣x ∈ [bj−1, bj]}.

For all ji−1 < j ≤ ji we have that [bj−1, bj] ⊆ [ai−1, ai] as the infimum on asmaller set is larger we have that ki ≤ mj for all ji−1 < j ≤ ji . Therefore,

σ(f,Q) =m∑j=1

(bj − bj−1)mj =

n∑i=1

((bji − bji−1)mji + (bji−1 − bji−2)mji−1 + · · ·+ (bji−1+1 − bji−1

)mji−1+1

)≥

n∑i=1

((bji − bji−1)ki + (bji−1 − bji−2)ki + · · ·+ (bji−1+1 − bji−1

)ki)=

n∑i=1

(bji − bji−1)ki =

n∑i=1

(ai − ai−1)ki = σ(f,P).

Similarly, Σ(f,P) ≥ Σ(f,Q).


Example: Let I = [0, 1], P ={

in

∣∣∣0 ≤ i ≤ n}, Q =

{i

mn

∣∣∣0 ≤ i ≤ mn}

and

f(x) = x. Then

σ(f,P) =1

2− 1

2n≤ 1

2− 1

2mn= σ(f,Q)

and

Σ(f,P) =1

2+

1

2n≥ 1

2+

1

2mn= Σ(f,Q).

Definition. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R be a bounded function. We define the lower Riemann integralto be ∫

−f = s(f) = sup

{σ(f,P)

∣∣∣ where P is a partition of I}

and upper Riemann integral to be

−∫f = S(f) = inf

{Σ(f,P)

∣∣∣ where P is a partition of I}.

If∫−f =

−∫f , then we say that f is Riemann integrable on I or just

integrable on I and we define its integral to be

b∫a

f =

b∫a

f(x)dx =

∫−f =

−∫f.

We also writea∫

b

f = −b∫

a

f.

Lemma 7.3. Let I = [a, b] ⊂ R be a closed and bounded interval, letf : I → R be a bounded function and let P and Q be any couple of partitions

of I. Then Σ(f,Q) ≥ σ(f,P). Furthermore,−∫f ≥

∫−f .


Proof. By Lemma 7.1 and Lemma 7.2 we have that

Σ(f,Q) ≥ Σ(f,P ∪Q) ≥ σ(f,P ∪Q) ≥ σ(f,P).

Fix ϵ > 0. We find a partition P such that σ(f,P) ≥∫−(f)−ϵ and a partition

Q such that Σ(f,Q) ≤−∫f + ϵ. Then

−∫f + ϵ ≥ Σ(f,Q) ≥ σ(f,P) ≥

∫−f − ϵ.

Since ϵ is arbitrary we can take it as small as we would like, we conclude that−∫f ≥

∫−f .

Examples:

1. Let I = [0, 1], f(x) = x and P ={

in

∣∣∣0 ≤ i ≤ n}. Then

1

2− 1

2n= σ(f,P) ≤

∫−f ≤

−∫f ≤ Σ(f,P) =

1

2+

1

2n.

By letting n → ∞ we obtain that∫−f =

−∫f = 1

2. Thus, f is Riemann

integrable on [0, 1] and1∫0

xdx = 12.

2. Let I = [0, 1] and

f(x) =

{1, x ∈ Q;0, Otherwise.

As any interval which is not a point contains both rational and ir-rational numbers we have that the supremum of f on such intervalis 1 and the infimum is 0. Thus, we deduce that Σ(f,P) = 1 and

σ(f,P) = 0 for any partition P of I. Hence,−∫f = 1 and

∫−f = 0 and

f is not integrable.


Lemma 7.4. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R be a Riemann integrable function on I. Suppose λ ∈ R, then λf

is Riemann integrable on I andb∫a

λf(x)dx = λb∫a

f(x)dx.

Proof. Suppose λ ≥ 0. Fix c, d ∈ I we notice that

inf{λf(x)

∣∣∣x ∈ [c, d]}= λ inf

{f(x)

∣∣∣x ∈ [c, d]}

andsup

{λf(x)

∣∣∣x ∈ [c, d]}= λ sup

{f(x)

∣∣∣x ∈ [c, d]}.

Then for any P partition of I it is obvious that σ(λf,P) = λσ(f,P) and

Σ(λf,P) = λΣ(f,P). Therefore,∫−λf = λ

∫−f and

−∫λf = λ

−∫f . Hence, if

∫−f =

−∫f , then

∫−λf = λ

∫−f = λ

−∫f =

−∫λf , so λf is Riemann integrable

andb∫a

λf(x)dx =−∫λf = λ

−∫f = λ

b∫a

f(x)dx.

If λ is negative, then we notice that multiplying by λ reverses the rolesof inf and sup. That is

inf{λf(x)

∣∣∣x ∈ [c, d]}= λ sup

{f(x)

∣∣∣x ∈ [c, d]}

andsup

{λf(x)

∣∣∣x ∈ [c, d]}= λ inf

{f(x)

∣∣∣x ∈ [c, d]}.

