Series of real numbers

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CALCULUS ON R

Mathematics and Computer-Science, February 2020

Coordinators: Dorel I. DUCA and Anca GRAD

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Series of real numbers

1. Defintions and Terminology

The notion of a series of real numbers is a natural extension for a finite summation.

The study of the series reduces to the study of particular sequences of real numbers, and determining the sum o a series is equivalent to computing the limit of a certain sequence.

1.1. General notions. This section contins the definitions for the following notions:

series of real numbers, convergent series, divergent series and the sum of a series.

Definition 1.1.1 Each ordered pair of two sequences of real numbers, ((u_n),(s_n))_n∈

N, where (u_n)_n∈

N is a given sequence, and

s_n=u₁+u₂+· · ·+u_n, for all n∈N is called a series of real numbers. ♦

The usual notation for a series of real numbers ((u_n),(s_n)) is

∞

X

n=1

u_n or X

n∈N

u_n or X

n≥1

u_n or u₁+u₂+...+u_n+...

and, when there is no confusion, shorter

Xu_n.

The real numberu_n,(n ∈N) is calledthe general termof the series

∞

P

n=1

u_n,and the sequence (u_n) is called the sequence of terms of the series

∞

P

n=1

u_n. The real number s_n, (n ∈N) is calledthe partial sum of degree n of the series

∞

P

n=1

u_n,while (s_n)is the sequence of the partial sums of the series

∞

P

n=1

u_n.

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Definition 1.1.2 The series

∞

P

n=1

u_n = ((u_n),(s_n))is said to beconvergentif the sequence of its partial sums (s_n), is convergent.

A series is called divergent if it is not convergent . ♦ Definition 1.1.3

If the sequence of the partial sums (s_n) of the series

∞

P

n=1

u_n= ((u_n),(s_n)) has the limit s ∈R∪ {+∞,−∞}, then this limit is caled the sum of the series

∞

P

n=1

u_n. Thus

∞

X

n=1

u_n= lim

n→∞s_n =s.

♦

Example 1.1.4 The series (1.1.1)

∞

X

n=1

1 n(n+ 1)

having as general term u_n = _n(n+1)¹ , (n∈N) and the partial sum of degree n ∈ N, equal to

s_n =u₁+· · ·+u_n = 1

1·2 +· · ·+ 1

n(n+ 1) = 1− 1 n+ 1.

Due to the fact that the sequence of the partial sums is convergent, the series is convergent as well, and it has the sum lim

n→∞ s_n= 1. Hence

∞

X

n=1

1

n(n+ 1) = 1. ♦

Example 1.1.5 A geometric series (of ratio q) has the following expression (1.1.2)

∞

X

n=1

qⁿ⁻¹,

where q is a fixed real number. The general term of (1.1.2) is u_n =qⁿ⁻¹,

(n ∈N), and its partial sum of degree n is for all n ∈N s_n = 1 +q+· · ·+qⁿ⁻¹ =







1−qⁿ

1−q , if q6= 1 n, if q= 1.

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Therefore, the geometric series (1.1.2) if and only if |q| < 1. In this case, its sum is 1/(1−q). Hence

∞

X

n=1

qⁿ⁻¹ = 1 1−q.

If q≥1, then the geometric seris (1.1.2) is divergent and has the sum +∞, so we write

∞

X

n=1

qⁿ⁻¹ = +∞.

If q≤ −1, the geometric series (1.1.2) is divergent and does not have any sum.♦ The study of a series reduces to two aspects:

1) Determining the nature of the series (convergent or divergnet).

2) If the series is convergent, determining the sum of the series.

In order to determine the nature of a series, there are several convergence/divergence criteria. The second aspect, with respect to the precise sum, is restricted to limited number of particul series, for which we can specify precisely the sum.

In the following we present some convergence/divergence criteria for series.

