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Cairo UniversityFaculty of Sciences

Department of Mathematics

Principles of MathematicalAnalysis

M 232

Mostafa SABRI

Contents

Local Study of Functions 10.1 Reminders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Taylor-Young Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.3 Estimation of the remainder . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1 Numerical Sequences 71.1 Limits of numerical sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Basic properties of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Limits of real sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Some standard limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Subsequences and the Bolzano-Weierstrass Theorem . . . . . . . . . . . . . 141.6 Cauchy criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.7 Recursive sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Numerical Series 232.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Series with nonnegative terms . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3 General series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Summation by packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Rearrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.6 Double series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.7 Cauchy product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 Sequences and Series of Functions 353.1 Pointwise convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 Uniform convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.3 Conservation of properties by uniform convergence . . . . . . . . . . . . . . 383.4 Uniform continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.5 Approximation theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Power Series 494.1 Power series and radius of convergence . . . . . . . . . . . . . . . . . . . . . 494.2 Properties of the sum function . . . . . . . . . . . . . . . . . . . . . . . . . 514.3 Problems on the boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.4 Usual functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

iv Contents

4.5 Analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.6 Power series and differential equations . . . . . . . . . . . . . . . . . . . . . 574.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5 Improper Integrals 615.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.2 Basic properties of improper integrals . . . . . . . . . . . . . . . . . . . . . 625.3 Integrals of nonnegative functions . . . . . . . . . . . . . . . . . . . . . . . . 635.4 Absolute convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.5 Cauchy criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Bibliography 69

Local Study of Functions

0.1 Reminders

Lemma 0.1. Suppose that f : ]a, b[→ R attains a maximum or minimum at c ∈ ]a, b[. Iff is differentiable at c, then f ′(c) = 0.

Proof. Suppose f has a maximum at c and let g(x) = f(x)−f(c)x−c . Then

– if x > c, we have g(x) ≤ 0, hence f ′(c) = limx↓c g(x) ≤ 0,– if x < c, we have g(x) ≥ 0, hence f ′(c) = limx↑c g(x) ≥ 0.

Hence, f ′(c) = 0. If f has a minimum at c, then −f has a maximum at c, hence −f ′(c) = 0and thus f ′(c) = 0.

Theorem 0.2 (Rolle’s Theorem). Let f : [a, b]→ R be continuous on [a, b] and differen-tiable on ]a, b[. If f(a) = f(b) = 0, then there exists c ∈ ]a, b[ such that f ′(c) = 0.

Proof. Since |f | is continuous on [a, b], |f | attains its maximum at a point c ∈ [a, b]. If|f(c)| = 0, then |f(x)| = 0 for all x (because 0 ≤ |f(x)| ≤ |f(c)|), hence f is identicallyzero and the theorem is clear. If |f(c)| > 0, then c 6= a, b, moreover f has a maximum orminimum at c. Hence, f ′(c) = 0 by the previous lemma.

Theorem 0.3 (Mean Value Theorem). Let f : [a, b] → R be continuous on [a, b] anddifferentiable on ]a, b[. Then there exists c ∈ ]a, b[ such that f(b)− f(a) = f ′(c)(b− a).

Proof. Let g(x) = (b−a)f(x)+(f(a)−f(b))x+af(b)−bf(a). Then g′(x) = (b−a)f ′(x)+f(a)− f(b). Moreover, g(b) = g(a) = 0, so by Rolle’s Theorem, there exists c ∈ ]a, b[ suchthat g′(c) = 0, i.e. f(b)− f(a) = f ′(c)(b− a).

Corollary 0.4. Let f : [a, b]→ R be continuous on [a, b] and differentiable on ]a, b[. Then

(i) f is constant iff f ′ = 0,

(ii) f is increasing (resp. decreasing) iff f ′ ≥ 0 (resp. f ′ ≤ 0),

(iii) if f ′ > 0 (resp. f ′ < 0), then f is strictly increasing (resp. strictly decreasing).

Proof. If f = C and x ∈ ]a, b[, then f(x+h)−f(x)h = C−C

h = 0, so f ′(x) = 0. Conversely, iff ′(x) = 0 for all x ∈ ]a, b[, let y > a, then by Theorem 0.3, we may find c ∈ ]a, y[ such thatf(y)− f(a) = (y − a)f ′(c) = 0. Hence, f(y) = f(a) for all y ∈ [a, b] and f is constant.

If f is increasing, then f(x+h)−f(x)h ≥ 0, hence f ′(x) ≥ 0. Conversely, if f ′ ≥ 0, then

given a ≤ x < y ≤ b, we have f(y) − f(x) = (y − x)f ′(c) ≥ 0 (if f ′ > 0, we also havef(y)− f(x) > 0). The case of decreasing functions is similar.

2 Local Study of Functions

0.2 Taylor-Young Formula

Definition 0.5 (Landau notation). Let f , g be two functions defined near x0 ∈ R. Assumeg(x) 6= 0 for x 6= x0. We say that

1. f = O(g) near x0 if there exists M > 0 such that |f(x)| ≤M |g(x)| near x0,

2. f = o(g) near x0 if limx→x0f(x)g(x) = 0,

3. f ∼ g near x0 if limx→x0f(x)g(x) = 1.

When f = O(g), we say that f is dominated by g, when f = o(g), we say that f isnegligible in front of g, when f ∼ g, we say that f is equivalent to g.

Recall that f is differentiable at x0 iff limx→x0f(x)−f(x0)

x−x0exists; in this case we have

by definition limx→x0f(x)−f(x0)

x−x0= f ′(x0). Hence, f is differentiable at x0 iff

f(x) = f(x0) + f ′(x0)(x− x0) + o(x− x0) near x0 . (?)

This means that x 7→ f(x0) + f ′(x0)(x − x0) is the best linear approximation of f nearx0. The aim of the Taylor-Young formula is to give the best polynomial approximation off near x0.

Lemma 0.6. Suppose g is differentiable near 0 and satisfies g′(h) = o(hk) near 0. Theng(h)− g(0) = o(hk+1) near 0.

Proof. Let ε > 0. Since g′(h)/hk → 0 as h → 0, we may find h0 > 0 such that |g′(h)| ≤ε|h|k for h ∈ ]−h0, h0[ \ {0}. Let h ∈ ]−h0, h0[ \ {0}. Then by Theorem 0.3, we may findx between 0 and h such that |g(h) − g(0)| = |g′(x)||h − 0| ≤ ε|x|k|h| ≤ ε|h|k+1. Hence,g(h)−g(0)hk+1 → 0 as h→ 0.

Theorem 0.7 (Taylor-Young formula). Let f be defined near x0 and assume f is differ-entiable n times at x0. Then near x0 we have

f(x) = f(x0) + f ′(x0)(x−x0) + f ′′(x0)2! (x−x0)2 + . . .+ f (n)(x0)

n! (x−x0)n + o((x−x0)n

).

Proof. We prove the result by induction. When n = 1, the theorem is true by (?).Suppose the result is true for n − 1 and apply it to f ′, which is differentiable n − 1

times at x0. Then taking h = x− x0, we have

f ′(x0 + h) =n−1∑j=0

(f ′)(j)(x0)j! hj + o(hn−1) =

n−1∑j=0

f (j+1)(x0)j! hj + o(hn−1) .

Let g(h) = f(x0 + h)− f(x0)−∑nk=1

f (k)(x0)k! hk. Then

g′(h) = f ′(x0 + h)−n∑k=1

f (k)(x0)(k − 1)!h

k−1 = f ′(x0 + h)−n−1∑j=0

f (j+1)(x0)j! hj = o(hn−1) .

Since g(0) = 0, we have g(h) = o(hn) by Lemma 0.6, as asserted.

0.2. Taylor-Young Formula 3

Corollary 0.8. The following expansions are valid near 0 :

ex = 1 + x+ x2

2! + . . .+ xn

n! + o(xn)

cos(x) = 1− x2

2! + x4

4! + . . .+ (−1)n x2n

(2n)! + o(x2n)

sin(x) = x− x3

3! + x5

5! + . . .+ (−1)n x2n+1

(2n+ 1)! + o(x2n+1)

(1 + x)a = 1 + ax+ a(a− 1)2 x2 + . . .+ a(a− 1) · · · (a− n+ 1)

n! xn + o(xn)

ln(1 + x) = x− x2

2 + x3

3 + . . .+ (−1)n+1xn

n+ o(xn)

Remark 0.9. In Corollary 0.8,(i) we may replace o(xn), o(x2n) and o(x2n+1) by O(xn+1), O(x2n+2) and O(x2n+3),

respectively. For example, ex = 1 + x + . . . + xn

n! + xn+1

(n+1)! + o(xn+1). Since xn+1

(n+1)! +o(xn+1) = O(xn+1), we have the assertion.

(ii) We may remove o(xn), o(x2n) and o(x2n+1) from the expansions, but replace = by ∼.For example, ln(1 + x) ∼ x and ln(1 + x) ∼ x− x2

2 near 0. Indeed, limx→0ln(1+x)

x =

limx→0x+o(x)

x = limx→0(1 + o(x)x ) = 1, and limx→0

ln(1+x)x−x2

2= limx→0

x−x2

2 +o(x2)x−x2

2=

limx→0(1 + o(x)1−x2

) = 1, since o(x)→ 0 as x→ 0 (see Remark 5 below).

Remark 0.10. 1. If f = o(g), then f = O(g).2. If f ∼ g, then f = O(g) and g = O(f).3. If j ≥ k, then near 0, o(xj) + o(xk) = o(xk). In particular, o(x) + o(x2) = o(x).4. x · o(xk) = o(xk+1) and o(xk) = xko(1).5. If k ≥ 0, then limx→0 o(xk) = 0. In particular, limx→0 o(1) = 0.6. o(g)− o(g) = o(g) and O(g)−O(g) = O(g).7. If c ∈ R, then c · o(g) = o(g) and c ·O(g) = O(g). 1

Proof. 1. If f = o(g), then choosing ε = 1, we may find δ > 0 such that |f(x)g(x) | ≤ 1, i.e.

|f(x)| ≤ |g(x)| if |x− x0| ≤ δ.2. If f ∼ g, then choosing ε = 1

2 , we may find δ > 0 such that for |x−x0| ≤ δ, |f(x)g(x)−1| ≤ 1

2 ,i.e. 1

2 ≤f(x)g(x) ≤

32 , so

12 ≤ |

f(x)g(x) | ≤

32 and thus |f(x)| ≤ 3

2 |g(x)| and |g(x)| ≤ 2|f(x)|.

3. If f/xj → 0 and g/xk → 0, then (f + g)/xk = (f/xj)xj−k + g/xk → 0 as x→ 0.4. If f/xk → 0, then xf/xk+1 → 0. By induction we get o(xk) = xko(1).5. If g = o(xk), then limx→0 g(x) = limx→0(g(x)/xk)xk = 0.6. If f1/g → 0 and f2/g → 0, then (f1 − f2)/g → 0. If |f1| ≤ C1|g| and |f2| ≤ C2|g| nearx0, then |f1 − f2| ≤ |f1|+ |f2| ≤ (C1 + C2)|g| near x0.

7. If f/g → 0, then cf/g → 0. If |f | ≤ C|g|, then |cf | ≤ C ′|g|.

1. O(g) and o(g) are sets of functions, so the equality f = O(g) is not very precise (it should be f ∈ O(g)and f ∈ o(g)). This is why we get some strange rules in Remark 0.10. Note also that the equality sign isnot symmetric: 0 = o(g), but o(g) 6= 0. Still, these notations are very helpful in practice.

4 Local Study of Functions

Example 0.11. (1) limx→0cos(x)−1x sin(x) = −1

2 . Indeed, cos(x) − 1 = −x2/2 + o(x2) andx sin(x) = x2 + o(x2), hence cos(x)−1

x sin(x) = −1/2+o(1)1+o(1) → −1/2.

(2) limx→0ex−2

√1+x+1x2 = 3

4 . Indeed,√

1 + x = 1+x/2−x2/8+o(x2), hence ex−2√

1 + x+1 = (1 + x+ x2/2 + o(x2))− (2 + x− x2/4 + o(x2)) + 1 = 3

4x2 + o(x2).

(3) Expand f(x) = x2 sin(x)1+x to the order 5.

Let g(x) = x2 sin(x) and h(x) = 1/(1 + x). Then g(x) = x3 − x5/6 + o(x5) andh(x) = 1− x+ x2 − x3 + x4 − x5 + o(x5). Hence, using remarks 3 and 4,

f(x) = g(x)h(x) = (x3 − x5/6 + o(x5))(1− x+ x2 − x3 + x4 − x5 + o(x5))= x3 − x4 + 5x5/6 + o(x5) .

0.3 Estimation of the remainder

Theorem 0.12 (Taylor-Lagrange formula). Let f : [a, b]→ R be continuous on [a, b] anddifferentiable (n+ 1) times on ]a, b[. Then there exists c ∈ ]a, b[ such that

f(b) = f(a) + f ′(a)(b− a) + f ′′(a)2! (b− a)2 + . . .+ f (n)(a)

n! (b− a)n + f (n+1)(c)(n+ 1)! (b− a)n+1 .

Proof. Let g(x) = f(x)− f(b) +∑ni=1

(b−x)ii! f (i)(x). Then

g′(x) = f ′(x)−n∑i=1

(b− x)i−1

(i− 1)! f (i)(x) +n∑i=1

(b− x)i

i! f (i+1)(x) = f (n+1)(x)(b− x)n

n! .

Let λ = −g(a) (n+1)!(b−a)n+1 and h(x) = g(x) + λ (b−x)n+1

(n+1)! . Then h(a) = h(b) = 0, so by Rolle’sTheorem, we may find c ∈ ]a, b[ such that 0 = h′(c) = (b−c)n

n! (f (n+1)(c) − λ). Hence,λ = f (n+1)(c). Inserting this in the equation h(a) = 0, we obtain Taylor’s formula.

Theorem 0.13 (Taylor formula with integral remainder). Let f : [a, b] → R be continu-ously differentiable (n+ 1) times on ]a, b[. Then

f(b) = f(a) + (b− a)f ′(a) + . . .+ (b− a)n

n! f (n)(a) +∫ b

a

(b− t)n

n! f (n+1)(t) dt .

Proof. We prove the result by induction. For n = 0, the theorem is true by the fun-damental theorem of integration (f(b) = f(a) +

∫ ba f′(t) dt). Now integrate by parts

with u = f (n+1)(t) and v = −(b−t)n+1

(n+1)! to get∫ ba

(b−t)nn! f (n+1)(t) dt = (b−a)n+1

(n+1)! f(n+1)(a) +∫ b

a(b−t)n+1

(n+1)! f(n+2)(t) dt. Hence, if the result is true for n, it is also true for n+ 1.

0.4 Exercises

1. Expand f(x) = 11+x+x2

2to the order 2 near 0.

2. Expand f(x) = ex+10 to the order 5 near 0.3. Let g(x) = 1

cosx on]−π

2 ,π2[. Expand g to the order 4 near 0.

4. Expand tan x to the order 5 near 0.

0.4. Exercises 5

5. Expand f(x) = x(1 + cosx) − 2 tan x to the order 3 near 0. Deduce a function whichis equivalent to f .

6. Calculate the following limits. Here a, b > 0.

• limx→0

x(1 + cosx)− 2 tan x2x− sin x− tan x • lim

x→0

(1x− 1

ln(1 + x))

• limx→0

ax − bx

x• lim

x→0

(ax + bx

2)1/x

7. Give an example of functions f, g, h such that f ∼ g near 0, but the relation h+f ∼ h+gis not true. Show that if h/g has a limit, then h+ f ∼ h+ g.

Chapter 1

Numerical Sequences

1.1 Limits of numerical sequencesDefinition 1.1. A numerical sequence is a map u : N → K, where K = R or C. Wedenote un := u(n). The map itself is denoted by (un)n∈N.

If K = R, (un)n∈N is real sequence. If K = C, (un)n∈N is complex sequence.

More generally, a sequence may start from an index k0 ∈ N. In this case, we denote itby (un)n≥k0 . For simplicity, we shall often denote sequences by (un).

Definition 1.2. We say that a sequence (un) is bounded if

∃M > 0 : |un| ≤M ∀n ∈ N .

Definition 1.3. We say that ` ∈ C is a limit for a numerical sequence (un) if

∀ε > 0 ∃n0 ∈ N : n ≥ n0 =⇒ |un − `| ≤ ε .

If (un) has a limit, we say that (un) is convergent. Otherwise, we say that (un) is divergent.

Lemma 1.4 (Uniqueness of the limit). Any sequence has at most one limit.

Proof. Suppose on the contrary that a numerical sequence (un) has two limits ` 6= `′ andchoose ε := |`−`′|

3 > 0. Then there exists an index n1 such that |un − `| ≤ ε ∀n ≥ n1, andan index n2 such that |un − `′| ≤ ε ∀n ≥ n2. Let n3 = max(n1, n2). Then

|`− `′| = |(`− un3) + (un3 − `′)| ≤ |`− un3 |+ |un3 − `′| ≤ 2ε = 23 |`− `

′| .

This contradiction shows that ` = `′.

The unique limit of a sequence, when it exists, is denoted by limn→∞

un, or more simplyby lim un. If (un) has a limit `, we say it converges to ` and denote un → `.

Remark 1.5. Given two sequences u and v, if ∃n0 : un = vn ∀n ≥ n0, and if u has a limit` ∈ C, then v converges to the same `. In other words, if a sequence converges, and if wechange a finite number of its terms, then this does not change the limit.

Lemma 1.6. Any convergent sequence is bounded.

Proof. Suppose (un) converges to ` ∈ C. There there exists n0 ∈ N such that |un −`| ≤ 1 ∀n ≥ n0. Hence, |un| ≤ |un − `| + |`| ≤ 1 + |`| ∀n ≥ n0. Take M :=max{|u0|, |u1|, . . . , |un0−1|, 1 + |`|}. Then |un| ≤M for all n, hence (un) is bounded.

8 Chapter 1. Numerical Sequences

Lemma 1.7. If lim un = ` ∈ C and lim vn = `′ ∈ C, then for any c ∈ C,

(i) limn→+∞

cun = c`, (ii) limn→+∞

(un + vn) = `+ `′,

(iii) limn→+∞

unvn = ``′ .

Proof. (i) Given ε > 0, there exists n0 such that

n ≥ n0 =⇒ |cun − c`| = |c| · |un − `| ≤ |c| · ε′ ≤ ε ,

where ε′ = ε/|c| if c 6= 0 and ε′ = 1 if c = 0.(ii) Given ε > 0, we have

|(un + vn)− (`+ `′)| = |(un − `) + (vn − `′)| ≤ |un − `|+ |vn − `′| .

We know there exist n1, n2 ∈ N such that |un − `| ≤ ε2 ∀n ≥ n1 and |vn − `′| ≤ ε

2∀n ≥ n2. Take n3 = max(n1, n2). Then |(un + vn)− (`+ `′)| ≤ ε ∀n ≥ n3.

(iii) Given ε > 0, we have

|unvn − ``′| = |unvn + un`′ − un`′ − ``′|

= |un(vn − `′) + (un − `)`′| ≤ |un||vn − `′|+ |un − `||`′| .

Since (un) converges, it is bounded, so there exists M > 0 such that |un| ≤M for alln. On the other hand, we may find n1, n2 ∈ N such that |vn − `′| ≤ ε

2M for n ≥ n1and |un − `| ≤ ε′ for n ≥ n2, where ε′ = ε

2|`′| if `′ 6= 0 and ε = 1 if `′ = 0. Take

n3 = max(n1, n2). Then |unvn − ``′| ≤ ε ∀n ≥ n3.

Lemma 1.8. Suppose un → ` 6= 0. Then there exists k1 such that un 6= 0 for all n ≥ k1.Moreover, the sequence (1/un)n≥k1 converges to 1/`.

Proof. Given ε > 0, we have | 1un− 1

` | = |un−``un|. We start by giving a lower bound on the

denominator. We have |un| ≥ |`| − |un − `|. Choosing ε′ = |`|2 , we know that |un − `| ≤ ε′

starting from an index k1. Hence, |un| ≥ |`| − ε′ = ε′ and we get∣∣∣ 1un− 1`

∣∣∣ =∣∣∣un − ``un

∣∣∣ ≤ 1|`|ε′|un − `| for all n ≥ k1 .

But we may find k2 such that |un − `| ≤ ε′′ for all n ≥ k2, where ε′′ = |`|ε′ε. Hence,| 1un− 1

` | ≤ ε for all n ≥ max(n1, n2).

Lemma 1.9. If (un) is bounded and (vn) converges to 0, then (unvn) converges to 0.

Proof. Let ε > 0. We may find M > 0 such that |un| ≤ M for all n. On the other hand,choosing ε′ = ε

M , we may find n0 such that |vn| = |vn − 0| ≤ ε′ for all n ≥ n0. Hence,|unvn| ≤Mε′ = ε for all n ≥ n0.

Definition 1.10. Let z0 ∈ K, where K = R or C, let Br(z0) := {z ∈ K : |z − z0| < r}and let Br(z0) = {z ∈ K : |z − z0| ≤ r}. If K = R, then Br(z0) = ]z0 − r, z0 + r[ andBr(z0) = [z0 − r, z0 + r]. If K = C, then Br(z0) is an open disk of radius r > 0, centeredat z0, while Br(z0) is a closed disk.

Definition 1.11. Let f : Br(z0)→ C and p ∈ Br(z0). We say that limz→p f(z) = ` if

∀ε > 0 ∃δ > 0 : ∀z ∈ Br(z0), |z − p| ≤ δ =⇒ |f(z)− `| ≤ ε .

1.2. Basic properties of R 9

Lemma 1.12. Let f : Br(z0)→ C and p ∈ Br(z0). Then

limz→p

f(z) = ` ⇐⇒(∀ {un} ⊆ Br(z0) : un → p =⇒ f(un)→ `

).

Proof. Let ε > 0 and suppose limz→p f(z) = `. Then we may find δ > 0 such that|f(z)− `| ≤ ε for any z ∈ Br(z0) satisfying |z − p| ≤ δ. But if un → p, then ∃n0 such that|un − p| ≤ δ for all n ≥ n0. Hence, |f(un)− `| ≤ ε for all n ≥ n0 and thus f(un)→ `.

Conversely, suppose that limz→p f(z) 6= `. Then

∃ε > 0 ∀δ > 0 : ∃z ∈ Br(z0), |z − p| ≤ δ but |f(z)− `| > ε .

Hence, for δ = 2−n we may find un ∈ Br(z0) such that |un− p| ≤ 2−n but |f(un)− `| > ε.Hence, un → p but f(un) 9 `. The assertion follows by contraposition.

Example 1.13. limx→0 sin( 1x) does not exist. Indeed, if xn = 2

(2n+1)π , then xn → 0 butsin( 1

xn) = sin( (2n+1)π

2 ) = (−1)n has no limit. Hence, sin( 1x) has no limit by Lemma 1.12.

Definition 1.14. We say that f : Br(z0)→ C is continuous at p ∈ Br(z0) if

∀ε > 0 ∃δ > 0 : ∀z ∈ Br(z0), |z − p| < δ =⇒ |f(z)− f(p)| ≤ ε .

Lemma 1.15. Let f : Br(z0) → C and p ∈ Br(z0). Then f is continuous at p iff∀ {un} ⊆ Br(z0), if un → p, then f(un)→ f(p).

Proof. Comparing Definition 1.11 and Definition 1.14, we see that f is continuous at p ifflimz→p f(z) = f(p). Hence, the assertion follows from Lemma 1.12.

Example 1.16. If un → `, then |un| → |`|, since x 7→ |x| is a continuous function.

1.2 Basic properties of RDefinition 1.17. We say that r ∈ R is an upper bound for a set A ⊂ R if

r ≥ a ∀a ∈ A .

We say that r ∈ R is a least upper bound for A ⊂ R if� r is an upper bound for A,� any s < r is not an upper bound for A.

The least upper bound is also called the supremum of A and is denoted by r = sup(A).

Theorem 1.18. R has the least-upper-bound property. That is, if a nonempty A ⊂ R hasan upper bound, then it has a least upper bound.

Proof. Admitted.

Theorem 1.19. Given x > 0 and y ∈ R, there exists n ∈ N such that nx ≥ y.

We refer to this as the archimedean property of the reals.

Proof. Suppose on the contrary that nx < y for all n ∈ N. Then A := {nx : n ∈ N} is anonempty set of real numbers which has an upper bound y. Hence, it has a least upperbound r = sup(A). Since r − x < r, r − x is not an upper bound for A, so there existsm ∈ N such that r − x < mx. Hence, r < (m + 1)x, which is impossible, since r is anupper bound for A. This contradiction proves the theorem.