Thus, for any P partition of I we have that σ(λf,P) = λΣ(f,P) andΣ(λf,P) = λσ(f,P). So∫

−λf = sup

{σ(λf,P)

∣∣∣ where P is a partition of I}=

sup{λΣ(f,P)


λ inf{Σ(f,P)

∣∣∣ where P is a partition of I}= λ

−∫f


and−∫λf = inf

{Σ(λf,P)


inf{λσ(f,P)


λ sup{σ(f,P)

∣∣∣ where P is a partition of I}= λ

∫−f.

Hence, if∫−f =

−∫f , then

∫−λf = λ

−∫f = λ

∫−f =

−∫λf , so λf is Riemann

integrable andb∫a

λf(x)dx =∫−λf = λ

−∫f = λ

b∫a

f(x)dx.

Theorem 7.5. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R and g : I → R be Riemann integrable functions on I. Suppose

λ, µ ∈ R, then λf + µg is Riemann integrable on I andb∫a

(λf + µg)(x)dx =

λb∫a

f(x)dx+ µb∫a

g(x)dx.

Proof. By the previous lemma it suffices to show that f + g is Riemann

integrable and thatb∫a

(f + g)(x)dx =b∫a

f(x)dx +b∫a

g(x)dx. Fix c, d ∈ I and

let

mf = inf{f(x)

∣∣∣x ∈ [c, d]},

mg = inf{g(x)

∣∣∣x ∈ [c, d]}

and

mf+g = inf{(f + g)(x)

∣∣∣x ∈ [c, d]}.

Given ϵ > 0 we can find y ∈ [c, d] such that (f + g)(y) < mf+g + ϵ. Thus,mf +mg ≤ f(y) + g(y) < mf+g + ϵ. As ϵ is arbitrarily small we deduce thatmf +mg ≤ mf+g.

Hence, for any P partition of I σ(f,P)+σ(g,P) ≤ σ(f+g,P). Fix ϵ > 0and take P a partition of I such that σ(f,P) >

∫−f − ϵ and Q a partition of


I such that σ(g,Q) >∫−g − ϵ. Then Lemma 7.1 and Lemma 7.2 imply that

σ(f,P ∪ Q) ≥ σ(f,P) >∫−f − ϵ and σ(g,P ∪ Q) ≥ σ(g,Q) >

∫−g − ϵ. We

deduce that∫−

(f + g) ≥ σ(f + g,P ∪Q) ≥ σ(f,P ∪Q) + σ(g,P ∪Q) >

∫−f +

∫−g − 2ϵ.

Since ϵ is arbitrary small we deduce that∫−(f + g) ≥

∫−f +

∫−g. Similarly,

−∫f +

−∫g ≥

−∫(f + g). As f and g are Riemann integrable we derive from

Lemma 7.3 that

b∫a

f(x)dx+

b∫a

g(x)dx =

∫−f +

∫−g ≤

∫−

(f + g) ≤

−∫(f + g) ≤

−∫f +

−∫g =

∫ b

a

f(x)dx+

∫ b

a

g(x)dx.

We conclude thatb∫a

f(x)dx+b∫a

g(x)dx =∫−(f + g) =

−∫(f + g) and therefore

f + g is Riemann integrable on I and

b∫a

(f + g)(x)dx =

∫−

(f + g) =

b∫a

f(x)dx+

b∫a

g(x)dx.

Theorem 7.6. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R and g : I → R be bounded functions. If f(x) ≤ g(x) for all

x ∈ I, then∫−f ≤

∫−g and

−∫f ≤

−∫g. In particular, if f and g are Riemann

integrable, thenb∫a

f(x)dx ≤b∫a

g(x)dx.


Proof. Let P = {ai}ni=0 be a partition of I. Since f(x) ≤ g(x) for all x ∈ I it

is not hard to show that inf{f(x)

∣∣∣x ∈ [ai−1, ai]}≤ inf

{g(x)

∣∣∣x ∈ [ai−1, ai]}

and sup{f(x)

∣∣∣x ∈ [ai−1, ai]}

≤ sup{g(x)

∣∣∣x ∈ [ai−1, ai]}

for all i. Hence,

σ(f,P) ≤ σ(g,P) and Σ(f,P) ≤ Σ(g,P). Therefore,

σ(f,P) ≤ σ(g,P) ≤ sup{σ(g,Q)

∣∣∣ where Q is a partition of I}=

∫−g

and

−∫f = inf

{Σ(f,Q)

∣∣∣ where Q is a partition of I}≤ Σ(f,P) ≤ Σ(g,P).

We deduce that∫−f = sup

{σ(f,P)

∣∣∣ where P is a partition of I}≤∫−g

and−∫f ≤ inf

{Σ(g,P)


−∫g.

If f and g are Riemann integrable, then

b∫a

f(x)dx =

∫−f ≤

∫−g =

b∫a

g(x)dx.