Theorem 1.1.6 (Cauchy’s general convergence criterion)The series

∞

P

n=1

u_nis convergent if and only if ε >0 exist˘a un num˘ar natural n_ε cu proprietatea c˘a oricare ar fi numerele naturale n ¸si p cu n ≥nε avem

∀ε∈R, ∃n_ε∈N s.t. ∀n ≥n_ε,∀p∈N, |u_n+1+u_n+2+· · ·+u_n+p|< ε.

Proof. Let s_n = u₁ + · · · +u_n, for all n ∈ N. Then the series

∞

P

n=1

u_n is convergent if and only if the sequence of the partial sums (s_n) is convergent, therefore, according to Cauchy’s theorem (with respect to fundamental-Cauchy sequnces), this is furthere equaivalent to(sn) being fundamental, meaning that

∀ε >0, ∃n_ε ∈N, s.t. ∀n≥n_ε, ∀p∈N, |s_n+p−s_n|< ε.

Due to the fact that

s_n+p−s_n=u_n+1+u_n+2+· · ·+u_n+p, ∀n, p∈N, we conclude the proof.

Example 1.1.7 The harmonic seris is stated as (1.1.3)

∞

X

n=1

1 n. It is divergent, and has the sum +∞.

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Solution. Assume by contradiction that the harmonic series (1.1.3) is convegent. This means, according to Cauchy’s general condensation criterion, (teorema 1.1.6),that for the particular ε= 1/2>0 there exists a natural number n₀, with the property that for all n and p natural number, such that n≥n₀ it holds

1

n+ 1 +· · ·+ 1 n+p

< 1 2. By choosing p=n=n0 ∈N, we get

(1.1.4) 1

n₀+ 1 +· · ·+ 1

n₀+n₀ < 1 2.

Moreover, from n₀+k≤n₀+n₀, for all k ∈N, k≤n₀ we deduce that 1

n₀+ 1 +· · ·+ 1

n₀ +n₀ ≥ n₀ 2n₀ = 1

2

hence the inequality (1.1.4) does not hold. This contradiction leads us to the conclusion that the harmonic series (1.1.3) is divergent. Due to the fact that the sequence of the partial sums (s_n) is increasing (in romanian este strict cresc˘ator), we have:

∞

X

n=1

1

n = +∞.

∞

X

n=1

sinn 2ⁿ is convergent.

Solution. Let u_n= (sinn)/2ⁿ,for all n∈N; then for each n, p∈N it hols

|u_n+1+u_n+2+· · ·+u_n+p|=

sin (n+ 1)

2ⁿ⁺¹ +· · ·+ sin (n+p) 2^n+p

≤

≤ |sin (n+ 1)|

2ⁿ⁺¹ +· · ·+|sin (n+p)|

2^n+p ≤ 1

2ⁿ⁺¹ +· · ·+ 1 2^n+p =

= 1 2ⁿ

1− 1

2^p

< 1 2ⁿ.

Letε >0 be randomly chosen. Due to the fact that the sequence (1/2ⁿ) has the limit 0, (by using the ε characterisation of the limi), there exists n_ε ∈ N such that 1/2ⁿ < ε, for all n ≥n_ε. Then

|u_n+1+u_n+2+· · ·+u_n+p|< 1 2ⁿ < ε,

for all n, p ∈Nwith n ≥nε. As a consequnce, the series (1.1.5) is convergent.

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Theorem 1.1.9 If the series

∞

P

n=1

u_n is convegent, then

n→∞lim u_n= 0.

Proof. Letε >0 be randomly chosen. Then, according to Cauchy’s general condensation criterion (teorema 1.1.6),there exists n_ε ∈Nsuch that

|u_n+1+u_n+2+· · ·+u_n+p|< ε, for all n, p∈Nwith n ≥n_ε. Choose p= 1. Then |u_n+1|< ε,for all n∈N, n≥n_ε, hence

|un|< ε, for all n∈N, n ≥nε+ 1;

therefore

n→∞lim u_n= 0.