Corollary 1.20. Given x > 1 and y ∈ R, there exists n ∈ N such that xn ≥ y.

Proof. Let x = 1 + h, h > 0. Then by the binomial theorem, (1 + h)n ≥ 1 + nh. By thearchimedean property, we may find n such that nh ≥ y − 1. Hence, xn ≥ y.

Lemma 1.21. Given x ∈ R and a > 0, there exists a unique n ∈ Z such that

na ≤ x < (n+ 1)a .

Proof. ♦ Uniqueness : If n and n′ satisfy the assertion, then

n ≤ xa < n′ + 1 =⇒ n− n′ < 1 =⇒ n− n′ ≤ 0

n′ ≤ xa < n+ 1 =⇒ n′ − n < 1 =⇒ n′ − n ≤ 0

}=⇒ n = n′ .

♦ Existence : By the archimedean property, we may find n1, n2 ∈ N such that an1 ≥ xand an2 ≥ −x. Thus, A := {k ∈ Z : ka ≤ x} is a nonempty subset of Z (−n2 ∈ A)which has an upper bound n1. Hence, A has a greatest element n ∈ A. 1 Sincen+ 1 /∈ A, we get na ≤ x < (n+ 1)a.

The case a = 1 in the previous lemma is of special interest :

Definition 1.22. If x ∈ R, the unique n ∈ Z satisfying n ≤ x < n+ 1 is called the integerpart of x, and denoted by [x]. The real x − [x] is called the fractional part of x, and isdenoted by {x}.

1.3 Limits of real sequences

We are now interested in certain notions that use the natural order of the reals. Letus first note the following result.

Lemma 1.23. (i) If a real sequence (un) has a limit `, then ` ∈ R.(ii) A complex sequence (un) converges iff the sequences (Reun) and (Im un) converge.

Proof. (i) Put ` = a + ib, where a, b ∈ R. Then for any ε > 0 we have |un − `| ≤ εstarting from an index n0. Thus, |b| ≤

√(un0 − a)2 + b2 = |un0 − `| ≤ ε for any

ε > 0, which means that b = 0.(ii) If (Reun) and (Im un) converge, then (un) = (Reun+Im un) converge by Lemma 1.7.

Conversely, suppose un → ` ∈ C and put ` = a+ ib, a, b ∈ R. Then

|Reun − a| ≤√

(Reun − a)2 + (Im un − b)2 = |un − `| ,

so the convergence of (un) to ` implies the convergence of (Reun) to a. Similarly,the sequence (Im un) converges to b.

Lemma 1.24 (Passing to the limit). Let u and v be two real convergent sequences. Ifthere exists k1 ∈ N such that un ≥ vn for all n ≥ k1, then lim un ≥ lim vn.

In particular, taking the constant sequence v ≡ 0, if un ≥ 0 for all n, then lim un ≥ 0.

1. This follows from the well-ordering principle, which says that any nonempty subset of N has a leastelement.

1.3. Limits of real sequences 11

Proof. Suppose un → `1 and vn → `2. By hypothesis, we have un − vn = |un − vn| for alllarge n. Hence,

`1 − `2 = lim(un − vn) = lim |un − vn| = |`1 − `2| ≥ 0 .

Lemma 1.25. Let (un) be a complex sequence and ` ∈ C. If there exists a real sequence(vn) converging to 0 such that |un − `| ≤ vn for all n, then un → `.

Proof. Let ε > 0. Since vn → 0, there exists n0 such that |vn| ≤ ε for all n ≥ n0, and thus|un − `| ≤ vn ≤ ε for all n ≥ n0.

Example 1.26. Let f be a bounded function defined on X ⊆ R. If M = supX f , thenwe may find a sequence (xn) of elements of X such that f(xn)→M .

Indeed,M− 1n is not an upper bound, so there exists xn ∈ X such thatM− 1

n < f(xn).But M is an upper bound, so f(xn) ≤ M . Hence, − 1

n < f(xn) − M ≤ 0 and thus|f(xn)−M | ≤ 1

n . Hence, f(xn)→M by Lemma 1.25.

Corollary 1.27 (Sandwich Theorem). Let (un), (vn) and (wn) be real sequences. If thereexists k such that un ≤ vn ≤ wn for all n ≥ k, and if lim un = limwn = ` ∈ R, then (vn)converges, and lim vn = `.

Proof. We have hypothesis 0 ≤ vn−un ≤ wn−un for any n ≥ k. Since lim(wn−un) = 0,the sequence (vn − wn) tends to 0 by Lemma 1.25. Since vn = un + (vn − un), we havelim vn = `+ 0 = `.

Definition 1.28. We say that a real sequence (un)� tends to +∞ if

∀M ∈ R ∃n0 ∈ N : n ≥ n0 =⇒ un ≥M ,

� tends to −∞ if (−un) tends to +∞, i.e. if

∀M ∈ R ∃n0 ∈ N : n ≥ n0 =⇒ un ≤M .

Note that if u and v are real sequences, and if there exists k ∈ N such that un ≤ vnfor all n ≥ k, then

(u → +∞ =⇒ v → +∞

), and

(v → −∞ =⇒ u → −∞

). This is

clear from the definitions, and may be seen as an appendix to the sandwich theorem.

Lemma 1.29. Let (un) be a real sequence. Then un → +∞ iff ∃k1 ∈ N : un > 0 ∀n ≥ k1,and the sequence ( 1

un)n≥k1 tends to 0.

Proof. Suppose un → +∞ and let ε > 0. Then we may find k1 such that un ≥ 1/ε for alln ≥ k1. Thus, for n ≥ k1, we have un > 0 and 0 < 1

un≤ ε. Thus, 1

un→ 0.

Conversely, let M > 0. If 1un→ 0, then we may find n0 ≥ k1 such that 0 < 1

un≤ 1

M ,since un > 0. Hence, un ≥M and un → +∞.

Definition 1.30. We say that a real sequence (un) is� bounded from above if there exists M ∈ R such that un ≤M for all n,� bounded from below if there exists m ∈ R such that un ≥ m for all n,� increasing if un+1 − un ≥ 0 for all n,� decreasing if un+1 − un ≤ 0 for all n,� monotone if it is either increasing or decreasing.

We say it is strictly increasing, strictly decreasing or strictly monotone if the correspondinginequality is strict.


Theorem 1.31. Let u be an increasing real sequence.1. If it is bounded from above, then it converges to ` = sup{un : n ∈ N}.2. If it is not bounded from above, then it tends to +∞.If v is decreasing and bounded from below, then it converges to `′ = inf{vn : n ∈ N}.

Proof. 1. If u is bounded from above, then A := {un : n ∈ N} has a least upper bound `by Theorem 1.18. Let ε > 0. Then `− ε is not an upper bound, so ∃n0 ∈ N such that`− ε < un0 . Since u is increasing and bounded from above by `, we get

n ≥ n0 =⇒ `− ε < un0 ≤ un ≤ ` ,

hence un → `.2. If u is not bounded from above, then given M > 0, we may find n0 such that un0 > M .

Since u is increasing, we thus have un ≥ un0 > M for all n ≥ n0. Thus, un → +∞.Finally, if u = −v, then u is increasing and bounded from above, hence converges to

supu = sup(−v) = − inf v, so that v = −u converges to − supu = inf v.

Definition 1.32. We say that two real sequences u and v are adjacent if� ∀n ∈ N, un ≤ vn,� u is increasing and v is decreasing,� (vn − un) tends to 0.

Lemma 1.33. If u and v are two adjacent sequences, then they converge to a commonlimit ` satisfying un ≤ ` ≤ vn for all n.

Proof. We have un ≤ vn ≤ v0 for all n, so u is increasing and bounded from above. Hence,u converges and un ≤ lim u by Theorem 1.31. Similarly, v is decreasing and bounded frombelow, hence converges and we have vn ≥ lim v. Finally, v = u+(v−u) and lim(v−u) = 0,so lim u = lim v.

Definition 1.34. We say that (In) is a sequence of nested intervals if each In is an intervalof R and In+1 ⊆ In.

Corollary 1.35 (The Nested Interval Theorem). If ([an, bn]) is a sequence of nested closedintervals whose length tends to 0, then the set ∩n∈N [an, bn] consists of a single point.

Proof. By hypothesis, the sequences (an) and (bn) are adjacent, so they possess a commonlimit ` satisfying an ≤ ` ≤ bn for all n. The point ` thus belongs to each [an, bn], hencethe intersection of these intervals contains ` and is nonempty.

On the other hand, if x ∈ ∩n∈N [an, bn], then an ≤ x ≤ bn for all n, so passing to thelimit we get ` = lim an ≤ x ≤ lim bn = `. Hence, x = ` and ∩n∈N [an, bn] = `.

1.4 Some standard limitsDefinition 1.36. Let (un) and (vn) be two sequences of numbers. We say that1. un = O(vn) if there exists n0 ∈ N and C > 0 such that |un| ≤ C|vn| for all n ≥ n0.2. un = o(vn) if there exists a sequence wn → 0 such that un = wnvn.3. un ∼ vn if there exists a sequence wn → 1 such that un = wnvn.Note that if v is never 0, then un = o(vn) iff un

vn→ 0 and un ∼ vn iff un

vn→ 1.

1.4. Some standard limits 13

Remark 1.37. If un = o(vn), then un = O(vn). If un ∼ vn, then un = O(vn) andvn = O(un). This is proved as in Chapter 0.

Theorem 1.38. (a) If α > 0, then limn→∞1nα = 0.

(b) If a ∈ C and |a| < 1, then limn→∞ an = 0.

(c) If p > 0, then limn→∞ n√p = 1.

(d) lim n√n = 1.

(e) If a > 1 and α ∈ R, then limn→∞nα

an = 0.(f) If a > 1, then limn→∞ a

n = +∞.

(g) For any α, β > 0, we have limn→∞(lnn)αnβ

= 0.

Proof. (a) Let un = 1nα . Given ε > 0, we use the archimedean property to find n0 ∈ N

such that n0ε1/α ≥ 1. Then 0 ≤ un ≤ 1

nα0≤ ε for any n ≥ n0.

(b) If a = 0, this is clear, otherwise 1|a| > 1, so given ε > 0 we may use Corollary 1.20 to

find n0 such that ( 1|a|)

n0 ≥ 1ε . Hence, |an| ≤ |a|n0 ≤ ε for any n ≥ n0.

(c) If p = 1, this is clear. If p > 1, put xn = n√p − 1. Then xn > 0 and by the binomial

theorem, 1 + nxn ≤ (1 + xn)n = p. Thus, 0 < xn ≤ p−1n and xn → 0.

If 0 < p < 1, we obtain the assertion by considering q = 1p .

(d) Put xn = n√n−1. Then xn ≥ 0 and n = (1+xn)n ≥ n(n−1)

2 x2n. Hence, 0 ≤ xn ≤

√2

n−1for n ≥ 2. Hence, xn → 0.

(e) Put a = 1 + h, h > 0. Let k be an integer such that k > α and k > 0. Then forn > 2k, we have n = n

2 + n2 >

n2 + k, so by the binomial formula,

an = (1 + h)n >(n

k

)hk = n(n− 1) · · · (n− k + 1)

k! hk >nkhk

2kk! .

Thus, 0 < nα

an <2kk!hknα−k for any n > 2k. Since α− k < 0, we have nα−k → 0.

(f) This follows from (e) by taking α = 0 and applying Lemma 1.29.(g) We first show that (lnn)α ≤ Cnβ/2 for some C > 0. Given γ > 0, let f(x) =

γ−1xγ − ln x. Then f ′(x) = xγ−1 − 1x = xγ−1

x ≥ 0 for any x ≥ 1. Furthermore,f(1) = γ−1 > 0. Hence, f(x) > 0 for any x ≥ 1. Taking γ = β

2α , we thus get therequired inequality for any n ∈ N∗. The value of the limit now follows from (a).

Lemma 1.39 (Decimal approximation of a real number). For any x ∈ R, the sequenceun = [10nx]

10n converges to x.

In particular, any real number is the limit of a sequence of rationals. We say that Qis dense in R.

Proof. By definition, [10nx] ≤ 10nx < [10nx] + 1, hence un ≤ x < un + 110n . Thus,

0 ≤ x− un < 110n and thus un → x.

Example 1.40. If x =√

2, then u0 = 1, u1 = 1.4, u2 = 1.41 and u3 = 1.414.


1.5 Subsequences and the Bolzano-Weierstrass Theorem

Definition 1.41. A sequence (vn)n∈N is said to be a subsequence of (un)n∈N if there existsa strictly increasing map ϕ : N→ N such that vn = uϕ(n) for all n ∈ N.

Example 1.42. (un+1)n∈N, (u2n)n∈N and (u2n+1)n∈N are subsequences of (un)n∈N.

Note that if ϕ : N→ N is strictly increasing, then we see by induction that

ϕ(n) ≥ n ∀n ∈ N .

Lemma 1.43. If un → `, then any subsequence (vn) of (un) also converges to `.

Proof. Given ε > 0, we may find n0 such that |un− `| ≤ ε for any n ≥ n0. But vn = uϕ(n)for some strictly increasing ϕ : N→ N. Hence, ϕ(n) ≥ n and

n ≥ n0 =⇒ ϕ(n) ≥ n0 =⇒ |uϕ(n) − `| ≤ ε .

Example 1.44. 1. The sequence (−1)n is divergent, since the subsequence (u2n) tendsto 1, while the subsequence (u2n+1) tends to −1.

2. The sequence un = cos(nπ/4) is divergent, since the subsequence u4n = (−1)n diverges.

Lemma 1.45. If the subsequences (u2n) and (u2n+1) of u converge to a common limit `,then u also converges to `.

Proof. Given ε > 0, we may find n1, n2 ∈ N such that

n ≥ n1 =⇒ |u2n − `| ≤ ε and n ≥ n2 =⇒ |u2n+1 − `| ≤ ε .

Take n0 = max(2n1, 2n2 + 1). Then for n ≥ n0, we have |un − `| ≤ ε, hence un → `.

Lemma 1.46 (cf. [12]). Any real sequence has a monotone subsequence.

Proof. Let (un) be a real sequence. Call un0 dominant if un0 > un for all n > n0.If (un) has an infinite number of dominant terms with indices n0 < n1 < . . ., then the

subsequence (unj )j∈N is decreasing, since unk > unk+1 . So the result is true in this case.If (un) has a finite number of dominant terms, let uN be the last one and put m1 =

N+1. Then um1 is not dominant, so there existsm2 > m1 such that um1 ≤ um2 . Similarly,um2 is not dominant since m2 > N , so there exists m3 > m2 such that um2 ≤ um3 . Byinduction, there exists a subsequence um1 ≤ um2 ≤ um3 ≤ . . .. This completes theproof.

Corollary 1.47 (Bolzano-Weierstrass Theorem). Any bounded sequence has a convergentsubsequence.

Proof. 1. Let u be a real bounded sequence. By Lemma 1.46, it has a monotone subse-quence v, which is of course bounded too. Hence, v converges by Theorem 1.31.

2. Let u be a complex bounded sequence and put un = xn + iyn, with xn, yn ∈ R. Since|xn| ≤

√x2n + y2

n = |un|, the sequence (xn) is real and bounded, so it has a convergentsubsequence (xϕ(n)) with limit a. Similarly, |yϕ(n)| ≤ |uϕ(n)|, so (yϕ(n)) has a convergentsubsequence (yϕ(ψ(n))) with limit b. Finally, (xϕ(ψ(n))) is a subsequence of (xϕ(n)), soxϕ(ψ(n)) → a by Lemma 1.43. Thus, uϕ(ψ(n)) = xϕ(ψ(n)) + iyϕ(ψ(n)) → a+ ib.

1.6. Cauchy criterion 15

1.6 Cauchy criterionDefinition 1.48. We say that (un) is a Cauchy sequence if

∀ε > 0 ∃n0 ∈ N : p, q ≥ n0 =⇒ |up − uq| ≤ ε .

Remark 1.49. Equivalently, (un) is a Cauchy sequence if

∀ε > 0 ∃n0 ∈ N : r ≥ n0, s ≥ 0 =⇒ |ur − ur+s| ≤ ε .

Indeed, to go from the first definition to the second, take p = r and q = r + s. For theconverse, take r = min(p, q) and s = max(p, q)−min(p, q).

Theorem 1.50. A sequence converges iff it is a Cauchy sequence.

Proof. Suppose un → ` and let ε > 0. Choose n0 such that |un − `| ≤ ε/2 for all n ≥ n0.Then for p, q ≥ n0 we have

|up − uq| = |(up − `)− (uq − `)| ≤ |up − `|+ |uq − `| ≤ ε .

Conversely, suppose (un) is Cauchy. We first show that (un) is bounded. Taking ε = 1,we may find n0 such that |up − uq| ≤ 1 for all p, q ≥ n0. In particular, |un| ≤ 1 + |un0 | forany n ≥ n0. Put M = max{|u0|, . . . , |un0−1|, 1 + |un0 |}. Then |un| ≤ M for all n, hence(un) is bounded.

It follows from the Bolzano-Weierstrass theorem that (un) has a convergent subse-quence (uϕ(n)) with limit `. So ∃n1 such that |uϕ(n) − `| ≤ ε/2 for all n ≥ n1. But (un) isCauchy, so ∃n2 such that |up − uq| ≤ ε/2 for all p, q ≥ n2. Since ϕ(n) ≥ n, we thus havein particular |un − uϕ(n)| ≤ ε/2 for all n ≥ n2. Hence, if n3 = max(n1, n2), we have

n ≥ n3 =⇒ |un − `| ≤ |un − uϕ(n)|+ |uϕ(n) − `| ≤ ε ,

hence (un) converges to `.

1.7 Recursive sequencesIn the following, un : N→ K, where K = R or C.

1.7.1 Arithmetico-geometric sequences

We say that (un) is an arithmetico-geometric sequence if there exist a, b ∈ K such thatthe following recurrence relation is satisfied :

∀n ∈ N : un+1 = aun + b .

If a = 1, we have an arithmetic sequence, hence

∀n ∈ N : un = u0 + nb .

Suppose a 6= 1, we show that

∀n ∈ N : un = an(u0 − r) + r , where r = b

1− a .

Indeed, put vn = un−r. Then vn+1 = aun+b−r = avn+ar+b−r = avn. Hence, (vn) is ageometric sequence with common ratio a. Hence, un = vn+ r = anv0 + r = an(u0− r) + r.


If u0 = r, the sequence is thus constant and converges to r. If |a| < 1, we also haveun → r. If u0 6= r and |a| > 1, then |un| ≥ |a|n|u0 − r| − |r| → +∞, so (un) is divergent(if (un) was convergent, (|un|) would also be convergent). We will see in Exercise 7 thatif |a| = 1, then (an) converges iff a = 1.

Conclusion : if |a| < 1, then un → r. If |a| ≥ 1, then (un) converges iff (u0 = r) or(a = 1 and b = 0), in which case (un) is constant.

1.7.2 Linear recursive sequences of order 2

We say that (un) is a linear recursive sequence of order 2 if

∀n ∈ N : un+2 = aun+1 + bun (?)

for some a, b ∈ K.

Lemma 1.51. Let P (X) = X2 − aX − b.(a) If P has two distinct roots r1 and r2, then the solutions of (?) are the sequences

(αrn1 + βrn2 )n∈N, with α, β ∈ K.(b) If P has a double root r, then the solutions of (?) are the sequences ((α+ βn)rn)n∈N,

with α, β ∈ K.(c) If K = R and P has no real roots, there exist ρ > 0 and θ /∈ {kπ : k ∈ Z} such that

the solutions of (?) take the form (ρn(λ cos(nθ) + µ sin(nθ)))n∈N, with λ, µ ∈ R.

Proof. Let E = {(un)n∈N : (un)n∈N satisfies (?)}. Using elementary linear algebra, we seethat E is a vector space of dimension 2 (the map f : E → K2 given by f : (un)n∈N 7→(u0, u1) is an isomorphism).

Hence, there are exactly two linearly independent sequences which satisfy (?).In case (a), the sequences (rn1 ) and (rn2 ) both satisfy (?), and they are not proportional.

Hence, they form a basis for E, and this proves (a).In case (b), the sequence vn = rn satisfies (?). Let us show that wn = nrn also satisfies

(?). Since P has a double root r, the discriminant ∆ = a2 + 4b is zero and we have r = a2 .

Thus,

wn+2 = (n+ 2)rn · r2 = (n+ 2)rn(ar + b) = a(n+ 1)rn+1 + b(nrn) + arn+1 + 2brn

= awn+1 + bwn + (a2

2 + 2b)rn = awn+1 + bwn

because r = a2 and ∆ = 0. Hence, (vn) and (wn) form a basis for E, and this proves (b).

Finally, in case (c), P has roots in C of the form r± = ρe±iθ, with ρ > 0 and θ /∈{kπ : k ∈ Z}. Since the sequences (rn+) and (rn−) both satisfy (?), their linear combination(ρn cos(nθ)) and (ρn sin(nθ)) also satisfy (?). Moreover, they are real and form a basis forE. This yields (c).

1.7.3 General recursive sequences of order 1

We conclude this chapter with a study of real sequences (un)n∈N satisfying a recurrencerelation of the form

∀n ∈ N : un+1 = f(un) ,

where f : I → R is a function defined on an interval I ⊆ R, with u0 ∈ I. We also supposethat I is stable by f , i.e. f(I) ⊆ I, to ensure that the sequence is well defined (indeed, ifsay u1 = f(u0) /∈ I, then u2 = f(u1) is undefined).

1.7. Recursive sequences 17

Definition 1.52. We say that an interval I ⊆ R is closed if it has the form

I = [a, b] , I = [a,+∞[ , I = ]−∞, b] or I = R .

Lemma 1.53. If I is closed, un ∈ I ∀n and un → ` ∈ R, then ` ∈ I.

Proof. This is trivial if I = R. If I = [a,+∞[, then un ≥ a ∀n, hence lim un ≥ a byLemma 1.24 and thus ` ∈ I. The other cases are proved similarly.

Lemma 1.54. Let f : I → I be a continuous function on a closed interval I. If thesequence un+1 = f(un), u0 ∈ I has a limit `, then ` ∈ I and ` = f(`).

Thus, the limit of such a sequence must be a fixed point of f .

Proof. Since un ∈ I ∀n and I is closed, then ` ∈ I by Lemma 1.53.On the other hand, un → ` implies f(un)→ f(`) by Lemma 1.15, hence un+1 → f(`).

But (un+1)n∈N is a subsequence of (un)n∈N, so un+1 → `. The uniqueness of the limit nowshows that ` = f(`).

Example 1.55. If un+1 = u2n + 2, then un diverges, because the equation x = f(x) has

no solution. Indeed, ∀x ∈ R, x < x2 + 2. Taking x = un, we see that (un) is strictlyincreasing. Since it diverges, it tends to +∞.

Lemma 1.56. If there exists ` ∈ I and k ∈ [0, 1[ such that the map f : I → I satisfies

∀x ∈ I : |f(x)− `| ≤ k|x− `| ,

then given u0 ∈ I, the sequence un+1 = f(un) converges to `.

Proof. We see by induction that |un − `| ≤ kn|u0 − `|, so un → `, since |k| < 1.

Example 1.57. Let u0 ≥ 0. We define the sequence un+1 = 5un+3un+5 , whose terms are

nonnegative because the interval R+ is stable by f : x 7→ 5x+3x+5 .

The equation x = 5x+3x+5 has two roots ±

√3. Since all the un are nonnegative, the limit

of u must be nonnegative. Hence, if u converges, it must tend to√

3 by Lemma 1.54.Given x ≥ 0, the inequality

|f(x)−√

3| = |5x+ 3− x√

3− 5√

3|x+ 5 = |(x−

√3)(5−

√3)|

x+ 5 ≤(5−

√3

5)|x−

√3|

proves that un →√

3 by Lemma 1.56.

Definition 1.58. A map f : I → R is a contraction if it is k-Lipschitz continuous on I,with k ∈ [0, 1[. That is, if ∃k ∈ [0, 1[ such that ∀x, y ∈ I : |f(x)− f(y)| ≤ k|x− y|.

Corollary 1.59. If ` ∈ I is a fixed point of f and if f : I → I is a contraction, then thesequence converges to `.

Remark 1.60. If f is differentiable on I, and if |f ′(x)| ≤ k on I, where k ∈ [0, 1[, then fis a contraction on I by the mean value theorem.


Example 1.61. Let u0 = 0 and un+1 = 14 sin un + 1

2 . Since the interval [0, 1] is stable byf(x) = 1

4 sin x+ 12 , this defines a sequence of elements in [0, 1].