Theorem 7.7. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R be a bounded function. Then f is Riemann integrable if and onlyif there exists a sequence Pn of partitions of I such that

limn→∞

(Σ(f,Pn)− σ(f,Pn)) = 0.

Moreover, if this is the case, then

b∫a

f(x)dx = limn→∞

Σ(f,Pn) = limn→∞

σ(f,Pn).


Proof. Suppose f is Riemann integrable. Given n ∈ N we find Q and Q′

partitions of I such that Σ(f,Q) ≤−∫f + 1

nand σ(f,Q′) ≥

∫−f − 1

n. Let

Pn = Q∪Q′. Then

0 ≤ Σ(f,Pn)− σ(f,Pn) ≤ Σ(f,Q)− σ(f,Q′) ≤−∫f +

1

n−∫−f − 1

n=

2

n.

We conclude that limn→∞

(Σ(f,Pn)− σ(f,Pn)) = 0.

On the other hand, suppose that there exists Pn a sequence of partitionsof I such that lim

n→∞(Σ(f,Pn)− σ(f,Pn)) = 0. Given ϵ > 0 take n such that

Σ(f,Pn)− σ(f,Pn) < ϵ. Then

0 ≤−∫f −

∫−f ≤ Σ(f,Pn)− σ(f,Pn) < ϵ.

As ϵ is arbitrary we conclude that−∫f =

∫−f and f is Riemann integrable.

Now suppose limn→∞

(Σ(f,Pn)− σ(f,Pn)) = 0, i.e., f is Riemann inte-

grable. Given ϵ > 0 we take N such that for all n ≥ N we have that

0 ≤ Σ(f,Pn)− σ(f,Pn) < ϵ.

Then by adding 0 =∫−f −

−∫f to the equation above we obtain that

0 =

∫−f −

−∫f ≤ Σ(f,Pn)−

−∫f +

∫−f − σ(f,Pn) <

∫−f −

−∫f + ϵ = ϵ.

Thus,

0 ≤

Σ(f,Pn)−−∫f

+

∫−f − σ(f,Pn)

< ϵ.


Since Σ(f,Pn)−−∫f ≥ 0 and

∫−f − σ(f,Pn) ≥ 0 we conclude that

0 ≤ Σ(f,Pn)−−∫f < ϵ

and

0 ≤∫−f − σ(f,Pn) < ϵ.

Hence,b∫

a

f(x)dx =

−∫f = lim

n→∞Σ(f,Pn)

andb∫

a

f(x)dx =

∫−f = lim

n→∞σ(f,Pn).

Example: Let I = [0, 1] and f(x) = x2. Let Pn = {ai}ni=0 be a partition ofI. Let δn = δ(Pn) = max

i{ai − ai−1} be the mesh of Pn. Then

Σ(x2,Pn)− σ(x2,Pn) =n∑

i=1

(ai − ai−1)(a2i − a2i−1) =

n∑i=1

(ai − ai−1)(ai − ai−1)(ai + ai−1) ≤n∑

i=1

2(ai − ai−1)(ai − ai−1) ≤

n∑i=1

2δn(ai − ai−1) = 2δn

n∑i=1

(ai − ai−1) = 2δn.

So we deduce that if δn tends to 0, then1∫0

x2dx = limn→∞

Σ(x2,Pn) = limn→∞

σ(x2,Pn).

So take Pn ={

in

∣∣∣0 ≤ i ≤ n}. Then δn = 1

nwhich tend to 0. Let us

compute

Σ(f,Pn) =n∑

i=1

(i

n− i− 1

n

)(i

n

)2

=1

n3

n∑i=1

i2 =


1

6n3n(n+ 1)(2n+ 1) −→

n→∞

1

3=

1∫0

x2dx.

Corollary 7.8. Let I = [a, b] ⊂ R be a closed and bounded interval, letf : I → R be a bounded function and let J = [c, d] ⊆ I be a closed intervalcontained in I. If f is Riemann integrable on I, then it is Riemann integrableon J .

Proof. Suppose f is integrable on I. Let Pn be a sequence of partitionsof I such that lim

n→∞(Σ(f,Pn)− σ(f,Pn)) = 0. From Lemma 7.2 we know

that adding points to each Pn will not change that property. Thus, we canassume that c, d ∈ Pn for all n. Clearly, Qn = Pn ∩ J are partitions of J .Now, it is easy to see that 0 ≤ Σ(f,Qn) − σ(f,Qn) < Σ(f,Pn) − σ(f,Pn)(the intervals in I that are not contained in J contribute their length timethe difference between the supremum and the infimum which is a positivequantity). Therefore, lim

n→∞(Σ(f,Qn)− σ(f,Qn)) = 0. We deduce that f is

integrable on J .