Remark 1.1.10 The mutual theorem for 1.1.9 (in Romanian, teorema reciproc˘a), is not usually true. This means that there are series

∞

P

n=1

u_n having

n→∞lim un= 0,

and being divergent at the same time. For example, consider the harmonic series (1.1.3) which, as we have proved is divergent, event though

n→∞lim 1 n = 0.

Remark 1.1.11 The power of the previous theorem lies in the fact

n→∞lim un6= 0 =⇒

∞

X

n=1

un is divergent .

This mean that if for a given series, it is easy to compute lim_n→∞u_n, one should do it. If that limit is 0, the series should be investigated by other means, but if it is different from 0 we conclude immediately that we are dealing with a divergent series.

Theorem 1.1.12 Let m ∈N be such that m > 1. Then the sereis

∞

P

n=1

un is convergent if and only if the series

∞

P

n=m

u_n is convergent.

Proof. We consider the sequnces of the partial sums corresponding to the two series: Let sn = u1+· · ·+un, for all n ∈ N and tn = um+· · ·+un, for all n ∈ N n ≥ m. Then the series

∞

P

n=1

un is convergent if and only if the sequence (sn)_n∈

N is convergent, which is

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equivalent to the sequence (t_n)_n≥m being convergent, fact further equivalent to the series

∞

P

n=m

un being convergent.

Theorem 1.1.13 If

∞

P

n=1

u_n and

∞

P

n=1

v_n are two convergent series and a and b are real numbers, then the series

∞

P

n=1

(au_n+bv_n) is convergent and has the sum a

∞

X

n=1

u_n+b

∞

X

n=1

v_n. Proof. For eachn ∈N it holds:

n

X

k=1

(au_k+bv_k) =a

n

X

k=1

u_k

! +b

n

X

k=1

v_k

! ,

so, according to the properties of the convergent sequences, we get the conclusion of the theorem.

Example 1.1.14 Due to the fact that the series

∞

X

n=1

1

2ⁿ⁻¹ and

∞

X

n=1

1 3ⁿ⁻¹

are convergent, having the sums 2 and 3/2, respectively, we deduce that the series

∞

X

n=1

1 2ⁿ − 1

3ⁿ

=

∞

X

n=1

1 2· 1

2ⁿ⁻¹ − 1 3· 1

3ⁿ⁻¹

is convergent and has the sum (1/2)·2−(1/3)·(3/2) = 1/2. ♦ Definition 1.1.15 Let

∞

P

n=1

u_n be a convergent series having the sum s, n be a natural number, and s_n =u₁+· · ·+u_n be the partial sum of degree n n of the series

∞

P

n=1

u_n. The real number r_n =s−s_n is called the remainder of degree n of the series

∞

P

n=1

u_n. ♦

Theorem 1.1.16 If a the series

∞

P

n=1

u_nconvergent, then the sequence(r_n)of the reminders of degree n, has the limit0.

Proof. Let s =

∞

P

n=1

u_n. Due to the fact the the sequence of the partial sums (s_n) of the series

∞

P

n=1

unis convergent as has the limits= lim

n→∞sn, and due to the fact thatrn =s−sn, for all n ∈N, we conclude that (rn) has the limit 0.

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2. Series with positive terms

Remark 1.2.1 If

∞

P

n=1

u_nis a convergent series of real numbers, then its attached sequence of partial sums (s_n) is bounded.

The mutual statement is not usually true, meaning that there are divergent series, having the sequence of the partial sums bounded. Such an example is the series

∞

P

n=1

(−1)ⁿ⁻¹ for which the general term of degree n for the partial sums iss_n, (n∈N) equal to

sn=

( 1, if n is even 0, dac˘an is odd.

Obviously, the sequence (s_n) is bounded (|s_n| ≤ 1, for all n ∈ N) even though the seires

∞

P

n=1

(−1)ⁿ⁻¹ is divergent, due to the fact that the sequnce (¸sirul (s_n) does not have a limit).