The function g(x) = x − f(x) satisfies g(0) < 0 and g(1) > 0. Since it is continuousand strictly increasing, there must exist a unique point ` such that g(`) = 0, i.e. ` = f(`).

The function f is 14 -Lipschitz continuous, since it is differentiable and |f ′(x)| ≤ 1

4 .Hence, the sequence converges to `.

We can actually estimate the speed of convergence :

|un − `| ≤(1

4)n|u0 − `| ≤

(14)n.

Sometimes the preceding criteria are not sufficient to conclude. We can then try tostudy the monotonicity of the sequence. For instance, if we show that (un) is increasingand bounded from above, then we can conclude that it converges (Theorem 1.31). Wehave the following criteria.

Lemma 1.62. If f(x) − x ≥ 0 ∀x ∈ I, then the sequence un+1 = f(un) is increasing. Iff(x)− x ≤ 0 ∀x ∈ I, then the sequence is decreasing.

Proof. We have un+1 − un = f(un)− un.

Lemma 1.63. If f is increasing, then the sequence un+1 = f(un) is monotone.

Proof. If u0 ≤ u1, then u1 = f(u0) ≤ f(u1) = u2 and by induction, un ≤ un+1 ∀n.Similarly, if u0 ≥ u1, then (un) is decreasing.

Remark 1.64. If f is increasing, we can use the fixed points of f to bound the sequence.For instance, if f(a) = a and un ≤ a, then un+1 = f(un) ≤ f(a) = a.

Example 1.65. Let u0 ≥ 0 and un+1 = u2n−un+1. Since f(x) = x2−x+1 = (x− 1

2)2 + 34 ,

we see that R+ is stable by f . The equation f(x) = x has a unique solution x = 1.The estimate |f(x)− 1| = |x||x− 1| gives no information. So we will argue by mono-

tonicity.We have ∀x ∈ R+, f(x) ≥ x, so the sequence (un) is increasing. Hence, if u0 > 1,

the sequence cannot converge to 1, the only fixed point of f , so it diverges to +∞. Ifu0 ≤ 1, the sequence is increasing and bounded above by 1 (use Remark 1.64). Hence, thesequence converges, and its limit must be 1.

We conclude this chapter with an important theorem. In contrast to Corollary 1.59,this theorem proves the existence of a fixed point under certain assumptions.

Theorem 1.66 (The Fixed Point Theorem 2, cf. [13]). If f : I → I is a contraction ona closed interval I, then f has a unique fixed point ` ∈ I. Moreover, for any c ∈ I, thesequence u0 = c, un+1 = f(un) converges to `. The speed of convergence may be estimatedby

|un − `| ≤kn

1− k |u1 − u0| ,

where k ∈ [0, 1[ is the Lipschitz constant of f .

2. of Banach or Picard.

1.8. Exercises 19

Proof. For any x, y ∈ I we have

|x− y| ≤ |x− f(x)|+ |f(x)− f(y)|+ |f(y)− y|≤ |x− f(x)|+ k|x− y|+ |f(y)− y|,

and thus|x− y| ≤ |f(x)− x|+ |f(y)− y|

1− k . (∗)

This already shows the uniqueness of the eventual fixed point : if x and y are two fixedpoints, then |x− y| = 0.

To prove its existence, let fn = f ◦ · · · ◦ f be the map f composed n times withitself, with the convention f0(x) = x. Let c ∈ I. We may write the sequence u0 = c,un+1 = f(un) in the form (fn(c))n∈N. Let us show that this is a Cauchy sequence. It willthen follow that it converges to a fixed point ` ∈ I by Lemma 1.54, which is the requiredresult.

We first note by induction that |fn(x) − fn(y)| ≤ kn|x − y| for any x, y ∈ I. Takingx = fn(c) and y = fm(c) in (∗), we thus get

|fn(c)− fm(c)| ≤ |f(fn(c))− fn(c)|+ |f(fm(c))− fm(c)|1− k

= |fn(f(c))− fn(c)|+ |fm(f(c))− fm(c)|

1− k

≤ kn|f(c)− c|+ km|f(c)− c|1− k = kn + km

1− k |u1 − u0| .

Since k ∈ [0, 1[, this tends to 0 as n,m→ +∞, hence (fn(c))n∈N is a Cauchy sequence.Keeping n fixed and lettingm→ +∞, we obtain the estimate on the speed of convergence.

1.8 Exercises

1. Calculate the limits of the following sequences, if they exist :

• 3n − (−2)n

3n + (−2)n •√n2 + n+ 1−

√n2 − n+ 1 • n−

√n2 + 1

n+√n2 − 1

•(

1 + 1n

)n• sin

1n

( 1n

)• sinn

n+ (−1)n

• n!nn

• en

nn•

(2 + (−1)n

) 1n

• (1 + 2n + 3n + 7n)1/n • n√n! • sin(n).

2. Let u = (un)n∈N, v = (vn)n∈N be two real sequences. Are the following assertionscorrect ? Justify your answer.(a) If u is increasing and bounded from above, then it converges.(b) If u is bounded from above and convergent, then it is increasing.(c) If u converges to l ∈ R, then it is bounded.(d) If u is decreasing and nonnegative, then it converges.(e) If u is decreasing and strictly positive, then it converges to a strictly positive limit.


(f) If u diverges to −∞ and v diverges to +∞, then u + v converges to 0.(g) If u tends to 0 and the un are nonzero, then 1/u tends to +∞ or −∞.(h) If u and v diverge to −∞, then u + v diverges to −∞.(i) If u converges and v diverges to +∞, then uv diverges.(j) If u tends to 0, then uv tends to 0.(k) If u is increasing and v converges to 0, then u− v is increasing.(l) If u tends to +∞ and v is bounded from below, then u + v tends to +∞.

(m) If u converges and v does not converge, then uv does not converge.(n) If u tends to +∞ and v is bounded from below beyond a certain index by a strictly

positive real number, then uv tends to +∞.3. Let (un)n≥1, (vn)n≥1 be two real sequences defined by

un := 1 + 11! + 1

2! + . . .+ 1n! and vn := un + 1

n · n! .

Show that (un)n and (vn)n are adjacent. Deduce that they converge to a commonlimit. We will admit that this limit is the real number e.We would now like to prove that e /∈ Q. For this, show that we can write un = an

n! withan ∈ N and deduce the inequality

an < en! < an + 1n

for all n ≥ 1.

Suppose that e = pq ∈ Q and deduce a contradiction by choosing n = q in the previous

inequality.4. Let u = (un)n∈N, v = (vn)n∈N be two real sequences which do not vanish beyond a

certain index. Are the following asserions correct ? Justify your answer.(a) If un ∼ vn, then for all k ∈ Z, we have ukn ∼ vkn ;(b) If un ∼ vn, then eun ∼ evn ;(c) If un ∼ vn, then ln un ∼ ln vn ;(d) If un = 1 + o(1), then eun ∼ e ;(e) If un ∼ vn, then |un| ∼ |vn| ;(f) If un ∼ vn, then unn ∼ vnn ;(g) If un = o(vn), then un = o(v2

n).5. Let a and b be strictly positive, a ≤ b, and let (un)n∈N and (vn)n∈N be two sequences

defined by

u0 = a and v0 = b

un+1 = 2unvnun + vn

and vn+1 = un + vn2 .

(1) Show that for all n ∈ N, un ≤ vn.(2) Show that (un)n and (vn)n converge and that their limits are equal.(3) Using the product unvn, determine the value of this limit.(4) Application: give rational approximations to

√2 and

√3.

1.8. Exercises 21

6. (Cesàro means). Let (un)n≥1 be a numerical sequence and (vn)n≥1 the sequence ofCesàro means defined by

vn := 1n

n∑k=1

uk .

(1) Show that if un → 0, then vn → 0.(2) Deduce that if un → l ∈ C then vn → l.(3) Does the convergence of vn imply the convergence of un?(4) Application: Let (xn)n≥1 be a sequence with strictly positive terms. Suppose the

sequence yn = xn+1xn

converges to l > 0. Show that the sequence (zn)n≥1 defined byzn = n

√xn converges to l.

(5) Application: Determine the limit of the sequence(2nn

)1/n.(6) Give an example of a sequence (xn)n≥1 such that n

√xn → l but the limit of xn+1

xndoes not exist.

7. Let z be a complex number with |z| = 1 such that the sequence (zn)n≥0 converge.Determine the limit of this sequence.Hint. You may use Exercise 6.

8. Let (un) be a real sequence such that, for all p ∈ N,

limn→∞

(un+p − un) = 0.

Does the sequence (un) converge ?9. Study the convergence of the following sequences, and give their limits if possible.

• u0 = 1 and un+1 =√

2un • u0 ≥ 0 and un+1 = un + 2un + 1

• u0 = 0 and un+1 =√

2− un • u0 = 0 and un+1 = 3un − 22un − 1

• u0, u1 and un+2 =√

2un+1 − un • u0 = 0, u1 = 1 and un+2 = 4un+1 − 4un + 2

• u0 = 13 and un+1 = 1− un

3 exp(un)

10. Let (un) be a real sequence and put

Sn =n∑k=1

uk and Tn =n∑k=1

ukk.

(1) Show that if Sn converges, then un → 0.(2) Show that if Tn converges, then un tends to 0 in the sense of Cesàro, i.e. 1

n

∑nk=1 uk →

0. Hint. Express uk as a function of Tj then use Exercise 6.11. Let n ∈ N and let fn : [0, 1]→ R be the function defined by fn(x) := xn + x2 + x− 1.

(1) Fix n ∈ N. Show that fn(x) is a continuous, strictly increasing function.(2) Deduce that fn has exactly one zero. We denote xn ∈ [0, 1] the zero of fn.(3) Fix x ∈ [0, 1]. Show that the sequence (fn(x))n is decreasing. Deduce that the

sequence (xn) is increasing.(4) Show that (xn) converges. We denote its limit by l.


(5) Show that for all n ∈ N, fn(

34

)> 0. Deduce that xn < 3

4 and that xnn → 0 asn→∞.

(6) Conclude that l2 + l − 1 = 0. What is the value of l ?12. (Fekete’s Lemma). Let (un)n≥0 be a real sequence satisfying un+m ≤ un + um for any

n,m ∈ N. Show that (unn )n≥1 tends to ` = infn≥1

unn ∈ R ∪ {−∞}.

13. Let (un) be a nonnegative, decreasing sequence that tends to 0. We put

vn :=( n∑k=1

uk)− nun.

(a) Let p ∈ N. Show that for all n ≥ p, we have vn ≥∑pk=1(uk − un).

(b) Deduce that if (vn) is a bounded sequence, then the sequence Sm =∑mk=1 uk

converges.(c) Study the converse.

14. Let (un)n be a real sequence. We say that ` ∈ R is a limit point of the sequence if thereexists a subsequence of (un) which converges to `.a) What are the limit points of the sequence un = (−1)n? the sequence un = sin(nπ3 )?b) Show that if a sequence converges, then it has a unique limit point. Is the converse

true ?c) Show that every bounded real sequence which has a unique limit point is convergent.d) Application: Let (un)n be a bounded real sequence such that un + u2n

2 converges.Show that (un)n converges.

e) Application: Let (an) and (bn) be two real sequences such that an + bn → 0 andean + ebn → 2. Show that the two sequences converge.

Chapter 2

Numerical Series

2.1 Generalities

Definition 2.1. Given a sequence (un), we call∑∞n=0 un = u0 + u1 + . . . a series with

general term un. We call Sk :=∑kn=0 un the partial sums of the series. We say that the

series converges if the sequence Sk converges. In this case, S = limk→∞ Sk =∑∞n=0 un is

called the sum of the series. Rn = S − Sn =∑∞k=n+1 uk is called the remainder of the

series.

Example 2.2. The geometric series∑λn converges iff |λ| < 1. Indeed, if λ = 1, the

series diverges. If λ 6= 1, we have Sk = 1 + λ+ . . .+ λk and λSk = λ+ . . .+ λk+1. Hence,Sk − λSk = 1 − λk+1, i.e. Sk = 1−λk+1

1−λ . We see that (Sk) converges iff |λ| < 1. In thiscase,

∑∞n=0 λ

n = limk→∞ Sk = 11−λ .

Remarks 2.3. � If∑un and

∑vn converge, then λ

∑un + µ

∑vn :=

∑(λun + µvn)

converges. The converse is clearly not true.� The convergence of a series

∑un is unchanged if we modify a finite number of un.

Lemma 2.4. If∑un converges, then un → 0.

Proof. If n ≥ 1, we have un = Sn − Sn−1. Hence, Sn → S =⇒ un → S − S = 0.

The converse is not true. For example, un = 1n → 0. However, using the identity

ln(1 + x) ≤ x with x = 1n , we have

Sn =n∑k=1

1k≥

n∑k=1

ln(k + 1

k

)= ln(n+ 1) ,

which tends to ∞ as n→∞.

Lemma 2.5. If∑un converges, then Rn → 0.

Proof. Assume Sn → S. Then Rn = S − Sn → S − S = 0.

Example 2.6.∑n≥1

1n(n+1) converges. Indeed, 1

n(n+1) = 1n −

1n+1 , hence

∑kn=1

1n(n+1) =

(1− 12) + (1

2 −13) + . . .+ ( 1

k −1

k+1) = 1− 1k+1 → 1 as k →∞.

Such series of the form∑

(vn − vn+1) are called telescoping series. In this case,∑kn=0(vn − vn+1) = v0 − vk+1.

24 Chapter 2. Numerical Series

2.2 Series with nonnegative termsLemma 2.7. Let

∑un be a series with un ∈ R+. Then

∑un converges iff (Sn) is bounded.

Proof. If un ≥ 0 for all n, then Sn is an increasing sequence of real numbers, whichconverges iff it is bounded.

Theorem 2.8 (Comparison Test). Let∑un and

∑vn be two series with nonnegative

terms.(1) Suppose un = O(vn). Then the convergence of

∑vn implies the convergence of

∑un,

and the divergence of∑un implies the divergence of

∑vn.

(2) Suppose un = o(vn). Then the convergence of∑vn implies the convergence of

∑un,

and the divergence of∑un implies the divergence of

∑vn.

(3) Suppose un ∼ vn. Then∑un converges iff

∑vn converges.

Proof. If un = O(vn), then there exists n0 ∈ N and C > 0 such that 0 ≤ un ≤ C vn for alln ≥ n0. Let Sn(u) =

∑nk=0 un and Sn(v) =

∑nk=0 vn. Suppose that

∑vn converges and

let Sv =∑∞n=0 vn. Then for n > n0, we have

Sn(u) = Sn0(u) +n∑

k=n0+1un ≤ Sn0(u) + C

n∑k=n0+1

vn ≤ Sn0(u) + C · Sv .

The RHS does not depend on n. Hence, (Sn(u)) is bounded and∑un converges by

Lemma 2.7.If un = O(vn) and

∑un does not converge, then

∑vn does not converge by contrapo-

sition.We thus proved (1). Result (2) follows because un = o(vn) implies un = O(vn). Result

(3) follows because un ∼ vn implies un = O(vn) and vn = O(un).

Example 2.9. 1. If∑un and

∑vn are two convergent series with nonnegative terms,

then∑

(max(un, vn)) converges. Indeed, max(un, vn) ≤ un + vn.2. Under the same hypotheses, if 0 ≤ α, β ≤ 1 are such that α + β = 1, then

∑uαnv

βn

converges. Indeed, uαnvβn ≤ max(un, vn)α max(un, vn)β = max(un, vn).

To get more concrete applications for the comparison test, we first need to have goodreference series with which to compare. A fundamental one is the following.

Theorem 2.10 (Riemann’s Rule). Let α ∈ R. Then∑n≥1

1nα converges iff α > 1.

Proof. We already saw that∑ 1

n diverges in the counter-example of Lemma 2.4.Suppose α 6= 1. Then 1

nα ∼1

α−1( 1nα−1 − 1

(n+1)α−1 ). Indeed,

nα

α− 1( 1nα−1 −

1(n+ 1)α−1

)= n

α− 1(1−

( n

n+ 1)α−1) = n

α− 1(1−

(1− 1

n+ 1)α−1)

= n

α− 1(1−

(1− α− 1

n+ 1 + o( 1n+ 1)

))= n

n+ 1 + n

n+ 1o(1)→ 1 .

But∑

( 1nα−1 − 1

(n+1)α−1 ) is a telescoping series, with∑kn=1( 1

nα−1 − 1(n+1)α−1 ) = 1− 1

(k+1)α−1 ,which converges iff α > 1. Hence,

∑n≥1

1nα converges iff α > 1.

2.2. Series with nonnegative terms 25

Here are two other proofs of the divergence of∑ 1

n :1) 1

n ∼ ln(n + 1) − ln(n). Indeed, ln(n+1)−ln(n)1/n = ln(1+1/n)

1/n = 1/n+o(1/n)1/n = 1 + o(1) → 1.

But∑(

ln(n + 1) − ln(n))is a telescoping series with

∑kn=1

(ln(n + 1) − ln(n)

)=

ln(k + 1)− ln(1)→∞ as k →∞, hence∑ 1

n diverges.2) Let Sk =

∑kn=1

1n . Then S2k−Sk = 1

k+1 + . . .+ 12k . But

1j ≥

12k for each k+1 ≤ j ≤ 2k.

Hence, S2k−Sk ≥ 12k + . . .+ 1

2k = k 12k = 1

2 . But if Sk → S, then S2k → S because (S2k)is a subsequence of Sk, so that S2k − Sk → 0, a contradiction. Hence,

∑ 1n diverges.

We now make an important remark :

Lemma 2.11. Let∑un be a series with nonnegative terms.

(i) If n√un ≥ 1 for infinitely many n, then

∑un diverges.

(ii) If there exists n0 and 0 ≤ λ < 1 such that n√un ≤ λ for all n ≥ n0, then

∑un

converges.

Proof. In case (i), un 9 0, so∑un diverges by Lemma 2.4.

In case (ii), un ≤ λn for all n ≥ n0, i.e. un = O(λn). But∑λn is a geometric series,

which converges because |λ| < 1, so∑un converges.

Corollary 2.12 (Root test). Let∑un be a series with nonnegative terms. Suppose that

n√un → µ 6= 1. Then

∑un converges if µ < 1, and

∑un diverges if µ > 1.

Proof. If µ < 1, then for ε = 1−µ2 , we may find n0 such that n

√un − µ ≤ ε ∀n ≥ n0, hence

n√un ≤ µ+ ε = 1+µ

2 < 1 ∀n ≥ n0, so∑un converges by Lemma 2.11.

If µ > 1, then ∃n1 ∈ N with n√un ≥ 1 ∀n ≥ n1, so un 9 0 and

∑un diverges.

Note that the root test fails when µ = 1. For example, n

√1n = 1

n√n → 1 and n

√1n2 =

( 1n√n)2 → 1, but

∑ 1n diverges while

∑ 1n2 converges.

Example 2.13. Study the convergence of∑n≥1 un, where un =

(1 + a2

n

)−n2

.

Solution : ln un = −n2 ln(1 + a2

n

)∼ −n2(a2

n ) = −na2. Hence, 1n ln un → −a2 and

thus n√un = eln n

√un → e−a

2< 1. Hence,

∑un converges.

Here is another criterion :

Lemma 2.14. Suppose there exists n0 such that for all n ≥ n0, we have un > 0, vn > 0and un+1

un≤ vn+1

vn. Then the convergence of

∑vn implies the convergence of

∑un, and the

divergence of∑un implies the divergence of

∑vn.

Proof. For n ≥ n0 we have un+1vn+1

≤ unvn, hence (unvn ) is decreasing. In particular, unvn ≤

un0vn0

and un ≤un0vn0

vn ∀n ≥ n0. Thus, un = O(vn) and the result follows from the comparisontest.

Corollary 2.15 (Ratio test). Let∑un be a series with strictly positive terms. Suppose

that un+1un→ λ 6= 1. Then

∑un converges if λ < 1, and

∑un diverges if λ > 1.

Proof. If λ < 1, then for ε = 1−λ2 , we may find n0 such that un+1

un− λ ≤ ε ∀n ≥ n0, hence

un+1un≤ λ+ ε = 1+λ

2 < 1. Let µ = 1+λ2 and vn = µn. Then

∑vn converges and vn+1

vn= µ.

Hence, un+1un≤ vn+1

vn∀n ≥ n0 and

∑un converges by Lemma 2.14.

If λ > 1, then for ε = λ−12 , we may find n1 such that un+1

un− λ ≥ −ε ∀n ≥ n0, hence

un+1un≥ λ − ε = 1+λ

2 > 1. In particular, un ≥ un0 for all n ≥ n0. But∑un0 = ∞, hence∑

un diverges by the comparison test.


Again, the ratio test fails if λ = 1.

Example 2.16. Study the convergence of∑un, where un = 1

(2n)!

(∏nk=1 (a+ k)2

), a ∈ R.

Solution : If a = −m for some m ∈ N∗, then un = 0 for all n ≥ m, hence∑un

converges and equals∑mn=0 un.

Otherwise, un+1un

= (a+n+1)2

(2n+1)(2n+2) →14 , hence

∑un is also convergent.

Theorem 2.17 (Integral test). Let f : [1,∞[ → R+ be continuous and decreasing. Putun = f(n). Then

∑n≥1 un converges iff the sequence

( ∫ n1 f(t) dt

)converges.

Proof. Since f decreases, we have∫ k+1k f(t) dt ≤ f(k) for all k ≥ 1, and

∫ kk−1 f(t) dt ≥ f(k)

for all k ≥ 2. Hence,∫ n+1

1f(t) dt =

n∑k=1

∫ k+1

kf(t) dt ≤

n∑k=1

f(k) ≤ f(1) +n∑k=2

∫ k

k−1f(t) dt = f(1) +

∫ n

1f(t) dt .

Hence, if∫ n

1 f(t) dt converges, Sn =∑nk=1 f(k) is bounded, hence

∑f(n) converges.

If∫ n

1 f(t) dt diverges, then∫ n+1

1 f(t) dt→∞, hence Sn →∞ and∑f(n) diverges.

Remarks 2.18. 1. In the previous proof, we established that∫ n+1

1f(t) dt ≤

n∑k=1

f(k) ≤ f(1) +∫ n

1f(t) dt.

Here is an application : study the convergence of an = 1√n

∑nk=1

1√k.

Solution : let f(x) = 1√x. Then

∫ n+1

1

1√t

dt = 2√t∣∣n+11 ≤

n∑k=1

1√k≤ 1 +

∫ n

1

1√k

= 1 + (2√t∣∣n1 ) .

Thus, 2(√n+1−1)√n

≤ un ≤ 1+2(√n−1)√n

and un → 2 by the Sandwich theorem.2. In case of convergence, if we sum instead over k ≥ n + 1 in the previous proof, we

obtain ∫ ∞n+1

f(t) dt ≤∞∑

k=n+1f(k) ≤

∫ ∞n

f(t) dt .

Here is an application : estimate the remainder of the series∑n≥1

1nα when α > 1.

Solution : let f(x) = 1xα and Rn =

∑∞k=n+1

1nα . Then

−1α− 1

1xα−1

∣∣∣∞n+1≤ Rn ≤

−1α− 1

1xα−1

∣∣∣∞n.

Thus, 1α−1

1(n+1)α−1 ≤ Rn ≤ 1

α−11

nα−1 , and Rn ∼ 1α−1

1nα−1 .

2.3 General seriesTheorem 2.19 (Cauchy criterion for series). A series of numbers

∑un converges iff it

satisfies the Cauchy criterion :

∀ε > 0 ∃N ∈ N : n ≥ N, p ≥ 0 =⇒ |Sn+p − Sn| = |un+1 + . . .+ un+p| ≤ ε .

2.3. General series 27

Proof. This follows immediately from the Cauchy criterion for sequences, since∑un con-

verges iff (Sn) converges.

Definition 2.20. A series of numbers∑un is absolutely convergent if

∑|un| converges.

Theorem 2.21. If∑un is absolutely convergent, then

∑un is convergent.

Proof. Suppose∑|un| converges and let ε > 0. By the Cauchy criterion, we may find N ∈

N such that∑n+pk=n+1 |uk| ≤ ε for all n ≥ N , p ≥ 0. Hence, |

∑n+pk=n+1 uk| ≤

∑n+pk=n+1 |uk| ≤ ε.

Hence,∑un converges by the Cauchy criterion.

Example 2.22.∑ sin(

√3n)

3n is absolutely convergent.If un = (−1)n( 1√

n+ 1√

n+1), then∑n≥1 un is not absolutely convergent, since |un| ∼

2√n. However,

∑un is convergent. Indeed, un = vn − vn+1, where vn = (−1)n√

n. Hence,∑n

k=1 uk =∑nk=1(vk − vk+1) = v1 − vn+1 = −1 + (−1)n√

n+1 → −1.