Corollary 7.9. Let I = [a, b] ⊂ R be a closed and bounded interval, letf : I → R be a bounded function and fix c ∈ I. If f is Riemann integrable on

[a, c] and [c, b], then it is Riemann integrable on I andb∫a

f(x)dx =c∫a

f(x)dx+

b∫c

f(x)dx.

Proof. From Theorem 7.7 we see that there exist Qn partitions of [a, c] suchthat lim

n→∞(Σ(f,Qn)− σ(f,Qn)) = 0 and Q′

n partitions of [c, b] such that

limn→∞

(Σ(f,Q′n)− σ(f,Q′n)) = 0. We define Pn = Qn ∪ Q′

n. We notice that

a, b ∈ Pn for all n, so Pn are partitions of [a, b]. We also notice that c ∈ Pn

and therefore it is easy to see that Σ(f,Pn) = Σ(f,Qn) + Σ(f,Q′n) and

σ(f,Pn) = σ(f,Qn) + σ(f,Q′n) for all n. We deduce that

0 ≤ Σ(f,Pn)− σ(f,Pn) = Σ(f,Qn) + Σ(f,Q′n)− σ(f,Qn) + σ(f,Q′

n) =

(Σ(f,Qn)− σ(f,Qn)) + (Σ(f,Q′n)− σ(f,Q′

n)) .

Since limn→∞

(Σ(f,Qn)− σ(f,Qn)) = 0 and limn→∞

(Σ(f,Q′n)− σ(f,Q′n)) = 0

we conclude that limn→∞

(Σ(f,Pn)− σ(f,Pn)) = 0. Thus, from Theorem 7.7 f


is Riemann integrable on I. Finally,

b∫a

f(x)dx = limn→∞

Σ(f,Pn) = limn→∞

(Σ(f,Qn) + Σ(f,Q′n)) =

limn→∞

Σ(f,Qn) + limn→∞

Σ(f,Q′n) =

c∫a

f(x)dx+

b∫c

f(x)dx.

Corollary 7.10. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R be a monotonic function on I, then f is Riemann integrable.

Proof. Without loss of generalization f is increasing. Let Pn = {ai}ni=0 be apartition of I, such that ai − ai−1 = b−a

nfor all 1 ≤ i ≤ n. Notice that

inf{f(x)

∣∣∣x ∈ [ai−1, ai]}

= f(ai−1) and sup{f(x)

∣∣∣x ∈ [ai−1, ai]}

= f(ai).

We deduce that

0 ≤ Σ(f,Pn)− σ(f,Pn) =

n∑i=1

(ai − ai−1) sup{f(x)

∣∣∣x ∈ [ai−1, ai]}−

n∑i=1

(ai − ai−1) inf{f(x)

∣∣∣x ∈ [ai−1, ai]}=

n∑i=1

b− a

nf(ai)−

n∑i=1

b− a

nf(ai−1) =

b− a

n

n∑i=1

(f(ai)− f(ai−1)) =(b− a)(f(b)− f(a))

n.

Then limn→∞

(Σ(f,Pn)− σ(f,Pn)) = 0 and f is integrable on I.

Theorem 7.11 (The Integral Test). Let f : [1,∞) → R be a non-negative

decreasing function. Set Sn =n∫1

f(t)dt. Then∞∑k=1

f(k) converges if and only

if (Sn) is a convergent sequence.


Proof. Fix n and let Pn = {1, 2, . . . , n} be a partition of In = [1, n]. As f is

decreasing f(i−1) = sup{f(x)

∣∣∣x ∈ [i− 1, i]}and f(i) = inf

{f(x)

∣∣∣x ∈ [i− 1, i]}.

Thus, Σ(f,Pn) =n−1∑k=1

f(k) and σ(f,Pn) =n∑

k=2

f(k). We notice thatn−1∑k=1

f(k) =

Σ(f,Pn) ≥ Sn ≥n∑

k=2

f(k) =n∑

k=2

f(k).

Suppose∞∑k=1

f(k) converges, then (Sn) is bounded above. As f(x) is non-

negative we have that (Sn) is increasing. From the Monotone ConvergenceTheorem (Sn) is convergent. On the other hand, suppose (Sn) is a convergent.

Thenn∑

k=2

f(k) is bounded. Since f(x) is non-negative from Theorem 3.3

∞∑k=2

f(k) converges. As∞∑k=1

f(k) = f(1) +∞∑k=2

f(k) it also converges.

Examples:

1. Take f(x) = 1x2 . Then

limn→∞

n∫1

f(t)dt = limn→∞

n∫1

1

t2(t)dt = lim

n→∞

−1

t

∣∣∣n1= lim

n→∞

(−1

n+ 1

)= 1.

Indeed,∞∑n=1

1n2 converges.

2. Take f(x) = 1x. Then

limn→∞

n∫1

f(t)dt = limn→∞

n∫1

1

tdt = lim

n→∞log t

∣∣∣n1= lim

n→∞log n

which is divergent. Indeed, the harmonic series∞∑n=1

1nis also divergent.