Remark 1.2.2 Each series with positive terms

∞

P

n=1

u_n has the property that its attached sequence of partial sums (s_n) is inscreasing; due to this, the boundedness of (s_n) is equivalent to its convergence.

This sections continues with convergence criteria for series with positive terms.

Definition 1.2.3 A series with positive terms is a series

∞

P

n=1

un with the property that un >0 for all n ∈N. ♦

Theorem 1.2.4 If

∞

P

n=1

u_n is a series with positive terms, then 1⁰ The series

∞

P

n=1

u_n has a sum, and

∞

X

n=1

u_n = sup ( _n

X

k=1

u_k : n∈N )

.

2⁰ The series

∞

P

n=1

u_n is convergent if and only if the sequence

s_n=

n

P

k=1

u_k

of the partial sums is bounded.

Proof. For eachn ∈N we set

s_n:=

n

X

k=1

u_k.

1⁰ The sequence (sn) is increasing, so, according to Weirstrass’ theorem with respect to monotonic sequences, statement 1⁰ is proved.

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2⁰ If the series

∞

P

n=1

u_n is convergent, than the sequence of the partial sumes (s_n) is convergent, and therefore bounded. The necessity is thus proved.

For the sufficiency, we know that the sequence (s_n) is bounded, (and we already know it is monotonic). Thus, it becomes convergent, consequently the series

∞

P

n=1

u_nis convergent.

Theorem 1.2.5 (the first comparison criterion) If

∞

P

n=1

un and

∞

P

n=1

vn are two series with positive terms with the property that there exists a >0 and n₀ ∈N such that

(1.2.6) u_n≤av_n for all n ∈N, n ≥n₀, then:

1⁰ If the series

∞

P

n=1

v_n is convergent, then the seires

∞

P

n=1

u_n is convergent too.

2⁰ If the series

∞

P

n=1

u_n is divergent, then the series

∞

P

n=1

v_n is divergent too.

Proof. For each n ∈N, let s_n =u₁+...+u_n and t_n=v₁+...+v_n; then, from (1.2.6) it holds

(1.2.7) s_n≤s_n₀ +a(v_n₀₊₁+...+v_n), for all n∈N, n≥n₀. 1⁰ If the series

∞

P

n=1

v_n is convergent, then the sequence (t_n) is bounded, consequently there exists a real numberM > 0 such thatt_n ≤M,for alln ∈N. From (1.2.7) we deduce that for all n∈N, n≥n₀ the following inequalities hold

s_n ≤s_n₀ +a(t_n−t_n₀)≤s_n₀ +at_n−at_n₀ ≤s_n₀ +at_n≤s_n₀ +aM,

implying that the sequence (s_n) is bounded. Then, according to Theorem 1.2.4, the series

∞

P

n=1

u_n is convergent.

2⁰ Assume that the series

∞

P

n=1

u_nis divergent. If the series

∞

P

n=1

v_n were convergent, then, according to statement 1⁰, the series

∞

P

n=1

u_n would be convergent, which contradicts the hypothesis that the series

∞

P

n=1

u_n is divergent. Therefore, the series

∞

P

n=1

v_n is divergent.

Example 1.2.6 The series

∞

P

n=1

n^−1/2 is divergent. Indeed, the inequality √

n ≤ n true for all n ∈ N, leads to the conclusionthat n⁻¹ ≤ n^−1/2, for all n ∈ N. Due to the fact that the harmonic series

∞

P

n=1

n⁻¹ is divergent, according to Theorem 1.2.4, statement 2⁰, we conclude that the series

∞

P

n=1

n^−1/2 is divergent. ♦

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Theorem 1.2.7 (the second comparison criterion for series) If

∞

P

n=1

u_n and

∞

P

n=1

v_n are series with positive terms with the property that there exists

(1.2.8) lim

n→∞

un

v_n ∈[0,+∞], then

1⁰ If

n→∞lim u_n

v_n ∈]0,+∞[, then the series

∞

P

n=1

u_n ¸si

∞

P

n=1

v_n have the same nature.