Remark 2.23. The comparison test of Section 2.2 fails when the reference sequencechanges sign. For example, if un = (−1)n( 1√

n+ 1√

n+1), then 1n = o(un). However,

∑un

converges, but∑ 1

n diverges.

Definition 2.24. A real series∑un is alternating if (−1)nun has a fixed sign.

For example,∑ (−1)n

n is alternating, since (−1)n (−1)nn = 1

n is always positive.

Theorem 2.25 (Aternating series criterion). Suppose∑un =

∑(−1)nvn is an alternating

series such that (vn) is nonnegative, decreasing and convergent to 0. Then∑un converges,

and its sum S satisfies S2n+1 ≤ S ≤ S2n for all n. Moreover,

|Rn| =

∣∣∣∣∣∣∞∑

k=n+1uk

∣∣∣∣∣∣ ≤ vn+1 .

Proof. We prove that S2n+1 and S2n are adjacent. We have

S2n+1 = S2n−1 + u2n + u2n+1 = S2n−1 + v2n − v2n+1 ≥ S2n−1,

S2n+2 = S2n + u2n+1 + u2n+2 = S2n − v2n+1 + v2n+2 ≤ S2n .

Thus, (S2n+1) is increasing, (S2n) is decreasing and S2n − S2n+1 = −u2n+1 = v2n+1 → 0.Thus, (S2n+1) and (S2n) converge to a common limit S satisfying S2n+1 ≤ S ≤ S2n for alln. Hence, 0 ≤ S2n − S ≤ S2n − S2n+1 = v2n+1, and 0 ≤ S − S2n−1 ≤ S2n − S2n−1 = v2n,which yields |R2n| ≤ v2n+1 and |R2n−1| ≤ v2n.

Remark 2.26. If un = (−1)n−1vn with (vn) nonnegative, decreasing and converging to0, then

∑un is also convergent, but now S2n ≤ S ≤ S2n+1. The estimate on Rn is still

valid.

Example 2.27. If α > 0, then∑ (−1)n

nα converges and |∑∞k=n+1

(−1)kkα | ≤

1(n+1)α .

In particular,∑ (−1)n√

nconverges.

If un = (−1)n√n+(−1)n , then un ∼

(−1)n√n

. However,∑un diverges. Indeed,

un = (−1)n√n

( 11 + (−1)n√

n

)= (−1)n√

n

(1− (−1)n√

n+ O( 1

n))

= (−1)n√n− 1n

+ O( 1n√n

) .


The first term gives a convergent series, the third one also yields a convergent series byRiemann’s rule. Hence

∑un diverges due to the divergence of

∑ 1n .

This example shows again that the comparison test fails when the reference sequence(here (−1)n√

n) changes its sign.

Remark 2.28. The condition that (vn) is decreasing in Theorem 2.25 is necessary. Forexample, if u2n = 1

3n and u2n+1 = − 1n+5 , then un = (−1)nvn with (vn) positive and

convergent to 0. However,∑un diverges, since S2n+1 =

∑nk=0

13k −

∑nk=1

1k+5 → −∞.

2.4 Summation by packetsTheorem 2.29. Let

∑un be a series of numbers and ϕ : N → N be strictly increasing.

Put v0 =∑ϕ(0)k=0 uk and vn =

∑ϕ(n)k=ϕ(n−1)+1 uk for n ≥ 1. Then

(1) If∑un converges, then

∑vn converges and has the same sum.

(2) If∑vn converges, then

∑un converges in the following cases :

(a) un → 0 and (ϕ(n+ 1)− ϕ(n)) is bounded,(b) (un) is real, and the uk of each packet {ϕ(n− 1) + 1 ≤ k ≤ ϕ(n)} have the same

sign.

Proof. Let Un =∑nj=0 uj and Vn =

∑nk=0 vk. Then

Vn =ϕ(0)∑k=0

uk +ϕ(1)∑

k=ϕ(0)+1uk + . . .+

ϕ(n)∑k=ϕ(n−1)+1

uk =ϕ(n)∑k=0

uk = Uϕ(n) .

Thus, (Vn) is a subsequence of (Un). So if Un →∑un, then Vn converges to the same

limit. This proves (1)Now suppose that

∑vn converges. Given n ∈ N, let p be the unique natural number

satisfying ϕ(p− 1) < n ≤ ϕ(p). Then

Vp − Un = Uϕ(p) − Un =ϕ(p)∑k=n+1

uk .

To prove (2.a), assume S =∑n≥0 vn and let K such that ϕ(n + 1) − ϕ(n) ≤ K for all

n. Now let ε > 0 and N ∈ N such that |un| ≤ ε2K for n ≥ N . Then for n ≥ N we have

|Vp − Un| ≤∑ϕ(p)k=n+1 |uk| ≤ (ϕ(p)− n) ε

2K ≤ε2 . But there exists N ′ such that |S − Vq| ≤ ε

2for q ≥ N ′. Given n ≥ max(N,ϕ(N ′)), we have ϕ(p) ≥ ϕ(N ′), so p ≥ N ′ and thus

|S − Un| ≤ |S − Vp|+ |Vp − Un| ≤ ε ,

so∑un converges.

To prove (2.b), note that if all uk have the same sign in each packet, then

|Vp − Un| =ϕ(p)∑k=n+1

|uk| ≤ϕ(p)∑

k=ϕ(p−1)+1|uk| =

∣∣∣∣∣∣ϕ(p)∑

k=ϕ(p−1)+1uk

∣∣∣∣∣∣ = |vp| .

Since∑vn converges, we may find N such that |vq| ≤ ε

2 for q ≥ N . Hence, for n ≥ ϕ(N),we have ϕ(p) ≥ ϕ(N), so p ≥ N and thus |Vp − Un| ≤ |vp| ≤ ε

2 . It now follows as beforethat

∑un converges.

2.5. Rearrangements 29

Remark 2.30. If un = (−1)n and we take packets of size 2, i.e. ϕ(n) = 2n + 1, thenvn = 0 for all n, hence

∑vn converges but

∑un diverges. So an additional condition like

(2.a) or (2.b) above is necessary to guarantee the convergence of∑un.

Example 2.31. Let un = (−1)nn+(−1)n . Then |un| is not decreasing, so we cannot apply the

criterion of alternating series. However, if we take ϕ(n) = 2n+ 1, then

vn = u2n + u2n+1 = 12n+ 1 −

12n = − 1

2n(2n+ 1) ∼ −1

4n2 ,

and since∑ 1

n2 converges, then∑vn converges. Since (2.a) is satisfied,

∑un is also

convergent.

Example 2.32. Let un = cos(2nπ/3)n . We take packets of size 3, i.e. ϕ(n) = 3n+ 2. Then

noting that cos(2π/3) = cos(4π/3) = −1/2, we get

vn = u3n + u3n+1 + u3n+2 = 13n −

12(3n+ 1) −

12(3n+ 2) ∼

16n2

as can be easily checked. Again, (2.a) is satisfied, so∑un converges.

2.5 RearrangementsTheorem 2.33. Suppose

∑un is absolutely convergent and let σ : N→ N be a permuta-

tion, i.e. a bijective map. Then∑uσ(n) is also convergent, and

∑uσ(n) =

∑un.

Proof. Let ε > 0. By the Cauchy criterion, we may find n0 ∈ N such that

n ≥ n0, p ≥ 1 =⇒ |un+1|+ . . .+ |un+p| ≤ ε .

Choose n1 such that the terms u0, . . . , un0 appear in the partial sum S′n1 =∑n1n=0 uσ(n).

Then S′n1 contains at least n0 + 1 terms, so n1 ≥ n0. Hence, for m ≥ n1, the termsu0, . . . , un0 disappear from the difference Sm − S′m. So there exists k ∈ N such that

|Sm − S′m| ≤ |un0+1|+ . . .+ |un0+k| ≤ ε .

Hence, Sm − S′m → 0, so S′m converges to the same limit of Sm.

Remark 2.34. Absolute convergence is necessary for the previous theorem to hold. Forexample, consider the series

∑un where un is given by

1,−1, 12 ,−12 ,

13 ,−13 ,

14 ,−14 , . . .

Then taking packets of size 2, we see that∑un converges since un → 0. However, consider

the following rearrangement :

1, 12 ,−1, 1

3 ,14 ,−12 ,

15 ,

16 ,

17 ,

18 ,−13 , . . .

12p−1 + 1 ,

12p−1 + 2 ,

12p−1 + 3 , . . . ,

12p ,−1p, . . .

Then this series diverges since it does not satisfy the Cauchy criterion. Indeed,

12p−1 + 1 + 1

2p−1 + 2 + 12p−1 + 3 + . . .+ 1

2p ≥ (2p − 2p−1) 12p = 1

2 .


2.6 Double series

Definition 2.35. A double sequence is a map u : N2 → C. We denote up,q := u(p, q). Themap itself is denoted by (up,q)(p,q)∈N2 .

A double series is a sequence Sn,m =∑np=0

∑mq=0 up,q.

Clearly, Sn,m =∑mq=0

∑np=0 up,q. But do we have

∑∞p=0

∑∞q=0 up,q =

∑∞q=0

∑∞p=0 up,q ?

Example 2.36. Consider the double sequence

1 -1 0 0 0 · · ·

0 1 -1 0 0 · · ·

0 0 1 -1 0 · · ·

...... 0 1 -1 0

up,q =

1 if q = p

−1 if q = p− 10 otherwise.

Then∑∞p=0

∑∞q=0 up,q = 1 + 0 + 0 + . . . = 1 but

∑∞q=0

∑∞p=0 up,q = 0 + 0 + 0 + . . . = 0.

So we cannot interchange the order of summation here.

Definition 2.37. We say that∑∞k=0 vk is a linear arrangement of

∑(p,q)∈N2 up,q if there

exists a bijection σ : N→ N2 such that vk = uσ(k).

Example 2.38. The series∑∞k=0

∑p+q=k up,q = (u0,0)+(u1,0+u0,1)+(u2,0+u1,1+u0,2)+. . .

is a linear arrangement of∑up,q. Here, σ(0) = (0, 0), σ(1) = (1, 0), σ(2) = (0, 1),

σ(3) = (2, 0) and so on.

Theorem 2.39. Suppose there exists B ≥ 0 such that∑mp=0

∑mq=0 |up,q| ≤ B for all

m ≥ 0. Then the series∑∞p=0

∑∞q=0 up,q and

∑∞q=0

∑∞p=0 up,q both converge to the same

sum. Moreover, every linear arrangement of∑

(p,q)∈N2 up,q converges to the same sum.

Proof. Let∑∞k=0 vk be a linear arrangement of

∑up,q. The sequence

(∑nk=0 |vk|

)is in-

creasing and bounded by B, hence∑|vk| and

∑vk both converge. Similarly, we see that

Rp =∑∞q=0 up,q and Lq =

∑∞p=0 up,q converge.

Let ε > 0. By the Cauchy criterion, we may find n0 such that

n ≥ n0, k ≥ 1 =⇒ |vn+1|+ . . .+ |vn+k| ≤ ε .

Choose n1 such that the terms v0, . . . , vn0 appear in the partial sum∑n1p=0

∑n1q=0 up,q.

Then for l ≥ n0 and n,m ≥ n1, the terms v0, . . . , vn0 disappear from the differenceDn,m,l =

∑mp=0

∑nq=0 up,q −

∑lk=0 vk and there exists k ∈ N such that

|Dn,m,l| ≤ |vn0+1|+ . . .+ |vn+k| ≤ ε .

Put S =∑∞k=0 vk. Then as l, n → ∞ we get |

∑mp=0Rp − S| ≤ ε. If we interchange∑m

p=0∑nq=0 =

∑nq=0

∑mp=0, we obtain similarly |

∑nq=0 Lq − S| ≤ ε. Hence,

∑∞p=0Rp and∑∞

q=0 Lq both converge to the same sum S.

If up,q ≥ 0 for all p, q, we have a much stronger result. Namely, we can alwaysinterchange the order of summation :

2.7. Cauchy product 31

Theorem 2.40. If up,q ≥ 0 for all p, q ≥ 0, then∞∑p=0

∞∑q=0

up,q =∞∑q=0

∞∑p=0

up,q .

This means that even if one of the double series diverges, then both will be +∞.

Proof. Admitted. See for example [15], page 23.

2.7 Cauchy productDefinition 2.41. The Cauchy product of two series

∑∞p=0 up and

∑∞q=0 vq is defined to

be the series∑∞n=0

(∑nq=0 un−qvq

)= u0v0 + (u0v1 + u1v0) + (u0v2 + u1v1 + u2v0) + . . ..

Example 2.42. The Cauchy product of∑n≥0

(−1)n√n+1 with itself is divergent. Indeed,∣∣∣∣∣∣

n∑q=0

un−qvq

∣∣∣∣∣∣ =n∑q=0

1√n− q + 1

√j + 1

≥ (n+ 1) 1n+ 1 = 1 .

Thus, the Cauchy product of two convergent series may diverge.Theorem 2.43. If

∑∞p=0 up and

∑∞q=0 vq are both absolutely convergent, then their Cauchy

product converges, and we have( ∞∑p=0

up)( ∞∑

q=0vq)

=∞∑n=0

( n∑q=0

un−qvq).

Proof. By hypothesis, there exist B1, B2 such that∑∞p=0 |up| ≤ B1 and

∑∞q=0 |vq| ≤ B2.

Hence,∑mp=0

∑mq=0 |up||vq| ≤ B1B2 for any m, and we may apply Theorem 2.39. Here we

have Rp = up∑∞q=0 vq and

∑∞p=0Rp =

(∑∞p=0 up

)(∑∞q=0 vq

). By the theorem, the Cauchy

product, which is a linear arrangement of∑

(p,q)∈N2 upvq, converges to the same sum.

Example 2.44. Let |a| < 1. The Cauchy product of 11−a =

∑∞p=0 a

p with 11+a =∑∞

q=0(−1)qaq is∞∑n=0

n∑q=0

an−q(−1)qaq =∞∑n=0

ann∑q=0

(−1)q =∞∑k=0

a2k = 11− a2 .

This is not surprising since we know that 11−a

11+a = 1

1−a2 .

Example 2.45. Let us calculate the Taylor series of 1(1−a)2 using the Cauchy product of

11−a with itself. We have

1(1− a)2 =

∞∑n=0

n∑q=0

an−qaq =∞∑n=0

(n+ 1)an .

Example 2.46. The series∑ zn

n! is absolutely convergent for any z ∈ C (use the ratiotest). The sum is by definition ez. Let us show that ez1+z2 = ez1ez2 for zj ∈ C. We have

ez1ez2 =∞∑n=0

n∑q=0

zn−q1(n− q)!

zq2q! =

∞∑n=0

n∑q=0

1n!

(n

q

)zn−q1 zq2 =

∞∑n=0

1n! (z1 + z2)n = ez1+z2 .

where we used the binomial theorem.Remark 2.47. Theorem 2.43 remains true if only one series is absolutely convergent. Fora proof, see e.g. [9].


2.8 Exercises1. Determine the nature of the following series, and calculate the sum in case of conver-

gence.

•∑n≥1

1n(n+ 1) •

∑n≥1

1n+ 1

(ln(

1 + 1n

)− ln(n)

n

)•

∑n≥2

1n3 − n

•∑n≥0

1√n+√n+ 1

•∑n≥1

sin(

π

4n2 − 1

)sin( 2nπ

4n2 − 1

).

Hint. Show that they are telescoping series.2. Let

∑n∈N un and

∑n∈N vn be series with un ≥ 0 and vn > 0 beyond some index. Are

the following assertions correct ? Justify your answer.(a) If the partial sums of the series

∑n∈N un are bounded, then the series converges.

(b) If un → 0, then the series∑n∈N un converges.

(c) If∑n∈N un converges, then un → 0.

(d) If∑n∈N un converges, then for all n0 ∈ N the series

∑n≥n0 un converges.

(e) Let n0 ∈ N. If∑n≥n0 un converges, then the series

∑n∈N un converges.

(f) If∑n≥0 un and

∑n≥0 vn converge, then

∑n≥0 unvn converges.

(g) If∑vn diverges and un ∼ vn, then

∑n∈N un diverges.

(h) If un = o(vn) and∑vn converges, then


(i) If un = o(vn) and∑vn diverges, then


(j) If un = O(vn), then∑n∈N un and

∑n∈N vn have the same nature.

(k) If un = o(

1n

), then


(l) If un = o(

1n

), then


(m) If un ∼ 1n2 , then


3. Study the convergence of the following series (here a > 0 and α ∈ R)

•∑n≥3

2n + 53n − 11 •

∑n≥1

n1+ 1n •

∑n≥1

(12 + 1

n

)n

•∑n≥3

1n · lnn · ln lnn •

∑n≥1

(n!)3

(3n)! •∑n≥1

(1− 1

n

)n

•∑n≥1

(1− 1

n

)n2

•∑n≥1

√n+ 1−

√n

n•

∑n≥1

nn

n!

•∑n≥1

1! + . . .+ (n− 1)!n! •

∑n≥1

2n + 5n

an•

∑n≥1

nlnn

(lnn)n

•∑n≥0

e−√n •

∑n≥1

ln(

cos 1n

)•

∑n≥1

n2

2n

•∑n≥1

n2 sin 2−n •∑n≥1

1− cos( 1n

)•

∑n≥2

1! + . . .+ (n− 2)!n!

•∑n≥1

(n6 + 3)α − (n2 + 2)3α •∑n≥1

n1

n2+1 − 1 •∑n≥1

nα

(1 + a) · · · (1 + an)

2.8. Exercises 33

4. (Cauchy condensation test). Suppose u1 ≥ u2 ≥ u3 ≥ . . . ≥ 0. Show that∑∞n=1 un

converges iff∑∞k=0 2ku2k converges.

Deduce new proofs about the nature of the series∑ 1

np and∑ 1

n(lnn)p .5. Let

∑n∈N un,

∑n∈N vn be series with un, vn ∈ R. Suppose ∃k0 such that vn 6= 0

∀n ≥ k0. Are the following assertions correct ? Justify your answer.(a) If un ∼ (−1)n

n2 , then∑n∈N un converges.

(b) If un ∼ (−1)nn , then


(c) If∑n≥0 un converges, then

∑n≥0 un is absolutely convergent.

(d) If∑n≥0 un is absolutely convergent, then

∑n≥0 un converges.

(e) If∑n≥0 un is absolutely convergent, then

∑n≥0 u

2n converges.

(f) If∑n≥0 un is absolutely convergent, then

∑n≥0

√|un| converges.

(g) If∑n≥0 un converges, then

∑n≥0 u

2n converges.

6. Study the convergence and absolute convergence of the following series:

•∑n≥1

√n+ (−1)n −

√n •

∑n≥2

(−1)n

2n+ (−1)n •∑n≥1

(−1)n

(2n− 1)3

•∑n≥1

(−1)n cosh( 1n

)sin( 1n

)•

∑n≥1

2−n sin(nθ) •∑n≥1

ln(

1 + (−1)n

n+ 1

)•

∑n≥1

sin(π√n2 + 1

)

7. (1) Let f(x) = ei√x

√x. Estimate f(n + 1) − f(n) using Taylor’s formula with integral

remainder. Deduce the nature of the series∑n≥1 f

′(n).(2) Determine a primitive F of f . Estimate F (n + 1) − F (n) by the same method.

Deduce the nature of the series∑n≥1 f(n).

8. (Raabe-Duhamel rule). Let∑un be a series with strictly positive terms.

(1) Show that if un+1un

= 1 − αn + O( 1

nβ) with β > 1, then the sequence (nαun) has a

finite non-zero limit, and� if α > 1, then

∑un converges;

� if α ≤ 1, then∑un diverges.

(2) Application: Study the convergence of the series∑un, where un = 1·3·5...(2n−1)

2·4·6...2n1√n.

(3) Application: Prove the Stirling approximation: there exists K > 0 such thatn! ∼ Knne−n

√n. One can show that K =

√2π.

(4) Show that if un+1un

= 1− αn + o( 1

n) with� α > 1, then

∑un converges;

� α < 1, then∑un diverges.

9. Let (an)n≥0 be a sequence with strictly positive terms. Suppose∑an diverges and put

Sn :=∑nk=0 ak. Study the nature of the series

∑ an+1Sn

.10. (Riemann rearrangement theorem). Let

∑k≥1 ak be a real series which converges, but

not absolutely. Suppose −∞ ≤ α ≤ ∞. Show that there exists a permutation σ of Nsuch that

∑∞k=1 aσ(k) = α.

11. Let (un) be a complex sequence and∑vn a series with nonnegative terms.

(1) Suppose that∑vn diverges. Show that


� if un = O(vn), then∑np=0 up = O

(∑np=0 vp

);

� if un = o(vn), then∑np=0 up = o

(∑np=0 vp

);

� if un ∼ vn, then∑np=0 up ∼

∑np=0 vp.

Applications: Show that if un → `, then un → ` in the sense of Cesàro.Show that

∑np=1

1k ∼ lnn.

(2) Suppose that∑vn converges. Show that

� if un = O(vn), then∑+∞p=n up = O

(∑+∞p=n vp

);

� if un = o(vn), then∑+∞p=n up = o

(∑+∞p=n vp

);

� if un ∼ vn, then∑+∞p=n up ∼

∑+∞p=n vp.

12. (Abel’s test). Let∑anvn be a series such that An :=

∑nk=0 ak is bounded and vn

decreases and tends to 0. Show that∑anvn converges.

Application: Study the convergence of the series∑ cosnθ

nα and∑ sinnθ

nα , where α > 0and θ is a real which is not an integer multiple of π.Variant: Show that if An converges and if vn is monotone and convergent, then

∑anvn

converges.13. Let (un) be a decreasing sequence of strictly positive reals such that the series

∑un

converges. Show that un = o( 1n).

Give an example of a sequence (un) of strictly positive reals such that the series∑un

converges but the sequence nun does not tend to 0.14. Let (an) be a nonnegative sequence such that the series

∑an converges. Show that if

α > 12 , then the series

∑ √annα converges. What can we say if α = 1

2?

Chapter 3

Sequences and Series of Functions

In this chapter, all functions f, fn are defined on a ball I = Br(z0) ⊆ K. Recall thatBr(z0) = ]z0 − r, z0 + r[ if K = R and Br(z0) is an open disk of radius of r if K = C.

We follow [14] closely in Sections 3.1 to 3.3, then we follow [2] in Sections 3.4 and 3.5.

3.1 Pointwise convergenceDefinition 3.1. We say that a sequence of functions (fn) converges pointwise on I if forany x ∈ I, the sequence (fn(x)) converges. In this case we define the limit function f by

f(x) = limn→∞

fn(x), x ∈ I .

We say that∑fn converges pointwise on I if the sequence of partial sums Sn(x) =∑n

k=0 fk(x) converges pointwise on I. The limit function of (Sn) is then called the sum ofthe series and is denoted by S(x) =

∑∞n=0 fn(x) for x ∈ I.

If the functions fn are continuous, differentiable, or integrable, is the same true for thelimit function ?

Recall that f is continuous at x if limt→x

f(t) = f(x). But limt→x

f(t) = limt→x

limn→∞

fn(t), andif fn are continuous at x, then f(x) = lim

n→∞fn(x) = lim

n→∞limt→x

fn(t). Hence, if (fn) arecontinuous, then limit function f will be continuous if

limt→x

limn→∞

fn(t) = limn→∞

limt→x

fn(t) ,

i.e. if we can interchange limits without affecting the result.Let us first illustrate that the order of the limits is important in general. Afterward,

we will give some conditions under which the limits can be safely interchanged.

Example 3.2. Form,n ∈ N∗, let sm,n = mm+n . Then for any fixed n, we have lim

m→∞sm,n =

1, hence limn→∞

limm→∞

sm,n = 1. However, for any fixedm, limn→∞

sm,n = 0, so limm→∞

limn→∞

sm,n =0. So the result changes when we interchange limits.

Example 3.3. Let fn(x) = x2

(1+x2)n for x ∈ R and n ∈ N. Since fn(0) = 0 for all n, we have∑∞n=0 fn(0) = 0. For x 6= 0, we have

∑∞n=0 fn(x) = x2∑∞

n=01

(1+x2)n = x2 11− 1

1+x2= 1 + x2.

Thus,∑fn converges pointwise on R to the function

S(x) ={

0 if x = 0 ,1 + x2 if x 6= 0 .

Thus, a convergent series of continuous functions may have a discontinuous sum.