7.2 The Fundamental Theorem of Calculus

Theorem 7.12 (The Fundamental Theorem of Calculus). Let I = [a, b] ⊂ Rbe a closed and bounded interval and let f : I → R be a bounded function.

7.3. RIEMANN INTEGRAL AND CONTINUOUS FUNCTIONS 69

If f is Riemann integrable and f = F ′ on I for some function F , thenb∫a

f(x)dx = F (b)− F (a).

Proof. Let P = {ai}ni=0 be a partition of I. By the Mean Value Theorem foreach i we have ti ∈ [ai−1, ai] such that F (ai) − F (ai−1) = (ai − ai−1)f(ti).Obviously

inf{f(x)

∣∣∣x ∈ [ai−1, ai]}≤ f(ti) ≤ sup

{f(x)

∣∣∣x ∈ [ai−1, ai]}.

Therefore,

F (b)− F (a) =n∑

i=1

(F (ai)− F (ai−1)) =n∑

i=1

(ai − ai−1)f(ti) ≤ Σ(f,P).

Hence, by taking the infimum on all the partitions of I we obtain that

F (b)− F (a) ≤−∫f =

b∫a

f(x)dx.

Similarly,

F (b)− F (a) ≥∫−f =

b∫a

f(x)dx.

We conclude thatb∫a

f(x)dx = F (b)− F (a).

Example:1∫

0

x2dx =x3

3

∣∣∣∣10

=1

3− 0 =

1

3.

7.3 Riemann Integral and Continuous

Functions

Theorem 7.13 (Continuous Function is Riemann Integrable). Let I =[a, b] ⊂ R be a closed and bounded interval. If f : I → R is a continuous


function on I, then it is Riemann integrable on I. Moreover, F (x) =x∫a

f(t)dt

is a differentiable function on (a, b) with F ′(x) = f(x) for all x ∈ (a, b).

Proof. Let Pn = {ai}ni=0 be a partition of I such that ai − ai−1 =b−an

for all1 ≤ i ≤ n. Write

Mi = sup{f(x)

∣∣∣x ∈ [ai−1, ai]}

andmi = inf

{f(x)

∣∣∣x ∈ [ai−1, ai]}.

As f is continuous by Theorem 5.3 we can find xi, yi ∈ [ai−1, ai] such thatMi = f(yi) and mi = f(xi). Since f is continuous on a closed and boundedinterval I it is uniformly continuous on I. Therefore, given ϵ > 0 we canchoose δ > 0 such that if |y− x| < δ with x, y ∈ I, then |f(y)− f(x)| < ϵ

b−a.

Let N be such that b−aN

< δ. Then for all n ≥ N we have that b−an

≤ b−aN

< δ.Hence, 0 ≤ Mi −mi = f(yi)− f(xi) <

ϵb−a

because |yi − xi| ≤ |ai − ai−1| =b−an

≤ b−aN

< δ. Thus, for all n ≥ N

0 ≤ Σ(f,Pn)− σ(f,Pn) =n∑

i=1

(ai − ai−1)(Mi −mi) =

n∑i=1

b− a

n(Mi −mi) <

n∑i=1

b− a

n

ϵ

b− a= ϵ.

We deduce that limn→∞


From Corollary 7.8 we have that f is integrable on [a, x] for all x ∈ (a, b).Fix c ∈ I and we will prove that F is differentiable at c. Then

F (x)− F (c) =

x∫a

f(t)dt−c∫

a

f(t)dt =

x∫c

f(t)dt.

Without loss of generalization c < x. Let

Mx = sup{f(y)

∣∣∣y ∈ [c, x]}

andmx = inf

{f(y)

∣∣∣y ∈ [c, x]}.

7.3. RIEMANN INTEGRAL AND CONTINUOUS FUNCTIONS 71

As f is continuous by Theorem 5.3 we can find r, s ∈ [c, x] such that Mx =f(r) and mx = f(s). Let P = {c, x} be a partition of [c, x]. Then

(x− c)mx = σ(f,P) ≤∫−f =

x∫c

f(t)dt =

−∫f ≤ Σ(f,P) = (x− c)Mx,

where the lower and upper integral here are on the interval [c, x]. Hence,

mx ≤ F (x)− F (c)

x− c≤Mx.

Because r, s ∈ [c, x], when x tends to c so does r and s. Since f is continuousat c we obtain that mx = f(s) and Mx = f(r) both tend to f(c) when xtends to c. We deduce from the Sandwich Theorem that

F ′(c) = limx→c

F (x)− F (c)

x− c= f(c).

Corollary 7.14. Let I = [a, b] ⊂ R be a closed and bounded interval. Iff : I → R is a bounded function which is continuous at every point in Iexcept finite number of points, then f is Riemann integrable on I.