2⁰ If

n→∞lim u_n v_n = 0, then:

a) If the series

∞

P

n=1

vn is convergent, then the series

∞

P

n=1

un is convergent too.

b) If the series

∞

P

n=1

∞

P

n=1

3⁰ If

n→∞lim u_n

v_n = +∞, then:

a) If the series

∞

P

n=1

u_n is convergent, then the series

∞

P

n=1

v_n is convergent too.

b) If the series

∞

P

n=1

v_n is divergent, then the series

∞

P

n=1

u_n is divergent too.

Proof. 1⁰ Let a := lim

n→∞ (u_n/v_n) ∈]0,+∞[; then there exists a natural number n₀ such that

u_n vn

−a

< a

2, for all n ∈N,n ≥n₀, implying that

(1.2.9) v_n≤(2/a)u_n, for all n ∈N,n ≥n₀ and

(1.2.10) u_n≤(3a/2)v_n, for all n ∈N,n ≥n₀. If the series

∞

P

n=1

un is converent, then, according to the first comparison criterion (teorema 1.2.5), which can be applied because (1.2.9), we get that the series

∞

P

n=1

v_n is convergent.

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If the series

∞

P

n=1

v_n is convergent, then, due to the fact that (1.2.10),according to the first comparion criterion (theorem 1.2.5) if follows that the series

∞

P

n=1

u_n is convergent.

2⁰ If lim

n→∞ (u_n/v_n) = 0,then there exists a natural number n₀ such that u_n/v_n<1, for all n∈N, n≥n0,implying that

u_n ≤v_n, for all n∈N, n≥n₀.

We apply then the first comparison criterion and the conclusion follows immediately.

3⁰ If lim

n→∞ (u_n/v_n) = +∞, then there exists a natural numbern₀ such that u_n/v_n>1, for all n ∈N, n≥n₀,implying that

vn≤un, for all n∈N, n≥n0.

We apply then the first comparison criterion and the conclusion follows immediately.

∞

P

n=1

n⁻² is convergent. Indeed

n→∞lim n²

n(n+ 1) = 1∈]0,+∞[, we deduce that the series

∞

P

n=1

n⁻² and

∞

P

n=1 1

n(n+1) have the same nature. Due to the fact that the series

∞

P

n=1 1

n(n+1) is convergent ( see example 1.1.4),we get that the series

∞

P

n=1

n⁻² is convergent. ♦

Theorem 1.2.9 (al the third comparison crieterion) If

∞

P

n=1

u_n and

∞

P

n=1

v_n are series with positive terms such that there exists n₀ ∈N such that:

(1.2.11) un+1

u_n ≤ vn+1

v_n , for all n∈N, n≥n₀, then:

1⁰ If the series

∞

P

n=1

v_n is converent, then the series

∞

P

n=1

u_n is converent too.

2⁰ If the series

∞

P

n=1

∞

P

n=1

Proof. Letn∈N, n ≥n₀+ 1; then from (1.2.11) we obtain succesively:

u_n₀₊₁

u_n₀ ≤ v_n₀₊₁ v_n₀

· · · u_n un−1

≤ v_n vn−1

,

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from where, by multiplying on both sides, we obtain u_n

un0

≤ v_n vn0

. Thus

u_n≤ un0

v_n₀v_n, for all n ∈N,n ≥n₀.

By applying the first comparison criterion (teorema 1.2.5). the theorem is porved.

Theorem 1.2.10 (Cauchy’s condensation criterion) Let

∞

P

n=1

u_n be a series with positive terms with the property that the sequence of the terms of the series, (u_n), is decreasing.

Then the series

∞

P

n=1

u_n and

∞

P

n=1

2ⁿu₂ⁿ have the same nature.