36 Chapter 3. Sequences and Series of Functions

Example 3.4. For x ∈ R and n ∈ N∗, let fn(x) = sinnx√n

and f(x) = limn→∞ fn(x) = 0.Then f ′(x) = 0 for all x. However, f ′n(x) =

√n cosnx, so f ′n(0) =

√n→∞. Hence, (f ′n)

does not converge pointwise to f ′.

Example 3.5. Let fn(x) = n2x(1− x2)n for x ∈ [0, 1] and n ∈ N∗. Then fn(0) = 0, andfor x > 0, limn→∞ fn(x) = 0. Thus, (fn) converges pointwise on [0, 1] to f ≡ 0. However,taking y = x2, we have

∫ 10 x(1 − x2)n dx = 1

2∫ 1

0 (1 − y)n dy = 12−(1−y)n+1

n+1 |10 = 12n+2 . So∫ 1

0 fn(x) dx = n2

2n+2 →∞, although∫ 1

0 f(x) dx = 0.Note that if gn(x) = nx(1 − x2)n, then gn still converges pointwise to g ≡ 0, but we

now have∫ 1

0 gn(x) dx = n2n+2 →

12 . Thus, even if

( ∫ ba gn(x) dx

)n∈N converges, its limit

may not be equal to∫ ba g(x) dx.

3.2 Uniform convergence

Definition 3.6. We say that a sequence of functions (fn) converges uniformly to a functionf on I if

∀ε > 0 ∃N ∈ N : n ≥ N =⇒ |fn(x)− f(x)| ≤ ε ∀x ∈ I .

We denote this by fn ⇒ f on I.We say that a series of functions

∑fn converges uniformly on I if the sequence of

partial sums Sn(x) =∑nk=0 fk(x) converges uniformly on I.

Remark 3.7. Recall that (fn) converges pointwise to f on I if

∀x ∈ I ∀ε > 0 ∃N ∈ N : n ≥ N =⇒ |fn(x)− f(x)| ≤ ε .

The difference is that in uniform convergence, there is one N which works for any x ∈ I.In pointwise convergence, for each x, we may find a corresponding N (which may dependon x). Hence, uniform convergence implies pointwise convergence.

Lemma 3.8. Let f, g be bounded functions on I and denote ‖f‖∞ := supx∈I |f(x)|. Then1. ‖f‖∞ ≥ 0, ‖f‖∞ = 0 ⇐⇒ f ≡ 0 on I,2. ‖αf‖∞ = |α| · ‖f‖∞ ∀α ∈ C,3. ‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.4. ‖fg‖∞ ≤ ‖f‖∞‖g‖∞.We call ‖f‖∞ the supremum norm of f .

Proof. Clearly, ‖f‖∞ ≥ 0. Moreover, ‖f‖∞ = 0 ⇐⇒ supx∈I |f(x)| = 0 ⇐⇒ |f(x)| =0 ∀x ∈ I ⇐⇒ f(x) = 0 ∀x ∈ I ⇐⇒ f ≡ 0 on I.

Next, ‖αf‖∞ = supx∈I |αf(x)| = supx∈I |α||f(x)| = |α| supx∈I |f(x)| = |α|‖f‖∞.Next, if x ∈ I we have |f(x) + g(x)| ≤ |f(x)| + |g(x)| ≤ ‖f‖∞ + ‖g‖∞. Taking the

supremum in the LHS we get ‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.Finally, if x ∈ I we have |f(x)g(x)| = |f(x)||g(x)| ≤ ‖f‖∞‖g‖∞, so taking the sup in

the LHS we get (4).

Remark 3.9. Note that

fn ⇒ f on I ⇐⇒ limn→∞

‖fn − f‖∞ = 0 .

3.2. Uniform convergence 37

Lemma 3.10. Suppose fn ⇒ f , gn ⇒ g on I and let α, β ∈ C. Then αfn+βgn ⇒ αf+βgon I. Moreover, if each fn and gn is bounded on I, then f and g are bounded on I, andfngn ⇒ fg on I.

Proof. By Lemma 3.8,

‖αfn + βgn − (αf + βg)‖∞ ≤ |α| · ‖fn − f‖∞ + |β| · ‖gn − g‖∞ → 0 ,

which proves that αfn + βgn ⇒ αf + βg on I.Next, note that if each fn is bounded on I, then f must be bounded on I since for ε = 1

we may find N such that ‖f‖∞ ≤ ‖fN − f‖∞ + ‖fN‖∞ ≤ 1 +C, where C = ‖fN‖∞ <∞because fN is bounded. Similarly, g is bounded on I. Hence,

‖fngn − fg‖∞ = ‖(fn − f)(gn − g) + g(fn − f)− f(g − gn)‖∞≤ ‖fn − f‖∞‖gn − g‖∞ + ‖g‖∞‖fn − f‖∞ + ‖f‖∞‖g − gn‖∞ → 0 .

Remark 3.11. The last result is not true if fn or gn is unbounded. For example, iffn(x) = 1

n and gn(x) = x on R, then fn ⇒ 0, gn ⇒ g, where g(x) = x, but fngn = xn does

not converge uniformly to 0 · g = 0. Indeed, ‖fngn − 0‖∞ ≥ fn(n2)gn(n2) = n→∞.

Lemma 3.12 (Cauchy criterion). A sequence of functions (fn) converges uniformly on Iiff

∀ε > 0 ∃N ∈ N : n,m ≥ N =⇒ |fn(x)− fm(x)| ≤ ε ∀x ∈ I . (?)

Proof. Suppose fn ⇒ f on I. Then there exists N ∈ N such that, if n ≥ N , then|fn(x)− f(x)| ≤ ε

2 for any x ∈ I. Hence,

n,m ≥ N =⇒ |fn(x)− fm(x)| ≤ |fn(x)− f(x)|+ |f(x)− fm(x)| ≤ ε

for any x ∈ I. Hence, (fn) satisfies the Cauchy condition.Conversely, suppose (?) holds. Then for any fixed x ∈ I, (fn(x)) is a Cauchy sequence

of numbers, hence converges to some `x ∈ C. Define f(x) := `x. Then, (fn) convergespointwise to f on I.

To see that fn ⇒ f on I, let ε > 0 and take N such that (?) holds. Fix n and letm→∞ in (?). Since fm(x)→ f(x), we get

n ≥ N =⇒ |fn(x)− f(x)| ≤ ε ∀x ∈ I .

Corollary 3.13. If∑fn converges uniformly on I, then fn ⇒ 0 on I.

Proof. Since∑fn is Cauchy, given ε > 0, we may findN ∈ N such that |Sn(x)−Sn−1(x)| ≤

ε for any n ≥ N and x ∈ I. Thus, |fn(x)| ≤ ε for any n ≥ N and x ∈ I, hence fn ⇒ 0.

Lemma 3.14. Let∑fn be a pointwise convergent series and put Rn(x) =

∑∞k=n+1 fn(x).

Then∑fn is uniformly convergent iff Rn ⇒ 0 on I.

Proof. If Sn ⇒ S on I, then Rn = S − Sn ⇒ 0 on I.Conversely, if

∑fn converges pointwise to S on I and Rn ⇒ 0 on I, then Sn =

S −Rn ⇒ S on I.

Corollary 3.15. Let∑

(−1)ngn be a series of functions on I such that� ∀x ∈ I, the numerical sequence (gn(x))n∈N is nonnegative and decreasing,� gn ⇒ 0 on I.

Then∑

(−1)ngn converges uniformly on I.


Proof. By the alternating series criterion, for any x ∈ I, the series∑

(−1)ngn(x) is con-vergent and |Rn(x)| ≤ gn+1(x). Hence, ‖Rn‖∞ ≤ ‖gn+1‖∞ → 0, so the series convergesuniformly by Lemma 3.14.

Definition 3.16. We say that∑fn converges absolutely on I if the numerical series∑

|fn(x)| converges for each x ∈ I.We say that

∑fn converges normally on I if the numerical series

∑‖fn‖∞ converges.

Lemma 3.17 (Weierstrass M-test). If∑fn is normally convergent, then it is absolutely

and uniformly convergent.

Proof. Given x ∈ I, |fn(x)| = O(‖fn‖∞

), hence

∑|fn(x)| converges. Thus,

∑fn is

absolutely convergent.Moreover, given ε > 0, we may find N ∈ N such that

∑mi=n ‖fi‖∞ ≤ ε for n,m ≥ N .

Hence,

n,m ≥ N =⇒∣∣∣ m∑i=n

fi(x)∣∣∣ ≤ m∑

i=n|fi(x)| ≤

m∑i=n‖fi‖∞ ≤ ε ∀x ∈ I .

Hence,∑fn satisfies the Cauchy criterion and thus converges uniformly.

Remark 3.18. The converse of the previous lemma is not true. For example, let

fn(x) =

0 if 0 ≤ x < 1

n+1 ,1n if 1

n+1 ≤ x <1n ,

0 if 1n ≤ x ≤ 1 .

Then∑fn converges absolutely on [0, 1] because for any x ∈ [0, 1], we have fn(x) = 0

for all large n.∑fn also converges uniformly on [0, 1] because ‖

∑n+pk=n+1 fk‖∞ ≤

1n+1 .

However,∑fn does not converge normally, since ‖fn‖∞ = 1

n and∑ 1

n diverges.

3.3 Conservation of properties by uniform convergence

Definition 3.19. A limit point of a ball I is a point in I or on its boundary.A neighborhood of a point x ∈ I is a ball V = Bs(x) for some s > 0.A neighborhood of +∞ and −∞ is an interval of the form ]M,+∞[ and ]−∞,M [,

respectively, for some M ∈ R.

Definition 3.20. Let a be a limit point of I. We say that limx→a f(x) = ` if for anyε > 0, we may find a neighborhood V of a such that x ∈ V ∩ I =⇒ |f(x)− `| ≤ ε.

Theorem 3.21. Suppose fn ⇒ f on I. Let x be a limit point of I and suppose that

limt→x

fn(t) = An .

Then (An) converges andlimt→x

f(t) = limn→∞

An .

Thus, we have limt→x limn→∞ fn(t) = limn→∞ limt→x fn(t).

3.3. Conservation of properties by uniform convergence 39

Proof. Given ε > 0, we may find by Cauchy’s condition N ∈ N such that

n,m ≥ N =⇒ |fn(t)− fm(t)| ≤ ε ∀t ∈ I .

Letting t → x we obtain |An − Am| ≤ ε for all n,m ≥ N , hence (An) is Cauchy andconverges, say to A. Now for any n ∈ N,

|f(t)−A| ≤ |f(t)− fn(t)|+ |fn(t)−An|+ |An −A| . (†)

Choose n0 ∈ N such that |f(t) − fn0(t)| ≤ ε/3 for any t ∈ I and |An0 − A| ≤ ε/3. Next,choose a neighborhood V of x such that |fn0(t)−An0 | ≤ ε/3 for all t ∈ V ∩I. Then takingn = n0 in (†) we get |f(t)−A| ≤ ε for any t ∈ V ∩ I.

Corollary 3.22. If (fn) is a sequence of continuous functions on I, and if fn ⇒ f on I,then f is continuous on I.

Proof. Let x ∈ I. Since fn is continuous at x, we have An = limt→x fn(t) = fn(x). UsingTheorem 3.21, we get limt→x f(t) = limn→∞ fn(x) = f(x). Hence, f is continuous atx.

Theorem 3.23. Let (fn) be a sequence of Riemann integrable functions on [a, b] such thatfn ⇒ f on [a, b]. Then f is Riemann integrable on [a, b] and∫ b

af(x) dx = lim

n→∞

∫ b

afn(x) dx .

Proof. We admit that f is Riemann integrable 1. Next,∣∣∣ ∫ b

afn(x) dx−

∫ b

af(x) dx

∣∣∣ =∣∣∣ ∫ b

a

(fn(x)− f(x)

)dx∣∣∣

≤∫ b

a|fn(x)− f(x)| dx ≤ (b− a)‖fn − f‖∞ → 0 .

Corollary 3.24 (Term by term integration). Let (fn) be a sequence of integrable functionson [a, b] such that

∑fn ⇒ S on [a, b]. Then S is integrable on [a, b] and∫ b

aS(x) dx =

∑∫ b

afn(x) dx .

Proof. Let Sn(x) =∑nk=0 fk(x). Since Sn ⇒ S, we have by Theorem 3.23,∫ b

aS(x) dx = lim

n→∞

∫ b

aSn(x) dx = lim

n→∞

n∑k=0

∫ b

afk(x) dx =

∞∑k=0

∫ b

afk(x) dx .

Theorem 3.25. Let (fn) be a sequence of differentiable functions on [a, b]. If (fn(x0))converges for some x0 ∈ [a, b], and if (f ′n) converges uniformly on [a, b], then (fn) convergesuniformly on [a, b] to a function f such that

f ′(x) = limn→∞

f ′n(x) ∀x ∈ [a, b] .

1. Here is a proof: Let X = (xk)nk=0 be a subdivision of [a, b], let s(f) = supX∑n−1

k=0 (xk+1 −xk) inf[xk,xk+1] f and S(f) = infX

∑n−1k=0 (xk+1 − xk) sup[xk,xk+1] f . Since fn is integrable, s(fn) = S(fn).

Let εn = ‖fn − f‖∞. Then fn − εn ≤ f ≤ fn + ε on [a, b], so∫ ba

(fn − εn) = s(fn − εn) ≤ s(f) ≤ S(f) ≤S(fn+εn) =

∫ ba

(fn+εn). Thus, 0 ≤ S(f)−s(f) ≤∫ ba

(fn+εn)−s(f) ≤∫ ba

(fn+εn−fn+εn) = 2εn(b−a).Taking n→∞, we get s(f) = S(f), so f is Riemann integrable.


Proof. Given ε > 0, choose N such that for n,m ≥ N , we have

|fn(x0)− fm(x0)| ≤ ε

2 and ‖f ′n − f ′m‖∞ ≤ε

2(b− a) .

Applying the mean value theorem to the function fn − fm, we get

|fn(x)− fm(x)− fn(t) + fm(t)| ≤ ε

2(b− a) |x− t| ≤ε

2 (‡)

for any n,m ≥ N and x, t ∈ [a, b]. Hence,

|fn(x)− fm(x)| ≤ |fn(x)− fm(x)− fn(x0) + fm(x0)|+ |fn(x0)− fm(x0)| ≤ ε

for any x ∈ [a, b] and n,m ≥ N . Hence, (fn) converges uniformly on [a, b]. Put f(t) =limn→∞ fn(t), fix x ∈ [a, b] and define

φn(t) = fn(t)− fn(x)t− x

and φ(t) = f(t)− f(x)t− x

for t ∈ [a, b] \ {x} .

Then φn converges pointwise to φ on [a, b] \ {x}. Moreover, by (‡),

|φn(t)− φm(t)| ≤ ε

2(b− a)

for any n,m ≥ N , so that (φn) converges uniformly on [a, b] \ {x}. Thus, φn ⇒ φ on[a, b] \ {x}.

Finally, apply Theorem 3.21 to φn with An = limt→x φn(t) = f ′n(x) to get

f ′(x) = limt→x

φ(t) = limn→∞

f ′n(x) .

Definition 3.26. We say that f ∈ Cp [a, b] if the derivatives f ′, f ′′, . . . f (p) exist and arecontinuous. If f ∈ Cp [a, b], we say that f is of class Cp on [a, b].

We say that f ∈ C∞ [a, b] if f (k) exists for all k ≥ 1.

Corollary 3.27. Let (fn) be a sequence of functions in Cp [a, b], where 1 ≤ p ≤ ∞. If(fn) converges pointwise on [a, b], and if for each k, 1 ≤ k ≤ p, the sequence (f (k)

n ) isuniformly convergent, then (fn) converges uniformly to a function f ∈ Cp [a, b] such that

f (k)(x) = limn→∞

f (k)n (x) ∀x ∈ [a, b] .

Proof. Follows by induction from Theorem 3.25.

Corollary 3.28 (Term by term differentiation). Let (fn) be a sequence of functions inCp [a, b], where 1 ≤ p ≤ ∞. If

∑fn is pointwise convergent on [a, b], and if for each k,

1 ≤ k ≤ p, the series∑f

(k)n converges uniformly on [a, b], then

∑fn converges uniformly

to a function S ∈ Cp [a, b] such that

S(k)(x) =∑

f (k)n (x) ∀x ∈ [a, b] .

3.4. Uniform continuity 41

3.4 Uniform continuityDefinition 3.29. Let I ⊂ R be an interval. We say that f is uniformly continuous on Iif

∀ε > 0 ∃δ > 0 : ∀x, y ∈ I, |x− y| ≤ δ =⇒ |f(x)− f(y)| ≤ ε .

Remark 3.30. Recall that f is continuous on I if

∀x ∈ I ∀ε > 0 ∃δ > 0 : ∀y ∈ I, |x− y| ≤ δ =⇒ |f(x)− f(y)| ≤ ε .

The difference is that the δ in uniform continuity depends only on ε, while in continuity,it may also depend on x. Hence, uniform continuity implies continuity.

Reminder 3.31. We say that f is Lipschitz continuous on I (Lip) if there exists K > 0such that ∀x, y ∈ I : |f(x)− f(y)| ≤ K|x− y|.

If f is Lip, then given ε > 0, if we choose δ = ε/K, we see that f is uniformlycontinuous.

Example 3.32. (a) f(x) = x2 is uniformly continuous on I = [0, 1], since it is Lip on[0, 1]. Indeed, given x, y ∈ I, |x2 − y2| = |x+ y||x− y| ≤ 2|x− y|.

(b) f(x) = x2 is not uniformly continuous on R. Indeed, take ε = 2 and any δ > 0. Nowchoose x = 1

δ and y = x+ δ. Then |x− y| ≤ δ, but |x2 − y2| = y2 − x2 = 2 + δ2 > ε.(c) f(x) = sin x is uniformly continuous on R, since it is Lip on R. Indeed, for any

x, y ∈ R, we may find z ∈ ]x, y[ such that | sin x− sin y| = | cos z| · |x− y| ≤ |x− y|.(d) f(x) =

√x is uniformly continuous on R+, but not Lip on R+.

Indeed, given K > 0, choose x = 1(2K)2 . Then |f(x) − f(0)| =

√x = 1√

x|x − 0| =

2K|x− 0| > K|x− 0|. Hence, f(x) =√x is not Lipschitz on R+.

On the other hand, note that if 0 ≤ x ≤ y, then x+y = (y−x)+2x ≤ |x−y|+2√xy.Similarly, if 0 ≤ y ≤ x, then x + y = (x − y) + 2y ≤ |x − y| + 2√xy. Thus, forany x, y ∈ R+ we have x − 2√xy + y ≤ |x − y|, i.e. (

√x − √y)2 ≤ |x − y|. Hence,

|√x−√y| ≤

√|x− y|. Given ε > 0, choosing δ = ε2, we thus have

∀x, y ∈ R+ : |x− y| ≤ δ =⇒ |√x−√y| ≤

√|x− y| ≤ ε ,

which shows that f is uniformly continuous on R+.

We thus have

f Lipshitz continuous =⇒ f uniformly continuous =⇒ f continuous,

and neither implication is an equivalence.

Theorem 3.33 (Heine-Cantor Theorem). Let [a, b] be a closed bounded interval. If f iscontinuous on [a, b], it is uniformly continuous on [a, b].

Proof. Suppose f is continuous on I = [a, b]. Suppose on the contrary that f is notuniformly continuous on I. Then

∃ε0 > 0 : ∀δ > 0, ∃x, y ∈ I, |x− y| ≤ δ but |f(x)− f(y)| ≥ ε0 .

Take this ε0. For δ = 2−n, we may thus choose xn, yn ∈ I such that |xn − yn| ≤ 2−n but|f(xn)− f(yn)| ≥ ε0. Hence,

lim(xn − yn) = 0 but ∀n ∈ N, |f(xn)− f(yn)| ≥ ε0 . (∗)


Since I is bounded, the sequence (xn) is bounded and has a subsequence (xϕ(n)) whichconverges to some α. Since I is closed, we have α ∈ I. Since yϕ(n) = xϕ(n) +(yϕ(n)−xϕ(n)),we have yϕ(n) → α.

Since f is continuous, we thus get lim(f(xϕ(n)) − f(yϕ(n))

)= f(α) − f(α) = 0. This

contradicts (∗). Hence, f is uniformly continuous on I.

3.5 Approximation theoremsDefinition 3.34. A map f : [a, b]→ C is piecewise continuous if there exists a subdivisiona = a0 < a1 < . . . < an = b such that the restriction of f to each ]ai−1, ai[ can be extendedto a continuous function on [ai−1, ai]. The set of such functions is denoted by CP [a, b].

Remark 3.35. If there is a subdivision a = a0 < a1 < . . . < an = b such that� g is continuous on each ]ai−1, ai[,� g has a right-sided limit at a0, . . . , an−1,� g has a left-sided limit at a1, . . . , an,

then g ∈ CP [a, b].

Definition 3.36. A map ϕ : [a, b] → C is a step function if there exists a subdivisiona = a0 < a1 < . . . < an = b of [a, b] such that the restriction of ϕ to each interval ]ai−1, ai[is constant. The set of such functions is denoted by E [a, b].

Lemma 3.37. For any f ∈ CP [a, b], we may find g ∈ C [a, b] and ϕ ∈ E [a, b] such thatf = g + ϕ.

Proof. We prove the lemma by induction over the number of points of discontinuity of f .If f has no point of discontinuity, take g = f and ϕ = 0.Suppose the result is true for piecewise continuous functions having n points of dis-

continuity and let f be discontinuous at n+ 1 points. Let c be one of them. We define

h(x) =

f(x) if x ∈ [a, c[limx→c− f(x) if x = c

f(x)− limx→c+ f(x) + limx→c− f(x) if x ∈ ]c, b] .

Then h is piecewise continuous on [a, b] and has at most n points of discontinuity. More-over, ϕ := f −h is a step function on [a, b]. By hypothesis, h may be written as h = g+ψfor some g ∈ C [a, b] and ψ ∈ E [a, b]. Hence, f = g︸︷︷︸

∈C[a,b]

+ (ϕ+ ψ)︸︷︷︸∈E[a,b]

.

Theorem 3.38. Any f ∈ CP [a, b] is the uniform limit on [a, b] of a sequence of stepfunctions.

Proof. By the preceding lemma, we may assume f ∈ C [a, b]. Indeed, if f = g + ϕ andϕn ⇒ g, then ϕn + ϕ⇒ f .

So let f ∈ C [a, b]. By Heine’s theorem, f is uniformly continuous on [a, b], so forε = 1

n , ∃δn > 0 such that ∀x, y ∈ [a, b] : |x− y| ≤ δn =⇒ |f(x)− f(y)| ≤ 1n .

Let a = a0 < a1 < . . . < ap = b be a subdivision of [a, b] such that ak+1 − ak ≤ δn andlet ψn be the unique step function on [a, b] such that ψn(ap) = f(ap) and ψn(t) = f(ak)for t ∈ [ak, ak+1[, 0 ≤ k ≤ p− 1. Then

∀t ∈ [ak, ak+1[ : |f(t)− ψn(t)| = |f(t)− f(ak)| ≤ 1/n , 0 ≤ k ≤ p− 1 .

Since the inequality is clearly true for t = ap, the proof is complete.

3.5. Approximation theorems 43

Definition 3.39. A function f : [a, b]→ C is piecewise linear if there exists a subdivisiona = a0 < a1 < . . . < an = b of [a, b] such that f is linear on each interval ]ai−1, ai[.

Theorem 3.40. Any f ∈ C [a, b] is the uniform limit on [a, b] of a sequence (fn) ofpiecewise linear continuous functions satisfying fn(a) = f(a) and fn(b) = f(b) for alln ∈ N.

Proof. Let f ∈ C [a, b], so f is uniformly continuous on [a, b]. Hence, for ε = 1n we may

find δn > 0 such that for any x, y ∈ [a, b], if |x− y| ≤ δn, then |f(x)− f(y)| ≤ 1n .

Let a = a0 < a1 < . . . < ap = b be a subdivision of [a, b] such that ak+1 − ak < δn andlet gn be the unique piecewise linear function on [a, b] such that gn(ak) = f(ak) for all k.

We may express gn as follows :

gn(t) = λt,kf(ak) + (1− λt,k)f(ak+1) if t ∈ [ak, ak+1] , 0 ≤ k ≤ p− 1 ,

where λt,k = ak+1−tak+1−ak . Then gn(a) = f(a), gn(b) = f(b) and

|f(t)− gn(t)| = |λt,kf(t) + (1− λtk)f(t)− gn(t)|

≤ λt,k|f(t)− f(ak)|+ (1− λt,k)|f(t)− f(ak+1)| ≤ 1n.