Proof. Without loss of generalization we assume f is discontinuous at onepoint c ∈ I. Let M > |f(x)| for all x ∈ I. Fix n ∈ N. As f is continuous on[a, c− 1

n] and [c+ 1

n, b] it is integrable on both. Therefore, from Theorem 7.7

there areQn andQ′n partitions of [a, c− 1

n] and [c+ 1

n, b] respectively such that

Σ(f,Qn) − σ(f,Qn) <1nand Σ(f,Q′

n)− σ(f,Q′n) <

1n. Let Pn = Qn ∪ Q′

n.Then Pn is a partition of I. Write

Mn = sup

{f(x)

∣∣∣x ∈ [c− 1

n, c+

1

n]

}and

mn = inf

{f(x)

∣∣∣x ∈ [c− 1

n, c+

1

n]

}.

Then

Σ(f,Pn) = Σ(f,Qn) +2

nMn + Σ(f,Q’n)


and

σ(f,Pn) = σ(f,Qn) +2

nmn + σ(f,Q′

n).

Hence,

0 ≤ Σ(f,Pn)−σ(f,Pn) = Σ(f,Qn)−σ(f,Qn)+2

n(Mn−mn)+Σ(f,Q′

n)−σ(f,Q′n)

<2

n+ 2M

2

n.

We conclude that limn→∞


Example: Let I = [0, 1] and let

f(x) =

{sin(1x

), x = 0;

0, x = 0.

Then f is bounded by 1 and continuous on (0, 1], but it is discontinuous at

0 (homework). Thus,1∫0

f(x)dx exists.

Theorem 7.15. Let I = [a, b] ⊂ R be a closed and bounded interval and letf : I → R and g : I → R be continuous functions. If f(x) ≥ g(x) for all x ∈ I

and if there exists c ∈ I such that f(c) > g(c), thenb∫a

f(x)dx >b∫a

g(x)dx.

Proof. As both f and g are continuous so is f−g and clearlyM = (f−g)(c) >0. Since f − g is continuous and (f − g)(c) =M we can find δ > 0 such thatfor all x ∈ (c− δ, c+ δ) ∩ I we have that (f − g)(x) ≥ M

2. By taking δ small

enough either (c− δ, c) ⊆ I or (c, c+ δ) ⊆ I. Wlog let us assume the latter.From continuity we also have that f(c+ δ) ≥ M

2. Then by Theorem 7.6

b∫a

f(x)dx−b∫

a

g(x)dx =

b∫a

(f − g)(x)dx =

c∫a

(f − g)(x)dx+

c+δ∫c

(f − g)(x)dx+

b∫c+δ

(f − g)(x)dx ≥

7.4. RIEMANN INTEGRAL AND UNIFORM CONVERGENCE 73

c∫a

0dx+

c+δ∫c

M

2dx+

b∫c+δ

0dx = δM

2> 0.

We deduce thatb∫a

f(x)dx >b∫a

g(x)dx.

Example: If you take I = [0, 1], g(x) = 0 and

f(x) =

{0, x = 0;1, x = 0.

Then f(x) ≥ g(x) for all x ∈ I and f(0) = 1 > 0 = g(0). However,1∫0

f(x)dx = 0 =1∫0

g(x)dx.

7.4 Riemann Integral and Uniform

Convergence

Theorem 7.16. Let I = [a, b] ⊂ R be a closed and bounded interval andlet fn : I → R be a sequence of functions uniformly converging to f : I →R on I. If fn are Riemann integrable, then f is Riemann integrable andb∫a

f(x)dx = limn→∞

b∫a

fn(x)dx.

Proof. Fix ϵ > 0. Take N such that for all n ≥ N and for all x ∈ Iwe have that |fn(x) − f(x)| < ϵ

b−a. Hence, for all n ≥ N we obtain that

fn(x)− ϵb−a

< f(x) < fn(x)− ϵb−a

. Theorem 7.6 implies that

∫ b

a

fn(x)dx− ϵ =

∫ b

a

(fn(x)−

ϵ

b− a

)dx =

∫−

(fn −

ϵ

b− a

)≤∫−f ≤

−∫f ≤

−∫ (fn +

ϵ

b− a

)=

∫ b

a

(fn(x) +

ϵ

b− a

)dx =

∫ b

a

fn(x)dx+ ϵ.


Thus, 0 ≤−∫f −

∫−f ≤ 2ϵ. As ϵ is arbitrary

−∫f =

∫−f and f is Riemann

integrable. We deduce that for all n ≥ N

b∫a

fn(x)dx− ϵ ≤b∫

a

f(x)dx ≤b∫

a

fn(x)dx+ ϵ.

Hence,b∫a

f(x)dx = limn→∞

b∫a

fn(x)dx.

7.5 Integration and Differentiation of Power

Series

Lemma 7.17. Let (an)∞n=0 be a sequence in R and let R be the radius

of convergence of∞∑n=0

anxn. Then R is the radius of convergence of both

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn and

∞∑n=0

ann+1

xn+1.