Proof. Letsn :=u1+u2+...+un be the partial sum of degreen ∈Nof the series

∞

P

n=1

un

and let S_n := 2u₂ + 2²u₂² +...+ 2ⁿu₂ⁿ be the partial sum of degree n ∈ N of the series

∞

P

n=1

2ⁿu₂ⁿ.

Assume that the series

∞

P

n=1

2ⁿu₂ⁿ is convergent; then the sequence (S_n) of the partial sums is bounde, consequently there exists a real nuber M >0 such that

0≤S_n ≤M, for all n∈N. In order to prove for the series

∞

P

n=1

u_n to be convergent, based on Theorem 1.2.4, if suffices to prove that the sequence (s_n) of the partial sums is bounded. Due to the fact that the series

∞

P

n=1

u_n is with positive terms, fromn ≤2ⁿ⁺¹−1, (n ∈N) we deduce that sn≤s₂ⁿ⁺¹−1 =u1+ (u2+u3) + (u4+· · ·+u7) +

+ (u₂ⁿ+u₂ⁿ₊₁+· · ·+u₂ⁿ⁺¹₋₁).

Due to the fact that the sequence (u_n) is decreasing, it follows that u₂^k > u₂^k₊₁ >· · ·> u₂^k+1₋₁, for all k∈N therefore s_n we can conlude that

sn ≤s₂ⁿ⁺¹−1 ≤u1+ 2·u2+ 2²·u₂² +· · ·+ 2ⁿ·u2ⁿ =

=u₁+S_n≤u₁+M.

Thus, the sequence (sn) is bounded, therefore the series

∞

P

n=1

un is convergent.

Assume now that the series

∞

P

n=1

u_nis convergent; then the sequence of the partial sums, (sn), of the series

∞

P

n=1

un is bounde, consequently, there exists a real number M > 0 such

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that 0≤s_n ≤M,for all n ∈N. In order to prove that the series

∞

P

n=1

2ⁿu₂ⁿ is convergent, it sufficies to prove that the sequence (S_n) is bounded. Let n∈N. Then

s₂ⁿ =u₁+u₂+ (u₃+u₄) + (u₅+u₆+u₇+u₈) +· · ·+

+ (u₂ⁿ⁻¹₊₁+· · ·+u₂ⁿ)≥

≥u₁+u₂+ 2u₂² + 2²u₂³ +· · ·+ 2ⁿ⁻¹u₂ⁿ ≥

≥u₁+1

2S_n ≥ 1 2S_n, consequently, the following inequalities hold

Sn≤2s2ⁿ ≤2M.

Thus the sequence (S_n) is bounded and thus the series

∞

P

n=1

2ⁿu₂ⁿ is convergent.

∞

X

n=1

1

n^a, with a ∈R,

calledthe generalized harmonic series,is divergent fora≤1 and convergent fora >1.

Solution. If a ≤ 0, then the sequence of the terms of the series (n^−a) does not have the limit 0, therefore the series

∞

P

n=1 1

n^a is divergent. If a > 0, then the sequence of the tersm of the series (n^−a) is decreasing and has the limit zero, thus we may apply the Cauchy’s condensation criterion, obtaining that the series

∞

P

n=1 1 n^a and

∞

P

n=1

2ⁿ₍₂n¹)^a have the same nature. Due to the fact that 2ⁿ₍₂¹n)^a = ₂a−1¹

n

, for all n ∈ N, we deduce that the series

∞

P

n=1

2ⁿ₍₂n¹)^a turns out to be a geometric series,

∞

P

n=1 1 2^a−1

n

, which is divergent for a≤1 and convergent for a >1. Consequently, the sereis

∞

P

n=1 1

n^a is divergent for a≤1 and convergent for a >1.

Theorem 1.2.12 (D’Alembert’s quotient criterion) Let

∞

P

n=1

u_n be a series with positive terms.

1⁰ If there exists a real number q∈[0,1[ and a natural number n₀ such that:

u_n+1

u_n ≤q for all n ∈N, n ≥n_0, then the series

∞

P

n=1

un is convergent.