Corollary 3.41. Any f ∈ CP [a, b] is the uniform limit on [a, b] of a sequence of piecewiselinear functions.

Theorem 3.42. Let f ∈ C [0, 1]. The sequence of polynomials (Bn) defined by

Bn(x) =n∑k=0

(n

k

)xk(1− x)n−kf

(kn

)converges uniformly to f on [0, 1].

We call Bn the nth Bernstein polynomial associated to f .

Proof. Let ε > 0. By Heine’s theorem, f is uniformly continuous, so there exists δ > 0such that for any x, y ∈ [0, 1] satisfying |x − y| ≤ δ, we have |f(x) − f(y)| ≤ ε

2 . On theother hand, if we put M = maxx∈[0,1] |f(x)|, we always have |f(x) − f(y)| ≤ 2M . Thus,for any x ∈ [0, 1] and n ∈ N,

– if k satisfies∣∣ kn − x

∣∣ ≤ δ, then ∣∣f(x)− f( kn)∣∣ ≤ ε

2 ,– if k satisfies

∣∣ kn − x

∣∣ ≥ δ, then ∣∣f(x)− f( kn)∣∣ ≤ 2M ≤ 2M (k−nx)2

n2δ2 , since in this case,(k−nx)2

n2δ2 ≥ 1.Hence, for any index k we have

∣∣f(x)− f( kn)∣∣ ≤ ε

2 + 2M (k−nx)2

n2δ2 . Thus,

∀x ∈ [0, 1] : |f(x)−Bn(x)| =∣∣∣ n∑k=0

(n

k

)xk(1− x)n−k

{f(x)− f

(kn

)}∣∣∣≤

n∑k=0

(n

k

)xk(1− x)n−k

(ε2 + 2M (k − nx)2

n2δ2

),

where in the first equality, we used the fact that∑nk=0

(nk

)xk(1−x)n−k = 1 by the binomial

theorem. We now note that

(k − nx)2 = k(k − 1) + (1− 2nx)k + n2x2 .


But k(nk

)= k n!

k!(n−k)! = n(n−1k−1)if k ≥ 1, and 0 if k = 0. Hence,

n∑k=0

k

(n

k

)xk(1− x)n−k =

n∑k=1

n

(n− 1k − 1

)xk(1− x)n−k = n

n−1∑j=0

(n− 1j

)xj+1(1− x)n−j−1

= nxn−1∑j=0

(n− 1j

)xj(1− x)(n−1)−j = nx .

Similarly, k(k − 1)(nk

)= n(n− 1)

(n−2k−2)if k ≥ 2 and 0 if k = 0, 1. Hence,

n∑k=0

k(k − 1)(n

k

)xk(1− x)n−k =

n∑k=2

n(n− 1)(n− 2k − 2

)xk(1− x)n−k

= n(n− 1)n−2∑j=0

(n− 2j

)xj+2(1− x)n−j−2 = n(n− 1)x2 .

We thus showed thatn∑k=0

(k−nx)2(n

k

)xk(1−x)n−k = n(n−1)x2 +(1−2nx)nx+n2x2 = −nx2 +nx = nx(1−x) ,

so

∀x ∈ [0, 1] : |f(x)−Bn(x)| ≤ ε

2 + 2Mn2δ2 · nx(1− x) = ε

2 + 2Mx(1− x)nδ2 ≤ ε

2 + 2Mnδ2 ,

since x ∈ [0, 1]. Choosing n0 such that 2Mnδ2 ≤ ε

2 for all n ≥ n0, we obtain the result.

Corollary 3.43 (Weierstrass approximation theorem). Any f ∈ C [a, b] is the uniformlimit on [a, b] of a sequence of polynomial functions.

Proof. Let ϕ : [0, 1] → R be given by ϕ(x) = a + (b − a)x. Then f ◦ ϕ ∈ C [0, 1], so it isthe uniform limit of a sequence (Bn) of associated Bernstein polynomials. Hence, f is theuniform limit of the sequence Bn ◦ ϕ−1, which is indeed a sequence of polynomials sinceBn ◦ ϕ−1(x) = Bn(x−ab−a ).

3.6 Exercises

1. Study the pointwise and uniform convergence of the following sequences of functions:

• fn(x) = xn on [0, 1] • fn(x) ={n2x if |x| ≤ 1

n ,1x if |x| > 1

n .• fn(x) = nαxe−nx on R+

• fn(x) = x

1 + nxon [0, 1] • fn(x) = 1

1 + nxon [0, 1] • fn(x) = ln(1 + nx)

1 + nxon R+

• fn(x) = (1− x)xn on [0, 1] • fn(x) = nx2e−nx on [0, π] • fn(x) = e−nx on [0, π]

• fn(x) = (1− |x|)n on ]− 1, 1[ • fn(x) = sinn(x) on [0, π] • fn(x) = 1n

sin(nx) on R

• fn(x) = x2 exp(− sin x

n

)on R • fn(x) = sin(nx)

n√x

on R+ • fn(x) = e−xn∑k=0

xk

k! on R+

3.6. Exercises 45

2. Let (fn)n∈N be a sequence of functions on I = [a, b]. Suppose that fn ⇒ f sur I, i.e. fnconverges uniformly to a function f on I. Are the following assertions correct? Justifyyour answer.(1) If f is discontinuous at a ∈ I, then ∀n ∈ N, fn is discontinuous at a.(2) If f is discontinuous at a ∈ I, then ∃n0 ∈ N such that, ∀n ≥ n0, the function fn is

discontinuous at a.(3) If f is continuous at a ∈ I, then ∀n ∈ N, fn is continuous at a.(4) If f is continuous at a ∈ I, then ∃n0 ∈ N such that, ∀n ≥ n0, the function fn is

continuous at a.(5) If fn is continuous and Fn is a primitive of fn, then Fn converges uniformly to a

primitive of f .(6) If fn : I → J and φ : J → R is continuous, then φ ◦ fn ⇒ φ ◦ f on I.(7) If fn : I → J and φ : J → R is uniformly continuous, then φ ◦ fn ⇒ φ ◦ f on I.

3. Let fn(x) =

0 if x < 1

n+1 ,

sin2 πx if 1

n+1 ≤ x ≤1n ,

0 if 1n < x.

Show that (fn) converges, but not uniformly, to a continuous function.Use the series

∑fn to show that absolute convergence, even at each point of R, does

not necessarily imply uniform convergence on R.4. (Continuous but differentiable nowhere). Put

ϕ(x) = |x|, −1 ≤ x ≤ 1

and extend the definition of ϕ(x) to all x ∈ R by defining

ϕ(x+ 2) = ϕ(x) .

Letf(x) =

∞∑n=0

(34)nϕ(4nx) .

Show that f is continuous on R but not differentiable at any point of R.5. (Dini Theorems).

(1) First Theorem: let fn : [a, b]→ R be a monotone sequence of continuous functions(i.e. fn+1 ≥ fn or fn+1 ≤ fn) which converge pointwise to a function f . If f iscontinuous, show that (fn)n∈N converges uniformly to f .

(2) Application: let P0(x) = 0 and Pn+1(x) = Pn(x) + x2−P 2n(x)

2 for n ∈ N. Show that(Pn)n∈N converges uniformly to the function |x| on [−1, 1].

(3) Second Theorem: let fn : [a, b] → R be a sequence of increasing functions (i.e.x ≤ y implies fn(x) ≤ fn(y)) which converge pointwise to a function f . If f iscontinuous, show that (fn)n∈N converges uniformly to f .

(4) Application: let fn(x) = (1 + xn)n. Show that (fn)n∈N converges uniformly to the

function ex on each interval [−a, a], where a > 0.6. Let fn(x) = x

1+nx2 for x ∈ R. Show that (fn) converges uniformly on R to a functionf , and that the equation f ′(x) = limn→∞ f

′n(x) is true for x 6= 0, but false for x = 0.


7. Let (fn)n∈N be a sequence of real-valued differentiable functions on [a, b]. Suppose thatfn converges pointwise to a function f and that there exists M > 0 such that ∀n ∈ N,∀x ∈ R, |f ′n(x)| ≤ M . Show that the convergence is uniform on [a, b]. Is the functionf necessarily differentiable on [a, b]?

8. Show that the series∑∞n=1(−1)n x2+n

n2 converges uniformly on any bounded interval ofR, but does not converge absolutely for any value of x.

9. Let fn(x) = 1x2+n2 for x ∈ R, n ∈ N.

(1) Show that∑n∈N fn(x) converges normally on R. Denote its sum by S(x).

(2) Show that S(x) is a continuous function on R.(3) Show that limx→∞ S(x) = limx→−∞ S(x) = 0.

10. Let fn(x) = x(1+x2)n for x ∈ R, n ∈ N.

(1) Show that∑n∈N fn(x) converges pointwise on R. Denote its sum by S(x).

(2) Show that∑n∈N fn(x) converges normally on [a,+∞[ and ]−∞,−a] for any a > 0.

Deduce that the function S(x) is continuous on ]−∞, 0[ ∪ ]0,∞[.(3) Calculate S(x) for each x ∈ R. Is the function S(x) continuous on R?

11. Consider∑fn, where fn is defined below. In each case,

(1) show that the series converges pointwise on I ;(2) determine the intervals on which the series converges normally ;(3) show that the limit function is continuous on I ;(4) study the series of derivatives ;(5) is the sum differentiable ?

• fn(x) = e−n(x2+1), I = R • fn(x) = x

n2 + x, I = R+

• fn(x) = x2e−nx, I = R∗+ • fn(x) = sin(x

2n), I = R

• fn(x) = 1n2(x+ n) , I = R+ • fn(x) = xe−n

2x

1 + n2x, I = R+

12. By considering the series of functions S(x) =∑n≥0(x2 )n, calculate

∑n≥0

n2n .

13. Put fn(x) = nxe−nx for x ∈ R+ et n ∈ N. Show that∑fn converges pointwise on R+.

Study the uniform convergence on R+ and [ε,+∞[, where ε > 0.14. Let fn(x) = (−1)n

n+x2 for x ∈ R, n ∈ N.(1) Show that

∑fn converges uniformly on R. We put S(x) =

∑∞n=1 fn(x).

(2) Deduce that limx→∞ S(x) = 0.(3) Show that

∑f ′n converges normally on any interval [−a, a], a > 0.

(4) Deduce that S is of class C1 on R.15. Are the following assertions correct ? Justify your answer.

(1) If f : R∗+ → R is continuous on [ε,+∞[ for any ε > 0, then f is continuous on R∗+.(2) If h : R∗+ → R is bounded on [ε,+∞[ for any ε > 0, then h is bounded on R∗+.(3) If

∑fn converges normally on [ε,+∞[ for any ε > 0, then

∑fn converges normally

on R∗+.16. Let fn(x) = 1

n(1+nx) for x ∈ R+ and n ≥ 1.

3.6. Exercises 47

(1) Show that∑fn converges pointwise on R∗+. Does it converge at 0?

For x > 0, we put S(x) =∑∞n=1 fn(x) and Sn(x) =

∑nk=1 fk(x).

(2) Show that∑fn is not uniformly convergent on R∗+.

(3) Show that S(x) is nonnegative, decreasing, and ∀n ≥ 1, x ∈ R∗+, Sn(x) ≤ S(x).(4) Show that if Hn =

∑nk=1

1k , we have Sn( 1

n) ≥ 12Hn.

(5) Deduce that limn→∞ S( 1n) = +∞. Deduce that limx→0 S(x) = +∞.

Thus, we see that we can sometimes interchange the limit with the summation, evenif we cannot apply the general theorem directly.

17. Study the pointwise, normal and uniform convergence of∑fn in the following cases:

(1) fn(x) = xe−nx

lnn on R+.(2) fn(x) = 1

n+n2(x−n)2 on R.

(3) fn(x) = (−1)nnx+(−1)n on [1,+∞[ (Hint: fn(x) =

(fn(x)− (−1)n

nx

)+ (−1)n

nx ).In each case, study the limit as x→ +∞.

18. Determine the domain of definition D of S(x) :=∑∞n=1(−1)n ln

(1+ x

n

). Is the function

of class C1 on D?19. Let (an) be nonnegative. Show that

∑an converges iff

∑an cos(nx) converges uni-

formly on ]0, 2π[.Deduce that if 0 < p ≤ 1, then

∑ cos(nx)np does not converge uniformly on ]0, 2π[.

However, it converges pointwise on ]0, 2π[ by Abel’s test.20. Show that for any a > 0,

∫ 10

11+xa dx =

∑∞n=0

(−1)n1+na .

21. Show that the function f(x) =∑∞n=1

sin(2nx)2n is not differentiable at 0. (Hint: near 0,

we may find C > 0 such that sin y ≥ Cy).22. (Sophomore’s dream). Show that

∫ 10 x−x dx =

∑∞n=1 n

−n and∫ 1

0 xx dx =

∑∞n=1(−1)n+1n−n.

Chapter 4

Power Series

4.1 Power series and radius of convergenceDefinition 4.1. A power series is a series of functions of the form

∑anz

n, where z is avariable in K = R or C, and (an) is a numerical sequence.

Definition 4.2. The radius of convergence of the power series∑anz

n is the value

R := sup Ia, where Ia = {r ∈ R+ : (anrn) is bounded } .

Note that Ia 6= ∅ because 0 ∈ Ia. So R exists; it is either finite or +∞. Moreover,Ia = [0, R[ or [0, R]. Indeed, if r ∈ Ia, then 0 ≤ r ≤ R, hence Ia ⊆ [0, R]. Conversely, if0 ≤ r < R, then r is not an upper bound for Ia, so ∃ r̃ ∈ Ia such that r < r̃, so (anr̃n) isbounded, hence (anrn) is bounded and r ∈ Ia. Hence, [0, R[ ⊆ Ia.

Example 4.3. 1. The power series∑anz

n,∑|an|zn and

∑λanz

n with λ ∈ C∗ have thesame radius of convergence.

2. The series∑ zn

n! has R = +∞, since rn

n! → 0 for any r ∈ R+.3. The series

∑nnzn has R = 0, since nnrn is unbounded for any r 6= 0.

4. The series∑ zn

n has R = 1, since rn

n is bounded for 0 ≤ r ≤ 1 and unbounded for r > 1.5. The series

∑n(−1)nzn has R = 1, because |n(−1)nrn| ≤ nrn → 0 if 0 ≤ r < 1, and for

r = 1, n(−1)n is unbounded. Hence, R = sup [0, 1[ = 1.

Lemma 4.4. If R is the radius of convergence of the power series∑anz

n, then�∑anz

n is absolutely convergent for any z such that |z| < R,�∑anz

n diverges for any z such that |z| > R.

Proof. � If |z| < R, choose r such that |z| < r < R. Since [0, R[ ⊆ Ia, we have r ∈ Ia,so (anrn) is bounded, say |anrn| ≤M . Hence,

|anzn| = |anrn|( |z|r

)n≤M

( |z|r

)n.

But∑M( |z|r

)n = M1−|z|/r converges, hence

∑|anzn| converges.

� If |z| > R, then |z| /∈ Ia, so (an|z|n) is unbounded. Hence, anzn 9 0 and∑anz

n

diverges.

Remark 4.5. The radius of convergence R of∑anz

n is thus the unique R ∈ R+ =R+ ∪ {+∞} such that

50 Chapter 4. Power Series

� if |z| < R, then∑|anzn| converges,

� if |z| > R, then∑|anzn| diverges.

(R satisfies this by Lemma 4.4. If there were two R1 < R2 satisfying this, then forR1 < |z| < R2, we would have

∑|anzn| convergent and divergent, a contradiction).

Remark 4.6. The behavior of the series is unpredictable when |z| = R < +∞. Forexample, the series S0(z) =

∑zn, S1(z) =

∑ zn

n and S2(z) =∑ zn

n2 have all a radius ofconvergence equal to 1. But S0(z) diverges at all points with |z| = 1, S1(z) diverges atz = 1 and converges at z = −1, and S2(z) converges at all points with |z| = 1.

Lemma 4.7. Let∑anz

n be a power series with radius of convergence R. If

n

√|an| → L ∈ R+ or

∣∣∣an+1an

∣∣∣→ L ∈ R+ ,

then R = 1L . Here, 1

L = 0 if L = +∞ and 1L = +∞ if L = 0.

Proof. Fix z ∈ K and put un = anzn. Then n

√|un| = n

√|an| · |z| → L |z|. If L = 0, then∑

|anzn| converges ∀z by the root test, so R = +∞. If L = +∞, then∑|anzn| diverges

∀z 6= 0, so R = 0. If 0 < L < +∞, then∑|anzn| converges if |z| < 1

L and diverges if|z| > 1

L . Hence, R = 1L . The second case is proved similarly using the ratio test.

Example 4.8. The series∑ n

3n zn has a radius of convergence equal to 3, since n

√n3n →

13 .

Lemma 4.9. Let∑anz

n and∑bnz

n be two power series with radii of convergence Raand Rb, respectively.1. If an = O(bn), then Ra ≥ Rb.2. If an = o(bn), then Ra ≥ Rb.3. If an ∼ bn, then Ra = Rb.4. If Ra+b is the radius of convergence of

∑(an + bn)zn, then Ra+b ≥ min(Ra, Rb). If

moreover Ra 6= Rb, then Ra+b = min(Ra, Rb).5. If cn =

∑nk=0 akbn−k, then

∑cnz

n =(∑

anzn)(∑

bnzn)for any |z| < min(Ra, Rb). If

Rab is the radius of convergence of∑cnz

n, then Rab ≥ min(Ra, Rb).

Proof. 1. Let Ia = {r ∈ R+ : (anrn) is bounded }, Ib = {r ∈ R+ : (bnrn) is bounded }and r ∈ Ib. Then (bnrn) is bounded, say |bnrn| ≤ M . But ∃n0 ∈ N, C > 0 such that|an| ≤ C|bn| ∀n ≥ n0. Hence, |anrn| ≤ M ′ := max{|a0|, |a1r|, . . . , |an0−1r

n0−1|, CM}.Hence, (anrn) is bounded and r ∈ Ia. Thus, Ib ⊆ Ia and Rb ≤ Ra.

2. If an = o(bn) then an = O(bn).3. If an ∼ bn, then an = O(bn) and bn = O(an).4. If |z| < min(Ra, Rb), then

∑anz

n and∑bnz

n converge absolutely, so∑

(an + bn)znconverge absolutely, so |z| ≤ Ra+b. Thus, min(Ra, Rb) = sup|z|<min(Ra,Rb) |z| ≤ Ra+b.If moreover Ra 6= Rb, say Ra < Rb, then for z satisfying Ra < |z| < Rb, the series∑

(an + bn)zn diverges since it is a sum of a divergent series and a convergent series.Hence, |z| ≥ Ra+b. Hence, Ra = inf |z|>Ra |z| ≥ Ra+b and thus min(Ra, Rb) = Ra ≥Ra+b.

5. If |z| < min(Ra, Rb), then∑anz

n and∑bnz

n converge absolutely. Let un = anzn

and vn = bnzn. The Cauchy product

∑wn, where wn =

∑nk=0 ukvn−k converges

absolutely and equals(∑

un)(∑

vn). But wn =

∑nk=0 akz

kbn−kzn−k = zncn, so

∑cnz

n

converges absolutely and equals(∑

anzn)(∑

bnzn). In particular, |z| ≤ Rab and thus

min(Ra, Rb) = sup|z|<min(Ra,Rb) |z| ≤ Rab.

4.2. Properties of the sum function 51

Example 4.10. The power series∑n≥1(1 + 1

2 + . . . + 1n)tn has a radius of convergence

equal to 1. Indeed, 1 ≤ (1 + 12 + . . . + 1

n) ≤ n, so n√

1 ≤ n

√(1 + 1

2 + . . .+ 1n) ≤ n

√n and

thus n

√(1 + 1

2 + . . .+ 1n)→ 1 by the Sandwich theorem. Actually, this is just the Cauchy

product of the series∑ant

n and∑tn, where a0 = 0 and an = 1

n for n ≥ 1. Hence, fort ∈ ]−1, 1[ we have

∑n≥1

(1 + 1

2 + . . .+ 1n

)tn =

(∑n≥0

tn)(∑

n≥1

1ntn)

= ln(1− t)t− 1 ,

where we used the Taylor series of ln(1− t) in the last equality.

Theorem 4.11. The power series∑n≥0 anz

n and the series of derivatives∑n≥1 nanz

n−1

have the same radius of convergence.

Proof. Let R and R′ be the radii of convergence of∑n≥0 anz

n and∑n≥1 nanz

n−1 and putI = {r ∈ R+ : (anrn) is bounded } and I ′ = {r ∈ R+ : (nanrn−1) is bounded }.

If r ∈ I ′, then for n ≥ 1, |anrn| ≤ n|anrn| = r|nanrn−1| ≤ rM for some M . Hence,|anrn| ≤M ′ ∀n, where M ′ = max(|a0|, rM). Hence, r ∈ I and I ′ ⊆ I. Thus, R′ ≤ R.

Suppose R′ < R and let r1, r2 such that R′ < r1 < r2 < R. Then |nanrn−11 | =

(nrn−11rn2

)|anrn2 | ≤ (nrn−11rn2

)M ′ for some M ′, since r2 < R. But nrn−11rn2

= nr1

( r1r2

)n → 0 since

r1 < r2. Hence, nrn−11rn2

is bounded by an M ′′. Thus, |nanrn−11 | ≤ M ′M ′′, which is absurd

since r1 > R′. Thus, R′ = R.

4.2 Properties of the sum functionWe shall suppose in this section that

∑anz

n is a power series with radius of convergenceR 6= 0 and sum function S(z).

Definition 4.12. We define the disk of convergence of∑anz

n as the open ball D =BR(0) = {z ∈ K : |z| < R}.

If the variable z is real, this disk is just the interval ]−R,R[.

Theorem 4.13. A power series converges normally, hence uniformly, on any closed ballwith center 0 contained in the open disk of convergence.

Proof. Let Br(0) ⊆ D. Then r < R. But supz∈Br(0) |anzn| = |an|rn and

∑|anrn| con-

verges because r < R. Hence,∑anr

n converges normally on Br(0).

Remark 4.14. 1. In general, the series does not converge uniformly on D. For example,∑zn has radius of convergence 1 and does not converge uniformly on its disk, since zn

does not converge uniformly to 0 on this disk.2. Actually, if the series converges uniformly on D, then it converges uniformly on D ={z ∈ K : |z| ≤ R}. Indeed, using the Cauchy criterion, uniform convergence on Dmeans that

∀ε > 0 ∃n0 ∈ N : n ≥ n0, p ≥ 0 =⇒ ∀z ∈ D : |anzn + . . .+ an+pzn+p| ≤ ε .

Fix n, p and z0 on the boundary of D, i.e. |z0| = R. By continuity of the mapz 7→ |anzn + . . . + an+pz

n+p|, by letting z → z0, we obtain the same inequality at z0.Hence, uniform convergence holds on all D.


3. If there exists z0 with |z0| = R at which the series is absolutely convergent, then we havenormal convergence on D. Indeed, supz∈D |anz

n| = |an||z0|n and∑|anzn0 | converges

by hypothesis.

Theorem 4.15. The sum function S(z) is continuous on its disk of convergence.

Proof. Let z0 ∈ D and r = R+|z0|2 . Then |z0| < r < R, so z0 ∈ Br(0) ⊆ D. But the series

converges uniformly on Br(0). Since each function anzn is continuous, it follows that Sis continuous on Br(0), in particular at z0. Since this holds for arbitrary z0 ∈ D, thefunction S is continuous on D.

Corollary 4.16. For all p ≥ 0, the sum function S(z) admits an expansion of order pnear 0, which takes the form S(z) = a0 + a1z + . . .+ apz

p + O(zp+1).

Proof. We have S(z) = a0 + a1z + . . . + apzp +

∑n≥p+1 anz

n. But∑n≥p+1 anz

n =zp+1∑

n≥p+1 anzn−p−1 and

∑n≥p+1 anz

n−p−1 = ap+1 + ap+2z + . . . is a power series, so itis continuous on its disk of convergence. Hence, there exists a neighborhood of 0 in which|∑n≥p+1 anz

n| ≤ C|zp+1|, where C = |ap+1|+ 1. Hence,∑n≥p+1 anz

n = O(zp+1).

For differentiation and integration, we suppose from now on that we have a powerseries S(x) =

∑anx

n of a real variable, i.e. K = R, with radius of convergence R > 0.

Theorem 4.17. If [a, b] ⊂ ]−R,R[, then∫ b

aS(t) dt =

∑n≥0

an

∫ b

atn dt, .