Proof. Let S be the radius of convergence of∞∑n=0

(n+ 1)an+1xn. Take y with

|y| < S. Then from Theorem 3.9∞∑n=0

(n + 1)an+1yn is absolutely conver-

gent. As |an+1yn| ≤ |(n+ 1)an+1y

n| from the Comparison Test∞∑n=0

an+1yn is

absolutely convergent. But then∞∑n=0

anyn = a0 + y

∞∑n=0

an+1yn is absolutely

convergent. We deduce that S ≤ R.

Take now y with |y| < R. Pick z with |y| < |z| < R. Then from

Theorem 3.9∞∑n=0

anzn is absolutely convergent. As z

y> 1 there is K such

that(

zy

)n> n for all n ≥ K. Then |anzn| ≥

∣∣∣anyn ( zy

)n∣∣∣ > |nanyn| for all

n ≥ K. From the Comparison Test we see that∞∑n=0

(n+1)an+1yn is absolutely

convergent. We conclude that S ≥ R, so S = R.

7.5. INTEGRATION AND DIFFERENTIATION OF POWER SERIES75

By the argument above∞∑n=0

ann+1

xn+1 has the same radius of converges as

∞∑n=0

(n + 1) ann+1

xn =∞∑n=1

anxn which has the same radius of convergence as

∞∑n=0

anxn.

Theorem 7.18 (Indefinite Integration of Power Series). Let (ai)∞i=0 be a

sequence in R and let R be the radius of convergence of f(x) =∞∑i=0

aixi.

Then g(x) =∞∑i=0

aii+1xi+1 is a differentiable function on (−R,R) such that

g′(x) = f(x). Moreover, if h(x) is another differentiable function such thath′(x) = f(x), then for all x ∈ (−R,R) we have that h(x) = g(x) + C forsome constant C.

Proof. From the lemma above R is also the radius of convergence of g(x).

We define fn(x) =n∑

i=0

aixi and gn(x) =

n∑i=0

aii+1xi+1. For 0 < r < R, we let

J = [−r, r]. From Corollary 5.10 both f and g are continuous. It is clearthat fn(x) = g′n+1(x) from the Fundamental Theorem of Calculus

x∫0

fn(t)dt =

x∫0

g′n+1(t)dt = gn+1(x)− gn+1(0) = gn+1(x)

for all x ∈ [−r, r].Theorem 3.11 says that fn converges uniformly to f on J and gn+1 con-

verges uniformly to g on J . We deduce from Theorem 7.16 that

x∫0

f(t)dt = limn→∞

x∫0

fn(t)dt = limn→∞

gn+1(x) = g(x).

As f is continuous from Theorem 7.13 we have that g is differentiableon (−r, r) and g′(x) = f(x) for all x ∈ (−r, r). Since r is arbitrary, givenx ∈ (−R,R), we can always take r such that |x| < r < R. The moreoverpart follows from Corollary 6.10.

Theorem 7.19 (Differentiation of Power Series). Let (an)∞n=0 be a sequence

in R and let R be the radius of convergence of f(x) =∞∑n=0

anxn. Then f is


differentiable function on the interval (−R,R) and f ′(x) =∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn.

Proof. For n ≥ 0 set bn = (n + 1)an+1 and let g(x) =∞∑n=0

bnxn. From

Lemma 7.17 g(x) has radius of convergence R. As bnn+1

= an+1 we obtainfrom Theorem 7.18 that

g(x) =

(∞∑n=0

bnn+ 1

xn+1

)′

=

(∞∑n=0

an+1xn+1

)′

=

(∞∑n=1

anxn

)′

= (f(x)−a0)′.

Since f(x) = (f(x) − a0) + a0 is the sum of two differentiable functionswe conclude that f is a differentiable function on the interval (−R,R) andf ′(x) = g(x) =

∞∑n=1

nanxn−1.

Theorem 7.20 (Uniqueness Theorem). Let (an)∞n=0 and (bn)

∞n=0 be sequences

in R. Suppose the radiuses of convergence of both∞∑n=0

anxn and

∞∑n=0

bnxn are

not zero. If∞∑n=0

anxn =

∞∑n=0

bnxn on some interval (−r, r), then an = bn for

all n.

Proof. This follows from the previous theorem as by induction for all n(Homework):

n!an =

(∞∑n=0

anxn

)(n)

(0) =

(∞∑n=0

bnxn

)(n)

(0) = n!bn.

7.6 The Exponential and Logarithmic

Functions

We recall that in the example following Theorem 3.8 we defined for x ∈ Rthat E(x) =

∞∑n=0

xn

n!.

7.6. THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS 77

Theorem 7.21 (The Exponential Function). Let E(x) =∞∑n=0

xn

n!. Then

1. E(0) = 1,

2. the radius of convergence of E(x) is ∞,

3. E ′(x) = E(x),

4. E(x+ y) = E(x)E(y),

5. E(x) > 0 for all x ∈ R,

6. E(x) is strictly increasing on R,

7. limx→∞

E(x) = ∞ and limx→−∞

E(x) = 0 (do not worry about the formal

definitions of these limits),

8. E(x) has an inverse function L(x) on (0,∞),

9. E(1) = e and

10. for any q ∈ Q we have that E(qx) = E(x)q and in particular E(q) = eq.