In particular, for any 0 < x < R,∫ x

0S(t) dt =

∑n≥0

ann+ 1x

n+1 .

Proof.∑anx

n converges uniformly on [a, b] and each anxn is continuous, so we may

integrate term by term by a theorem in Chapter 3.

Example 4.18. We may thus recover the Taylor series of ln and arctan :

ln(1 + x) =∫ x

0

11 + t

dt =∫ x

0

∑n≥0

(−1)ntn dt =∑n≥0

(−1)n

n+ 1 xn+1 =

∑n≥1

(−1)n−1

nxn ,

arctan(x) =∫ x

0

11 + t2

dt =∫ x

0

∑n≥0

(−1)nt2n dt =∑n≥0

(−1)n

2n+ 1x2n+1 ,

both expansions are valid on ]−1, 1[.

Theorem 4.19. The sum function S : ]−R,R[ → C is of class C∞, and its derivativesare given by

S(k)(x) =∑n≥k

n(n− 1) . . . (n− k + 1)anxn−k .

In particular,S(k)(0) = k!ak .

4.3. Problems on the boundary 53

Proof. Let fn(x) = anxn. The series

∑fn converges pointwise on ]−R,R[, each fn is of

class C∞ and∑f

(k)n , which also has radius of convergence R by Theorem 4.11, converges

uniformly on any [−r, r] ⊆ ]−R,R[. Let x0 ∈ ]−R,R[. Then x0 ∈ [−r, r] for r = R+|x0|2 .

Hence, S is of class C∞ on [−r, r] by the theorem. Since any point in ]−R,R[ has aneighborhood [−r, r] on which S is of class C∞, it follows that S is of class C∞ on ]−R,R[with the given formula for S(k).

Finally, we have S(k)(x) = k!ak+ (k+1)!1! ak+1x+ (k+2)!

2! ak+2x2+. . ., so S(k)(0) = k!ak.

4.3 Problems on the boundary

Theorem 4.20 (Abel). If∑n≥0 an converges, then the function S defined by

S(x) =∑n≥0

anxn, −1 < x < 1 ,

satisfieslimx→1−

S(x) =∑n≥0

an .

This result is immediate if an ≥ 0 for all n, since in this case∑anx

n converges normallyon [−1, 1]. In the general case, the proof goes as follows:

Proof. Put Sn =∑nk=0 ak if n ≥ 0 and S−1 = 0. Then

m∑n=0

anxn =

m∑n=0

(Sn − Sn−1)xn =m∑n=0

Snxn −

m∑n=1

Sn−1xn

=m∑n=0

Snxn −

m−1∑n=0

Snxn+1 = Smx

m + (1− x)m−1∑n=0

Snxn .

Taking m→ +∞ we get

S(x) = (1− x)∑n≥0

Snxn (?)

because Sm converges and xm → 0 for |x| < 1. Let ε > 0 and put S =∑n≥0 an. There

exists n0 such that |S − Sn| ≤ ε/2 for all n ≥ n0. Hence, noting that (1− x)∑n≥0 x

n = 1for |x| < 1, we obtain by (?) that for any 0 ≤ x < 1,

|S(x)− S| =∣∣∣(1− x)

∑n≥0

(Sn − S)xn∣∣∣ ≤ (1− x)

n0−1∑n=0|Sn − S|xn + (1− x)

∑n≥n0

ε

2xn

≤ (1− x)n0−1∑n=0|Sn − S|xn + ε

2(1− x)∑n≥0

xn = (1− x)n0−1∑n=0|Sn − S|xn + ε

2 .

As x→ 1−, this tends to ε/2, so ∃δ > 0 : x ∈ ]1− δ, 1[ =⇒ |S(x)− S| ≤ ε.

Corollary 4.21. If S(x) =∑anx

n has radius of convergence R, and if∑anR

n converges,then S is continuous on ]−R,R].


Proof. Let bn = anRn and f(t) =

∑bnt

n. By Theorem 4.20, limt→1− f(t) =∑bn. Let

ε > 0. There is thus δ > 0 such that

t ∈]1− δ, 1[ =⇒∣∣∣∑ bnt

n −∑

bn∣∣∣ ≤ ε =⇒

∣∣∣∑ an(Rt)n −∑

anRn∣∣∣ ≤ ε .

Put x = Rt. Then

x ∈]R−Rδ,R[ =⇒ t ∈]1− δ, 1[ =⇒∣∣∣∑ anx

n −∑

anRn∣∣∣ ≤ ε .

We thus showed that limx→R− S(x) = S(R), so S is continuous at R. Since we alreadyknow it is continuous on ]−R,R[, the corollary is proved.

4.4 Usual functionsDefinition 4.22. The power series

∑n≥0

zn

n! has radius of convergence R = +∞. Its sumis called the complex exponential function, and denoted exp : z 7→ ez.

Theorem 4.23. Starting from the series ez =∑ zn

n! , we may recover the following iden-tities, without using any geometric arguments :(1) ∀z1, z2 ∈ C, ez1+z2 = ez1ez2.(2) ez 6= 0 and 1

ez = e−z.(3) ez = ez and |ez| = eRe(z).(4) d

dz ez = ez.

(5) The restriction of exp to R is positive, increasing and satisfies ex → ∞ as x → ∞and ex → 0 as x→ −∞.

(6) There exists a number π such that eπi/2 = i and ez = 1 iff z2πi ∈ Z.

(7) exp is periodic, with period 2πi.(8) Let S1 be the unit circle. The map f : R→ S1 defined by f(t) = eit is surjective.(9) The map exp : C→ C∗ defined by exp(z) = ez is surjective.(10) There exists a unique continuous function L defined on C\R− such that L(1) = 0 and

∀z ∈ C \ R−, eL(z) = z. This function is called the principal value of the logarithmand coincides with the function ln on R∗+.

Proof. Admitted, see the prologue of [15]. For (10), see [2].

Definition 4.24. We call complex cosine and sine, denoted respectively by cos and sin,the maps from C to C defined by

∀z ∈ C, cos z = eiz + e−iz

2 =∑n≥0

(−1)n z2n

(2n)! ,

∀z ∈ C, sin z = eiz − e−iz

2i =∑n≥0

(−1)n z2n+1

(2n+ 1)! .

We call complex hyperbolic cosine and sine, denoted respectively by cosh and sinh (orch and sh), the maps from C to C defined by

∀z ∈ C, cosh z = ez + e−z

2 =∑n≥0

z2n

(2n)! ,

∀z ∈ C, sinh z = ez − e−z

2 =∑n≥0

z2n+1

(2n+ 1)! .

4.5. Analytic functions 55

We call complex tangent and cotangent, denoted respectively by tan and cot, the mapsdefined by

∀z /∈ π2 + πZ, tan z = sin zcos z , ∀z /∈ πZ, cot z = cos z

sin z = tan(π

2 − z).

Lemma 4.25. For all z, a, b ∈ C we have

cos(iz) = cosh z cosh(iz) = cos zsin(iz) = i sinh z sinh(iz) = i sin zeiz = cos z + i sin z ez = cosh z + sinh ze−iz = cos z − i sin z e−z = cosh z − sinh zcos2 z + sin2 z = 1 cosh2 z − sinh2 z = 1cos(a+ b) = cos a cos b− sin a sin b cosh(a+ b) = cosh a cosh b+ sinh a sinh bsin(a+ b) = sin a cos b+ cos a sin b sinh(a+ b) = sinh a cosh b+ cosh a sinh b

Proof. Exercise.

Lemma 4.26. (1) cos is even, periodic, of period 2π. Moreover, ∀z ∈ C, cos(z + π) =− cos z and cos(π2 − z) = sin z.

(2) sin is odd, periodic, of period 2π. Moreover, ∀z ∈ C, sin(z + π) = − sin z and sin(π2 −z) = cos z.

(3) cosh is even, periodic, of period 2πi. Moreover, ∀z ∈ C, cosh(z + πi) = − cosh z andcosh(z + iπ2 ) = i sinh z.

(4) sinh is odd, periodic, of period 2πi. Moreover, ∀z ∈ C, sinh(z + πi) = − sinh z andsinh(z + iπ2 ) = i cosh z.

Proof. Exercise. Use Theorem 4.23.

4.5 Analytic functions

Definition 4.27. Let X ⊆ K. We say that f : X → C is analytic near 0 if there exists apower series

∑anz

n with radius of convergence R > 0 and an open ball Br(0) ⊆ X withradius 0 < r ≤ R such that f(z) =

∑anz

n on Br(0).We say that f is analytic near z0 ∈ C if the function g(z) = f(z + z0) is analytic near

0.We say that f is analytic on X if f is analytic near any point in X.

For example, the function f(z) = 11−z , defined on C \ {1}, is analytic near 0 because

f(z) =∑zn on B1(0). More generally, we have

Lemma 4.28. The function f(z) = 1(1−z)k+1 is analytic near 0 for any k ∈ N. Moreover,

1(1− z)k+1 =

∑n≥0

(n+ k

k

)zn on B1(0) .

This lemma is easy to prove if the variable is real, by differentiating 11−x =

∑xn and

using Theorem 4.19. For complex variables, we argue by induction.


Proof. The coefficient(n+kk

)is a polynomial of degree k in n, so

(n+kk

)rn → 0 if 0 ≤ r < 1,

and(n+kk

)rn →∞ if r ≥ 1. Hence, the radius of convergence of the above series is 1; it is

thus well defined on B1(0).We already know the result for k = 0; this is just a geometric series. Suppose the

result is true for k. Write 1 − z =∑anz

n with a0 = 1, a1 = −1 and ak = 0 for k ≥ 2,and put bn =

(n+k+1k+1

). Then the Cauchy product of

∑anz

n with∑bnz

n has generalterm cn = a0bn + a1bn−1 =

(n+k+1k+1

)−(n+kk+1

)= (n+k+1)!

(k+1)!n! −(n+k)!

(k+1)!(n−1)! = (n+k+1)!−n(n+k)!(k+1)!n! =

(n+k+1)(n+k)!−n(n+k)!(k+1)!n! = (n+k)!

k!n! =(n+kk

). Hence, (1 − z)

∑(n+k+1k+1

)zn+1 =

∑(n+kk

)zn =

1(1−z)k+1 , which proves the result for k + 1.

For the rest of this section, we suppose that f is a function of a real variable, i.e.X ⊆ R.

Lemma 4.29. If f is analytic near x0, then f is of class C∞ on an interval ]x0 − r, x0 + r[.Moreover, the expansion of f near x0 is given by

∑ f (n)(x0)n! (x− x0)n.

We call this the Taylor series of f at x0.

Proof. Let g(x) = f(x + x0). By hypothesis, there exists a power series∑anx

n withradius R > 0 and an interval ]−r, r[ with 0 < r ≤ R such that g(x) =

∑anx

n on ]−r, r[.But x ∈ ]x0 − r, x0 + r[ iff x − x0 ∈ ]−r, r[. Hence, f(x) = g(x − x0) =

∑an(x − x0)n

on ]x0 − r, x0 + r[. By Theorem 4.19, g is of class C∞ on ]−r, r[, so f is of class C∞ on]x0 − r, x0 + r[. Moreover, an = g(n)(0)

n! = f (n)(x0)n! .

This lemma ensures the uniqueness of the series expansion : if f(x) =∑an(x−x0)n =∑

bn(x− x0)n, then an = bn = f (n)(x0)n! . Here is an application of this :

Lemma 4.30. Let f be analytic near 0 and∑anx

n its power series.� If f is even, then ∀p ∈ N, a2p+1 = 0.� If f is odd, then ∀p ∈ N, a2p = 0.

Proof. Suppose that f(x) =∑anx

n on ]−r, r[. Then g(x) = f(−x) satisfies g(x) =∑an(−x)n =

∑(−1)nanxn. If f is even, then g(x) = f(x), so by the uniqueness of the

expansion, an = (−1)nan and thus a2p+1 = 0 ∀p. Similarly, if f is odd, g(x) = −f(x), soan = (−1)n+1an and thus a2p = 0 ∀p.

Note that the converse of Lemma 4.29 is incorrect : we shall see in Exercise 8 somefunctions of class C∞ which are not analytic near 0. The following results give us criteriathat ensure the converse is true.

Lemma 4.31. Let f : ]−a, a[ → C be of class C∞. Let Sn(x) =∑nk=0

f (k)(0)k! xk and

Rn(x) = f(x) − Sn(x). Then f is analytic near 0 iff there exists an interval ]−r, r[ onwhich Rn converges pointwise to the zero function.

Proof. If f is analytic near 0, then there exists an interval ]−r, r[ on which f(x) =∑k≥0

f (k)(0)k! xk, so for any x ∈ ]−r, r[, Sn(x)→ f(x), hence Rn(x)→ f(x)− f(x) = 0.

Conversely, suppose Rn converges pointwise to the zero function on ]−r, r[ and letx ∈ ]−r, r[. Then f(x) = limn→∞(f(x) − Rn(x)) = limn→∞ Sn(x) =

∑∞k=0

f (k)(0)k! xk, so f

is analytic near 0.

4.6. Power series and differential equations 57

Corollary 4.32. If f : ]−a, a[ → C is of class C∞ and if there exist C ≥ 0 and ω > 0such that for all x ∈ ]−a, a[ and n ∈ N we have |f (n)(t)| ≤ Cn!ωn, then f is analytic near0.

Proof. By the Taylor-Lagrange formula, we have |Rn(x)| ≤ |x|n+1

(n+1)! supθ∈[0,x] |f (n+1)(θ)| ≤C|x|n+1ωn+1. Hence, Rn converges pointwise to 0 on ]−r, r[, where r = min(a, ω−1).

Remark 4.33. The hypothesis of the corollary is satisfied in particular if the successivederivatives of f are uniformly bounded, i.e. bounded independently of x.

Remark 4.34. The converse of the corollary is also true; see Exercise 11.

4.6 Power series and differential equations

To show that a function f is analytic near a point, and to determine the expansion, itis often useful to use differential equations. Here is the general argument :1. Find a differential equation that f satisfies, and indicate initial conditions.2. Suppose y(x) =

∑anx

n is a solution of the differential equation and deduce the an andthe radius R. If R 6= 0, we will have f(x) = y(x) =

∑anx

n on ]−R,R[ by uniquenessof the solution of the differential equation.Before we give an example to clarify this method, let us state a general result on

differential equations.

Theorem 4.35 (Cauchy-Lipschitz Theorem). Let a1, . . . , an, b be continuous functions onan interval I with values in C. Let x0 ∈ I and α1, . . . , αn ∈ C. Then there exists a uniquesolution f , defined on I, to the differential equation

y(n) + a1(x)y(n−1) + . . .+ an(x)y = b(x)

which satisfies

f(x0) = α1, f ′(x0) = α2, . . . , f (n−1)(x0) = αn .

Proof. Admitted. See [1] for a proof.

Actually, we shall only need the cases n = 1 and n = 2, and we only use the uniquenessof the solution. Let us now give an example to illustrate the method described above.

Lemma 4.36. The function f(x) = (1 + x)α is analytic near 0 for any α ∈ R. Moreover,

(1 + x)α =∑n≥0

(α

n

)xn on ]−1, 1[ ,

where(α

0)

= 1 and(αn

)= α(α− 1) . . . (α− n+ 1)/n! for n ≥ 1.

Proof. We note that f satisfies on ]−1, 1[ the differential equation

(E){

(1 + x)y′ = αy,

y(0) = 1.


Suppose (E) has a solution y(x) =∑n≥0 anx

n. Then y′(x) =∑n≥1 nanx

n−1 =∑n≥0(n+

1)an+1xn. Hence,

0 = (1 + x)y′ − αy =∑n≥0

(n+ 1)an+1xn +

∑n≥0

(n+ 1)an+1xn+1 − α

∑n≥0

anxn

But∑n≥0(n+ 1)an+1x

n+1 =∑n≥1 nanx

n =∑n≥0 nanx

n. Hence,∑n≥0

((n+ 1)an+1 + (n− α)an

)xn = 0 .

By the uniqueness of the expansion, we thus have((n+ 1)an+1 + (n−α)an

)= 0 for all n,

in other words,an+1 = α− n

n+ 1 an n ≥ 0 .

By a simple calculation we deduce that an = a0(αn

). But a0 = y(0) = 1, so an =

(αn

).

Hence, y(x) =∑(α

n

)xn. If α ∈ N, then

(αn

)= 0 for all n ≥ α + 1, so the radius of

convergence of this series is R = +∞. Otherwise, |an+1||an| = |α−n|

n+1 → 1, so R = 1. In anycase, y(x) =

∑(αn

)xn is a solution of (E) which is well defined on ]−1, 1[. Since f is also

a solution of (E), we have f(x) = y(x) =∑(α

n

)xn on ]−1, 1[ by Theorem 4.35.

To conclude, note that conversely, given a differential equation, we can try to find itssolution by supposing that it takes the form of a power series

∑anx

n, then calculate thean and the radius R.

4.7 Exercises1. Let P and Q be nonzero polynomials and let α ∈ R. Calculate the radius of convergence

of the power series∑n≥2 anz

n, for

• an = αn • an = 100 cos(n) • an = e−2n

• an = coshnn

• an =(

1 + 12n

)n• an =

(2n− 1n+ 1

)2n

• an = 1nα

• an = (−1)n

nπn• an = sin

(e−(n!)

)• an = 1

n+ e−n• an = e−

√n • an = lnn

• an = en2 • an = e

√n

n3 • an = n

lnn

• an = n! • an = n!(2n)! • an = n!

nn

• an = 134n−1 • an = −n

3 − n+ 2n2 + 5 • an = lnn+ e−n

n3 + ln(1 + sin(3−n))

• an = lnn23n−2 • an = P (n)

Q(n) • a2k = 2k et a2k+1 = 0

2. Let a ∈ R and k ∈ N∗. Calculate the radius of convergence of the following powerseries :

•∑n≥0

25nzn •∑n≥0

25nz2n •∑n≥0

z3n

27n •∑n≥0

anzkn •∑n≥0

anzn,

where

4.7. Exercises 59

1. an is the sum of the squares of the divisors of n.2. an = 1 if n is prime, an = 0 otherwise.3. an is the number of couples (x, y) ∈ Z2 such that x2 + y2 ≤ n2.

3. Let∑anz

n be a power series with radius of convergence R and let k ∈ N∗ and α ∈ R.Calculate the radius of convergence of the power series with general term:

• anzkn • n

n2 + n+ 2anzn • nαanz

n

• aknzn • ane

√nzn • anz

n2

(Hint: For the last series, first consider the case where R is finite, nonzero, then studythe cases R = +∞ and R = 0).

4. Let an be a real sequence, an → a. Calculate the radius of convergence of f(x) =∑n≥0

ann! x

n, then calculate limx→∞ e−xf(x).

5. Determine the radius of convergence and the sum of the following real power series:

•∑n≥0

n2xn •∑n≥1

(n+ 2)2

(n+ 2)! xn •

∑n≥1

(−1)n

2n− 1x2n

•∑n≥1

(−1)n

2n x2n •∑n≥1

xn

1 + . . .+ n•

∑n≥0

sin(nα)n! xn

•∑n≥0

(−1)n+1nx2n+1 •∑n≥0

(coshn)xn •∑n≥1

xn

n(n+ 2)

•∑n≥0

( 1n!

n∑k=0

k · k!)xn •

∑n≥0

( n∑k=0

1k!)xn •

∑n≥1

n(−1)nxn

6. Show that if∑an,

∑bn and

∑cn converge to A, B and C, and if cn =

∑nk=0 akbn−k,

then C = AB. (Hint: use Abel’s continuity theorem).Thus, we do not need

∑an and

∑bn to converge absolutely if we already know that

their Cauchy product converges.7. Give a power series expansion near 0 for the following functions:

• 1(x− 1)(x− 2) • ln(x+ a), a > 0 • 1 + x

(1− x)3

• 1(1 + x2)(1− x) •

∫ x

0

arctan(t2)t

dt • arctan(1− x2

1 + x2

)• ex cosx • ln(x2 − 5x+ 6) • cos3 x

• 1√1− x

• arcsin x • ln(1 + x+ x2)

8. In this exercise we show that a function of class C∞ is not necessarily equal to itsTaylor series.1. Give an example of a function f of class C∞ on a domain D such that the equalityf(x) =

∑ f (n)(0)n! xn is only true on an interval which is strictly smaller than D.

2. Show that if f(x) =

e− 1x2 if x 6= 0,

0 if x = 0, then

∑ f (n)(0)n! xn converges, but f(x) 6=

∑ f (n)(0)n! xn for any x 6= 0.


3. Show that if f(x) =∑e−n sinn2x, then f is of class C∞ but

∑ f (n)(0)n! xn diverges

for all x 6= 0.9. (Bernstein Theorem). Show that if f is of class C∞ on ]−a, a[, and if f and all its

derivatives are nonnegative on ]−a, a[, then f is analytic near 0.Deduce a new proof of the fact that for −1 < x < 1 we have (1 + x)α =

∑n≥0

(αn

)xn

for any α ∈ R, where(α

0)

= 1 and(αn

)= α(α− 1) . . . (α− n+ 1)/n!

(Hint: Consider g(y) = (1− y)−β with β > 0, then deduce the general case).

10. Calculate∑∞n=1

12nn ,

∑∞n=1

(−1)n−1

n and∑∞n=0

(−1)n2n+1 . (Hint: Use the standard Taylor

series. The 2nd and 3rd series will need some justifications).11. (Converse of Corollary 4.32). Show that if f is analytic near 0, then there exist strictly

positive reals C, ω and r such that for all n ∈ N and x ∈ ]−r, r[, |f (n)(x)| ≤ Cn!ωn.12. Expand ex2−2x as a power series near 1.13. Let f(x) =

∑anx

n be a power series defined for |x| < R. Show that f is analytic on]−R,R[, i.e. f has a power series expansion near any a ∈ ]−R,R[.

14. Let α ∈ R and f(x) = cos(α arcsin x).1) Show that f is the unique solution on ]−1, 1[ of the differential equation

(E){

(1− x2)y′′ − xy′ + α2y = 0,y(0) = 1, y′(0) = 0

2) Show that if y(x) =∑anx

n is a solution of (E) on ]−R,R[, 0 < R ≤ 1, then

∀n ∈ N : an+2 = n2 − α2

(n+ 1)(n+ 2) an .

Deduce the value of an.3) Calculate the radius of convergence of

∑anx

n. Deduce a power series expansionfor f .

15. Show that the differential equation

4xy′′ + 2y′ + y = 0

has, up to a multiplicative constant, a unique solution which is analytic near 0. Givethis solution explicitly.

16. Show that the differential equation

(1 + x2)y′′ − 2y = 0

with the conditions y(0) = 0 and y′(0) = 1 has a unique solution f which is analyticnear 0. Calculate the coefficients and the radius of convergence of the power series,then make it explicit as much as possible.

17. (Tauberian theorem). Let f(x) =∑anx

n for −1 < x < 1 and suppose nan → 0. Showthat if f(x)→ ` as x→ 1−, then

∑an converges, and is equal to `.

Show that this partial converse of Abel’s theorem is false without the condition nan → 0(consider an = (−1)n.)

18. Calculate, for x ∈ R and in case of convergence, the sum∑∞n=1

xn

n(n+1) .

Chapter 5

Improper Integrals

5.1 GeneralitiesReminder 5.1. If f is a bounded function on [a, b] and if X = (xk)0≤k≤n is a subdivisionof [a, b], we put

ck(f) = infx∈[xk,xk+1]

f(x), Ck(f) = supx∈[xk,xk+1]

f(x) ,

define the Darboux sums

s(X, f) =n−1∑k=0

(xk+1 − xk)ck(f), S(X, f) =n−1∑k=0

(xk+1 − xk)Ck(f) ,

and denote s = supX s(X, f) and S = infX S(X, f). Then� A bounded function on [a, b] is said to be Riemann integrable on [a, b] if s = S. We

then denote s = S =∫ ba f(t) dt.

� Any continuous function on [a, b], and any monotone function on [a, b] is Riemannintegrable on [a, b].� If f is continuous on [a, b], it admits a primitive F , and

∫ ba f(t) dt = F (b)− F (a).

Definition 5.2. We say that a function f is locally integrable on an interval I if it isRiemann integrable on any [c, d] ⊆ I.Remark 5.3. If f is continuous or monotone on [a, b[, then it is locally integrable on [a, b[by Reminder 5.1.Definition 5.4. 1. Let f be locally integrable on [a, b[. If limx→b−

∫ xa f(t) dt exists, we

denote it by∫ ba f(t) dt, and say that the improper integral

∫ ba f(t) dt is convergent.