Proof. 1. This is immediate.

2. We fix x and use the Ratio Test:

xn+1

(n+1)!

xn

n!

=x

n+ 1−→n→∞

0 < 1.

Therefore, E(x) is convergent for all x ∈ R and we deduce from Theo-rem 3.9 that its radius of convergence is ∞.

3.

E ′(x) =

(∞∑n=0

xn

n!

)′

=∞∑n=0

nxn−1

n!=

∞∑n=1

xn−1

(n− 1)!=

∞∑n=0

xn

n!= E(x).

4. This is proved in the example following Theorem 3.8.


5. As all the coefficients in E(x) are positive it follows that for all x ≥ 0we have that E(x) ≥ E(0) = 1. We notice that 1 = E(0) = E(x−x) =E(x)E(−x), so E(−x) = 1

E(x). We deduce that E(x) > 0 for all x ∈ R.

6. Since E ′(x) = E(x) > 0 it follows from the Mean Value Theorem thatE(x) is strictly increasing.

7. As all the coefficients in E(x) are positive for x ≥ 0 we have thatE(x) ≥ 1 + x. Thus, lim

x→∞E(x) = ∞. Now, E(−x) = 1

E(x)so we

deduce that limx→−∞

E(x) = 0.

8. Since E(X) is strictly increasing it is injective. Since limx→∞

E(x) =

∞, limx→−∞

E(x) = 0 and E(x) is continuous the Intermediate Value

Theorem implies that E(x) is surjective on (0,∞). (Note that all weneed for the previous part is that E(X) is not bounded above and itis not bounded below by a positive number and these are obvious fromthe proof.) Therefore, E(x) is a bijective function from R onto (0,∞)and thus has an inverse.

9. (1 +

1

m

)m

=m∑

n=0

(m

n

)1

mn=

m∑n=0

m(m− 1)(m− 2) · · · (m− n+ 1)

n!

1

mn=

m∑n=0

m(m− 1)(m− 2) · · · (m− n+ 1)

mn

1

n!

≤m∑

n=0

1

n!≤

∞∑n=0

1

n!= E(1).

We deduce that

e = limm→∞

(1 +

1

m

)m

≤ E(1).

7.6. THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS 79

We will now show that e ≥m∑

n=0

1n!

for all m. Fix m and take k ≥ m.

Because 1 ≥ k−jk

for all 0 ≤ j < m we have for any n ≤ m that

k(k − 1) · · · (k − n+ 1)

kn≥ k(k − 1) · · · (k −m+ 1)

km.

We deduce that(1 +

1

k

)k

=k∑

n=0

(k

n

)1

kn=

k∑n=0

k(k − 1) · · · (k − n+ 1)

n!

1

kn=

k∑n=0

k(k − 1) · · · (k − n+ 1)

kn1

n!≥

m∑n=0

k(k − 1) · · · (k − n+ 1)

kn1

n!≥

m∑n=0

k(k − 1) · · · (k −m+ 1)

km1

n!=k(k − 1) · · · (k −m+ 1)

km

m∑n=0

1

n!.

As k ≥ m we can let it tend to infinity with m still fixed and obtainthat:

e = limk→∞

(1 +

1

k

)k

≥ limk→∞

k(k − 1) · · · (k −m+ 1)

km

m∑n=0

1

n!=

(m∑

n=0

1

n!

)limk→∞

k(k − 1) · · · (k −m+ 1)

km=

m∑n=0

1

n!.

We now let m tend to infinity and obtain that

e ≥ limm→∞

m∑n=0

1

n!=

∞∑n=0

1

n!= E(1).

We conclude that E(1) = e.

10. From part 4 and induction it follows that for any integer n ≥ 0 wehave that E(nx) = E(x)n. The proof of part 5 implies that E(−x) =E(x)−1. Therefore, E(−nx) = E(−x)n = (E(x)−1)

n= E(x)−n. If

q = nm

with n,m ∈ Z,

E(x)n = E(nx) = E(mqx) = E(qx)m.


Because E(x) > 0 so is E(x)n and therefore we can take an m-th rootof it. We conclude that

E(qx) = (E(x)n)1m = E(x)

nm = E(x)q.

In particular, E(q) = E(1)q = eq.

Definition. We call E(x) the exponential function and we write ex =E(x). We call L(x), the inverse of ex on (0,∞), the logarithmic functionand write log(x) for it.

Corollary 7.22. For a > 0 and q ∈ Q we have that aq = eq log a.

Proof. We see that eq log a =(elog a

)q= aq.

Definition. Given a > 0 and x ∈ R we can now define ax = ex log a.

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Notes on Introductory Real Analysis