2. Let f be locally integrable on ]a, b]. If limx→a+∫ bx f(t) dt exists, we denote it by∫ b

a f(t) dt, and say that the improper integral∫ ba f(t) dt is convergent.

3. Let f be locally integrable on ]a, b[ and let c ∈ ]a, b[. If the two improper integrals∫ ca f(t) dt and

∫ bc f(t) dt are convergent, we denote their sum by

∫ ba f(t) dt, and say that

the improper integral∫ ba f(t) dt is convergent.

Remark 5.5. The last definition does not depend on c. Indeed, if c < c′ ∈ ]a, b[, then∫ xc f(t) dt =

∫ c′c f(t) dt +

∫ xc′ f(t) dt, where both integrals are well-defined Riemann inte-

grals. Hence,∫ xc f(t) dt has a limit as x→ b− iff

∫ xc′ f(t) dt has a limit as x→ b−. Moreover,

if this limit exists, we have∫ bc =

∫ c′c +

∫ bc′ . The same can be done on the left side, so the

improper integrals∫ ca f(t) dt and

∫ bc f(t) dt converge iff

∫ c′a f(t) dt and

∫ bc′ f(t) dt converge,

and in this case∫ ca +

∫ bc =

∫ c′a −

∫ c′c +

∫ c′c +

∫ bc′ =

∫ c′a +

∫ bc′ .

62 Chapter 5. Improper Integrals

Example 5.6. (1)∫ x

0 e−t dt = −e−t|x0 = 1 − e−x → 1 as x → +∞. Hence,

∫+∞0 e−t dt

converges and equals 1.(2) If x > 1, then

∫ x1

1t dt = ln x→ +∞ as x→ +∞. Hence,

∫+∞1

1t dt diverges.

(3) If x ∈ [0, 1[, then∫ x

01√1−t dt = −2

√1− t|x0 = 2(1 −

√1− x) → 2 as x → 1−. Hence,∫ 1

01√1−t dt converges and equals 2.

(4) Arguing as in (1),∫ 0−∞ e

t dt = 1. Hence,∫+∞−∞ e−|t| dt converges and equals 2.

Remark 5.7. If f is defined on ]a, b[, the two sides must be treated independently. Forexample,

∫ x−x t dt = 0 for any x, but

∫+∞−∞ t dt diverges because

∫+∞0 tdt diverges.

Theorem 5.8 (Riemann’s rule). Let α ∈ R.1.∫ 1

01tα dt converges iff α < 1.

2.∫+∞

11tα dt converges iff α > 1.

Proof. If α = 1, then∫ 1ε

1t dt = − ln ε → +∞ as ε → 0. Also,

∫ x1

1t dt = ln x → +∞ as

x→ +∞. Hence,∫ 1

01t dt and

∫+∞1

1t dt diverge.

If α 6= 1, then∫ 1ε

1tα dt = 1

1−α(1 − 1εα−1 ) has a finite limit as ε → 0 iff α < 1. Also,∫ x

11tα dt = 1

1−α( 1xα−1 − 1) has a finite limit as x→ +∞ iff α > 1.

Hence,∫+∞

01tα dt diverges for all α ∈ R.

5.2 Basic properties of improper integrals

Theorem 5.9. 1. (Linearity). Let f, g be locally integrable on ]a, b[ and let λ, µ ∈ R.If∫ ba f(t) dt and

∫ ba g(t) dt converge, then

∫ ba (λf(t) + µg(t)) dt converges and equals

λ∫ ba f(t) dt+ µ

∫ ba g(t) dt.

2. (Integration by parts). Let f, g ∈ C1( ]a, b[). Suppose fg has finite limits la at a and

lb at b. Then∫ ba f(t)g′(t) dt converges iff

∫ ba g(t)f ′(t) dt converges. In this case,∫ b

af(t)g′(t) dt = lb − la −

∫ b

ag(t)f ′(t) dt .

3. (Change of variables). Let f be continuous on ]a, b[ and φ : ]c, d[ → ]a, b[ be C1,bijective, with a continuous inverse. Then

∫ ba f(t) dt converges iff

∫ dc f(φ(s))|φ′(s)| ds

converges. In this case, ∫ b

af(t) dt =

∫ d

cf(φ(s))|φ′(s)|ds .

Proof. In all properties we assume a is finite and f, g are locally integrable on [a, b[. Thecase ]a, b] is similar, and the case ]a, b[ is deduced by considering ]a, c] and [c, b[.1. Given x ∈ [a, b[, by the linearity of the Riemann integral,

∫ xa (λf(t) + µg(t)) dt =

λ∫ xa f(t) dt + µ

∫ xa g(t) dt. If each integral on the RHS has a limit when x → b−, then

the LHS has a limit as asserted.2. Given x ∈ [a, b[, integrate by parts on [a, x] to get

∫ xa f(t)g′(t) dt = f(x)g(x)−f(a)g(a)−∫ x

a g(t)f ′(t) dt. Assuming limx→b f(x)g(x) = lb, then limx→b∫ xa f(t)g′(t) dt exists iff

limx→b∫ xa g(t)f ′(t) dt exists, and we have the stated equality.

5.3. Integrals of nonnegative functions 63

3. Since φ is a continuous injection, it is strictly monotone 1. Suppose φ is strictly increas-ing. Then φ(c) = a, limy→d− φ(y) = b and limx→b− φ

−1(x) = d.Let y ∈ [c, d[. By a change of variables in [a, φ(y)], we have

∫ yc f(φ(s))φ′(s) ds =∫ φ(y)

a f(t) dt. Hence, limy→d−∫ yc f(φ(s))φ′(s) ds exists iff limy→d−

∫ φ(y)a f(t) dt exists iff

limx→b−∫ xa f(t) dt exists, in which case we have the stated equality.

If φ is strictly decreasing, then φ(c) = b, limy→d− φ(y) = a and the same argumentapplies (the limits of the integral now have to be inverted, so φ′ becomes |φ′|).

Example 5.10.∫ 1

0 ln tdt converges. Indeed, f(t) = t and g(t) = ln t are C1 on ]0, 1] withf(1)g(1) = 0 and limt→0 f(t)g(t) = 0. Integrating by parts and noting that

∫ 10 1 dt = 1

converges, we get that∫ 1

0 ln t dt converges and equals 0− 0− 1 = −1.

5.3 Integrals of nonnegative functionsLemma 5.11. Let f be nonnegative, locally integrable on [a, b[. Then

∫ ba f(t) dt converges

iff the function F (x) =∫ xa f(t) dt is bounded from above on [a, b[.

Proof. Since f is nonnegative, F (x) is increasing, so it has a limit 2 at b which is finite or+∞. This limit is finite iff F is bounded from above on [a, b[.

Corollary 5.12 (Comparison test). Let f, g be nonnegative and locally integrable on [a, b[.(1) If f = O(g) near b, then the convergence of

∫ ba g(t) dt implies the convergence of∫ b

a f(t) dt, and the divergence of∫ ba f(t) dt implies the divergence of

∫ ba g(t) dt.

(2) If f = o(g) near b, then the convergence of∫ ba g(t) dt implies the convergence of∫ b

a f(t) dt, and the divergence of∫ ba f(t) dt implies the divergence of

∫ ba g(t) dt.

(3) If f ∼ g near b, then∫ ba g(t) dt converges iff

∫ ba f(t) dt converges.

Proof. (1) By hypothesis, ∃M, δ > 0 such that 0 ≤ f(x) ≤ Mg(x) for all x ∈ ]b− δ, b[.Integrating we get

∀x ∈ ]b− δ, b[ :∫ x

af(t) dt =

∫ b−δ

af(t) dt+

∫ x

b−δf(t) dt ≤

∫ b−δ

af(t) dt+M

∫ x

b−δg(t) dt .

If∫ ba g(t) dt converges, then

∫ xb−δ g(t) dt has a finite limit as x→ b. Hence, the RHS is

bounded by a C > 0 independent of x, so F (x) =∫ xa f(t) dt converges by Lemma 5.11.

If∫ ba f(t) dt does not converge, then

∫ ba g(t) dt does not converge by contraposition.

(2) If f = o(g), then f = O(g), so this follows from (1).(3) If f ∼ g, then f = O(g) and g = O(f), so this follows from (1).

Remark 5.13. Analogous results hold for functions f, g locally integrable on ]a, b[. Theresults are also true if f and g are negative functions, by considering −f and −g. Therestriction is that g cannot change sign.

1. If φ is not strictly monotone, there exist t1 < t2 < t3 such that φ(t1) < φ(t2) > φ(t3) or φ(t1) >φ(t2) < φ(t3). Suppose the first case holds; the other case is similar. The first case has two sub-cases :

If φ(t1) < φ(t3) < φ(t2), then by the intermediate value theorem there exists c ∈ ]t1, t2[ such thatφ(c) = φ(t3), so φ is not injective, a contradiction.

If φ(t3) < φ(t1) < φ(t2), there exists c ∈ ]t2, t3[ with φ(c) = φ(t1), again a contradiction.2. This is proved exactly like sequences; we only indicate the bounded case. Let ε > 0 and ` = sup[a,b[ F .

Then ` − ε is not an upper bound, so there exists x0 such that ` − ε < F (x0). But F is increasing, sox0 ≤ x < b =⇒ `− ε < F (x0) ≤ F (x) ≤ `. Hence, |F (x)− `| ≤ ε and F converges to `.


Example 5.14. Consider I =∫+∞

2π

ln(cos( 1x)) dx. If x > 2

π , then 0 < cos( 1x) < 1, so

x 7→ ln(cos( 1x)) is continuous and strictly negative on

]2π ,+∞

[. To study the convergence

near 2π , put x = 2

π + h. Then

1x

= 12π (1 + πh

2 )= π

2(1− πh

2 + o(h))

= π

2 −π2h

4 + o(h) .

Hence, cos( 1x) = sin(π2h

4 + o(h)) = π2h4 + o(h) = π2h

4 (1 + o(1)) and thus ln(cos( 1x)) =

ln(π2h4 ) + o(1) = ln(π2

4 ) + ln(x− 2π ) + o(1). But

∫ 1− 2π

0 ln x dx converges, so∫ 1

2π

ln(x− 2π ) dx

converges and thus∫ 1

2π

ln(cos( 1x)) dx converges.

Near +∞, cos( 1x) = 1 − 1

2x2 + o( 1x2 ), so ln(cos( 1

x)) = ln(1 + (cos( 1

x) − 1))∼ − 1

2x2 .Since

∫∞1

1x2 dx converges, then

∫∞1 ln(cos( 1

x)) dx converges. Conclusion : I converges.

5.4 Absolute convergence

Definition 5.15. Let f be locally integrable on ]a, b[. We say that∫ ba f(t) dt is absolutely

convergent if∫ ba |f(t)|dt is convergent.

Lemma 5.16. Let f be locally integrable on ]a, b[. If∫ ba f(t) dt is absolutely convergent,

then it is convergent.

Proof. Let f+ = 12(|f | + f) and f− = 1

2(|f | − f). Then 0 ≤ f+ ≤ |f | and 0 ≤ f− ≤ |f |.But

∫ ba |f(t)| dt converges, so

∫ ba f+(t) dt and

∫ ba f−(t) dt converge by the comparison test.

But f = f+ − f−, so∫ ba f(t) dt converges.

Corollary 5.17 (Bounded interval criterion). Let [a, b] ⊂ R be a bounded interval andsuppose f is locally integrable and bounded on ]a, b[. Then

∫ ba f(t)dt converges.

In particular, if f is locally integrable on [a, b[ and has a limit at b, then∫ ba f(t) dt

converges.

Proof. Assume |f(t)| ≤ M for all t ∈ ]a, b[. Then∫ ba M dt = M(b − a) converges, hence∫ b

a |f(t)|dt converges by the comparison test, so∫ ba f(t) dt converges.

Example 5.18. (a)∫ 1

0 cos(1t ) dt converges by Corollary 5.17, since cos(1

t ) is continuousand bounded by 1 on ]0, 1].

(b) Making the change of variables t = 1s in the previous integral, we deduce that∫+∞

1cos(s)s2 ds converges.

(c) Consider∫+∞

0sin tt dt. The function sin t

t is continuous on ]0,+∞[ and tends to 1 ast → 0. Hence,

∫ 10

sin tt dt converges by the bounded interval criterion. We now study∫+∞

1sin tt dt. For this, we integrate by parts, taking f(t) = 1

t and g(t) = − cos t,which are C1 on [1,+∞[ and satisfy f(1)g(1) = − cos(1) and limt→+∞ f(t)g(t) = 0.Hence,

∫+∞1

sin tt dt and

∫+∞1

cos tt2 dt have the same nature. We just saw that the latter

converges, hence∫+∞

1sin tt dt converges. Conclusion :

∫+∞0

sin tt dt converges.

(d) Let us show that∫+∞

0| sin t|t dt diverges. Consider un =

∫ nπ0| sin t|t dt.

We have un =∑n−1k=0

∫ (k+1)πkπ

| sin t|t dt ≥

∑n−1k=0

∫ kπ+ 2π3

kπ+π3

| sin t|t dt.

5.5. Cauchy criterion 65

But if t ∈[kπ + π

3 , kπ + 2π3

], then | sin t| ≥ 1

2 and 1t ≥

1(k+1)π . Hence, un ≥∑n−1

k=01

2(k+1)π ·π3 = 1

6∑nk=1

1k → +∞ as n → ∞. Thus, un → +∞ and

∫+∞0

| sin t|t dt

diverges.

Example 5.19 (Bertrand integrals). Given α, β ∈ R, we prove that

Iα,β :=∫ ∞

2

1xα(ln x)β dx converges ⇐⇒ α > 1 or (α = 1 and β > 1) ,

Jα,β :=∫ 1/2

0

1xα(| ln x|)β dx converges ⇐⇒ α < 1 or (α = 1 and β > 1) ,

Kα,β :=∫ 2

1/2

1xα(| ln x|)β dx converges ⇐⇒ β < 1 .

Let fα,β(x) = 1xα(lnx)β . Then fα,β is continuous and nonnegative on ]2,+∞[.

If α < 1, then 1/xfα,β

= xα−1(ln x)β = (lnx)βx1−α → 0 as x→∞, so 1

x = o(fα,β) near ∞. But∫∞2

1x dx diverges, so Iα,β diverges.If α > 1, take γ = 1+α

2 . Then fα,β1/xγ = 1

xα−γ(lnx)β → 0 as x → ∞ since α − γ > 0.Hence, fα,β = o( 1

xγ ) near ∞. Since γ > 1,∫∞

21xγ dx converges, hence Iα,β converges.

If α = 1, making the change of variables u = ln x, we get that I1,β converges ⇐⇒∫∞ln 2

1uβ

du converges ⇐⇒ β > 1.To study Jα,β, consider the change of variables u = 1

x . Then Jα,β converges ⇐⇒∫∞2

uα

(lnu)βduu2 = I2−α,β converges ⇐⇒ 2 − α > 1 or (2 − α = 1 and β > 1), i.e. α < 1 or

(α = 1 and β > 1).To study Kα,β, let Lα,β =

∫ 11/2

1xα(| lnx|)β dx. Near 1 we have ln x = ln(1 + (x − 1)) ∼

x − 1, so 1xα(| lnx|)β ∼

1|x−1|β . But

∫ 11/2

1|x−1|β dx converges ⇐⇒

∫ 1/20

1uβ

du converges⇐⇒ β < 1. Hence, Lα,β converges iff β < 1. Similarly, Mα,β =

∫ 21 fα,β(x) dx converges iff

β < 1, so Kα,β = Lα,β +Mα,β converges iff β < 1.

5.5 Cauchy criterion

Theorem 5.20 (The continuous Cauchy criterion). Let A = Br(z0) ⊆ K, where K = Ror C. Let a be a limit point of A. Then f has a limit at a iff for any ε > 0 there exists aneighborhood V of a such that x, y ∈ V ∩A =⇒ |f(x)− f(y)| ≤ ε.

Proof. Suppose limx→a f(x) = `. Then given ε > 0, there exists a neighborhood V of asuch that x ∈ V ∩ A =⇒ |f(x) − `| ≤ ε

2 . Hence, x, y ∈ V ∩ A =⇒ |f(x) − f(y)| ≤|f(x)− `|+ |f(y)− `| ≤ ε.

Conversely, suppose f satisfies the continuous Cauchy condition. Let ε > 0 and let{an} ⊆ A with an → a. Then there exists N ∈ N such that an ∈ V ∩A for all n ≥ N . Soby Cauchy’s condition, |f(an)− f(am)| ≤ ε for any n,m ≥ N . Hence, (f(an)) is Cauchy,so it converges.

Suppose (an) and (bn) are two sequences converging to a. We showed that (f(an)) and(f(bn)) converge. The sequence (cn) defined by c2n = an and c2n+1 = bn also convergesto a, so (f(cn)) converges and thus the subsequences (f(an)) and (f(bn)) converge to thesame limit.

We showed that there exists ` such that for any sequence {an} ⊆ A, an → a, thesequence f(an)→ `. Hence, limx→a f(x) = `.


Corollary 5.21 (Cauchy criterion for integrals). 1. Let f be locally integrable on [a, b[,b ∈ R. Then

∫ ba f(t) dt converges iff

∀ε > 0 ∃δ > 0 : ∀x, x′ ∈ [a, b[ , x, x′ ∈ ]b− δ, b[ =⇒∣∣∣ ∫ x′

xf(t) dt

∣∣∣ ≤ ε .2. Let f be locally integrable on [a,+∞[. Then

∫+∞a f(t) dt converges iff

∀ε > 0 ∃A ∈ R : ∀x, x′ ∈ [a,+∞[ , x, x′ > A =⇒∣∣∣ ∫ x′

xf(t) dt

∣∣∣ ≤ ε .Proof. Apply the continuous Cauchy criterion to g(x) =

∫ xa f(t) dt, noting that neighbor-

hoods of b have the form ]b− δ, b[, while neighborhoods of +∞ have the form ]A,+∞[.

A similar criterion holds of course for integrals which are improper on the left.

5.6 Exercises

1. Let α < β < γ ∈ R. Calculate the following integrals, then deduce the convergenceand the value of the improper integrals when α→ −∞, β → 0 and γ → +∞

•∫ γ

0

dx(x+ 1)(x+ 2)(x+ 3) •

∫ γ

1

ln xxa

dx, a > 1 •∫ γ

0e−√x dx

•∫ γ

α

dxx2 + 2x+ 3 •

∫ γ

β

ln x(x+ 1)2 dx •

∫ γ

β

(1− x arctan 1

x

)dx

•∫ γ

0

dx√1 + ex

•∫ γ

0

dxx3 + 1 •

∫ γ

0

x arctan x(1 + x2)2 dx

2. (Gamma Function). We define the Γ function on ]0,+∞[ by

Γ(x) =∫ ∞

0tx−1e−t dt .

1) Show that Γ is well defined, i.e. the integral converges for all x > 0.

2) Show that for all x > 0, Γ(x+ 1) = xΓ(x) and for all n ∈ N, Γ(n+ 1) = n!.

3. (Gauss integral).

1) Let f(x) =∫ 1

0e−x(1+t2)

1+t2 dt. Show that f is differentiable on R and that f ′(x) =−∫ 1

0 e−x(1+t2) dt.

2) Put g(x) = f(x2). Show that for all x ∈ R,

g′(x) = −2e−x2∫ x

0e−t

2 dt, then that g(x) +( ∫ x

0e−t

2 dt)2

= π

4 .

3) Deduce the value of∫∞

0 e−t2 dt, then that of Γ(1

2).

5.6. Exercises 67

4. Study the convergence of the following improper integrals:

•∫ ∞

0(x+ 2)−

√x2 + 4x+ 1 dx •

∫ ∞1

e−x1/3+sinx dx •

∫ ∞0

xae−x dx, a ∈ R

•∫ ∞

1e−√

lnx dx •∫ ∞

1

e−√

lnx

xdx •

∫ ∞1

dxxcos(1/x)

•∫ ∞

2

xlnx

(ln x)x dx •∫ ∞

0e−(lnx)2 dx •

∫ ∞0

ln xx2 − 1 dx

•∫ 1

0arctan

(1x

sin( 1x

))

dx •∫ 1

0

arctan xx2 dx •

∫ ∞0

dxxa + xb

, a, b ∈ R

•∫ ∞

1xa(1− e−

1√x ) dx, a ∈ R •

∫ ∞0

x− 1(x+ 2)7/3 dx •

∫ ∞0

dx(x+ 3)2

√|x− 2|

•∫ 1

0|1− xa|b dx, a, b ∈ R •

∫ 1

0

xa − 1ln x dx, a ∈ R •

∫ ∞−∞

dxex + x2e−x

•∫ e

1

dxln x •

∫ ∞e

dxln x •

∫ ∞0

e−x√x

dx

•∫ ∞

0

sin(5x)− sin(3x)x5/3 dx •

∫ ∞0

xa

1 + xbdx, a, b ∈ R •

∫ ∞0

( 3√x+ 1− 3√x)√xdx

5. (Improper integrals and limits). We learned that if a numerical series∑an converges,

then an → 0.1. Find a function f which is nonnegative, continuous and such that

∫∞0 f(x) dx con-

verges, but limx→∞

f(x) does not exist.

2. Show that if f is locally integrable on [a,+∞[, if∫∞a f(x) dx converges and if

limx→∞

f(x) exists, then limx→∞

f(x) = 0.

3. Show that if f is locally integrable on [a,+∞[, if∫∞a f(x) dx converges and if f is

uniformly continuous, then limx→∞

f(x) = 0.

4. Application: study the convergence of∫∞

0 sin(sin x) dx.6. Study the convergence of the following improper integrals:

•∫ ∞

0sin(x) dx •

∫ 1

0

sin xxa

dx, a > 0 •∫ ∞

1

sin xxa

dx, a > 0

•∫ ∞

0sin(ex) dx •

∫ 1

0

1x2 sin( 1

x2 ) dx •∫ ∞

1

sin2 x

xdx

•∫ ∞

1

sin x√x+ sin x dx

7. (Abel’s criterion). Let f and g be two functions on [a, b[, b ∈ R ∪ {+∞}, such that� f is of class C1, monotone and tends to 0 at b,� g is continuous and ∃M > 0 : ∀c ∈ [a, b[,

∣∣ ∫ ca g(x) dx

∣∣ ≤M .Show that

∫ ba f(x)g(x) dx converges 3.

Application: study the convergence of∫∞

0(√x+ cosx−

√x)

dx.

8. Show that∫ π/2

0 ln(sin x) dx =∫ π/2

0 ln(cosx) dx = −π ln 22 .

9. Show that∫∞

01

1+x4 dx =∫∞

0x2

1+x4 dx = π2√

2 .

3. The criterion remains true if f is not of class C1 and if g is not continuous, but the proof becomesmore complicated.


10. Let a, b > 0 and let f be a continuous function on R+ such that∫∞

1f(x)x dx converges.

1. Show that for α > 0, we have∫ ∞α

f(bx)− f(ax)x

dx =∫ aα

bα

f(x)x

dx .

2. Deduce that the integral∫∞

0f(bx)−f(ax)

x dx converges and is equal to f(0) ln ab .

3. Show the existence of∫∞

0e−bx−e−ax

x dx and give its value.4. Deduce from the previous question the convergence and value of

∫ 10x−1lnx dx.

5. Calculate∫∞

0sin3 xx2 dx using the relation sin x = eix−e−ix

2i .11. Calculate the following integrals after establishing their convergence (n ∈ N∗):

•∫ ∞

0

1(x2 + 1)n dx •

∫ ∞−∞

1(x2 + 2x+ 3)n dx .

Bibliography

[1] E. A. Coddington. An Introduction to Ordinary Differential Equations. Dover Pub-lications, 1989.

[2] C. Deschamps and A. Warusfel. Mathématiques tout-en-un [2e année] MP, PC, PSI.Dunod, 2001.

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[4] C. Deschamps and A. Warusfel. Mathématiques tout-en-un [2e année] ECS. Dunod,nouvelle edition, 2008.

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exercices corrigés. Pearson, 2007.[9] J-P. Marco, P. Thieullen, and J-A. Weil. Mathématiques L2 : Cours complet avec

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95:44–45, 1988.[13] R. Palais. A simple proof of the Banach contraction principle. J. fixed point theory

appl., 2:221–223, 2007.[14] W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, third edition, 1976.[15] W. Rudin. Real and Complex Analysis. McGraw-Hill, third edition, 1986.

Download - PrinciplesofMathematical Analysis - Unistrairma.math.unistra.fr/~sabri/Lectures_M232.pdfLocalStudyofFunctions 0.1 Reminders Lemma0.1. Supposethatf: ]a,b[ →R attainsamaximumorminimumatc∈]a,b[.If

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