2⁰ If there exists a natural number n₀ such that:

u_n+1

u_n ≥1 for all n∈N, n≥n0,

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then the series

∞

P

n=1

u_n is divergent.

Proof.1⁰ We apply the third comparison criterion for series with positive terems (teorema 1.2.9, statement 1⁰), by choosing v_n :=qⁿ⁻¹, for all n ∈N. It holds

u_n+1

u_n ≤q = v_n+1

v_n , for all n∈N, n≥n₀, and the series

∞

P

n=1

v_n is convergent, consequently, the series

∞

P

n=1

u_n is convergent.

2⁰ From un+1/un ≥ 1 it follows that un+1 ≥ un, for all n ∈N, n ≥n0, therefore, the sequence (un) does not have the limit 0; thus, according to Theorem 1.1.9, the series

∞

P

n=1

un is divergent.

Theorem 1.2.13 (the consequence of the quotient criterion) Let

∞

P

n=1

u_n be a series with positive terms, for which there exists lim

n→∞

un+1

un . 1⁰ If

n→∞lim u_n+1

u_n <1, then the series

∞

P

n=1

u_n is convergent.

2⁰ If

n→∞lim u_n+1

u_n >1, then the series

∞

P

n=1

u_n is divergent.

Proof. Leta:= lim

n→∞

un+1

un . It is clear thata ≥0.

1⁰ Since a ∈ [0,1[ it follows that there exists a real number q ∈]a,1[. Then, from a∈]a−1, q[ it follows that there exists a natural number n₀ such that

u_n+1

u_n ∈]a−1, q[, for all n∈N, n ≥n₀. It follows that

u_n+1 un

≤q, for all n∈N, n≥n₀. By applying the quotient criterion, we get that the series

∞

P

n=1

u_n is convegent.

2⁰ If 1< a, then there exists a natural number n₀ such that u_n+1

u_n ≥1, for all n∈N, n≥n₀. By applying the quotient criterion, we get that the series

∞

P

n=1

un is divergent.

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∞

X

n=1

(n!)³ (3n)!

is convergent.

Solution. We have that

n→∞lim u_n+1

u_n = 1 27 <1,

thus, accordingly to the consequence of the quotient criterion, the series (1.2.12) is convergent.

Remark 1.2.15 Consider the series

∞

P

n=1

u_n. If there exists lim

n→∞

un+1

un and it is equal to 1, then the consequence of the quotient criterion cannot be applies. The series

∞

P

n=1

u_n could be either convergent or divergent. There are series either convergent or divergent with the property that lim

n→∞

un+1

un = 1. For example, consider the series

∞

P

n=1

n⁻¹ and

∞

P

n=1

n⁻². In both cases lim

n→∞

un+1

un = 1, but the first one is divergent ( see example 1.1.4) while the second one is convergent (see example 1.2.8). ♦

Theorem 1.2.16 (Cauchy’s root criterion) Let

∞

P

n=1

1⁰ If there exists a real number q∈[0,1[ and a natural number n₀ such that

(1.2.13) √ⁿ

un≤q, for all n∈N, n≥n0, then the series

∞

P

n=1

un is convergent.

2⁰ If there exists the natural number n₀ such that

(1.2.14) √ⁿ

u_n≥1, for all n ∈N, n ≥n₀, then the series

∞

P

n=1

u_n is divergent.

Proof. 1⁰ Assume that there exists q∈[0,1[ andn0 ∈N such that (1.2.13) holds. Then u_n≤qⁿ, for all n ∈N, n ≥n₀.

We apply the first comparison criterion ( theorem 1.2.5, statement 1⁰), by considering v_n:=qⁿ⁻¹,for alln ∈Nand a:=q. Due to the fact that the series

∞

P

n=1

qⁿ⁻¹ is convergent, we get that the series

∞

P

n=1

un is convergent.

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2⁰ From (1.2.14) we deduce that u_n ≥ 1, for all n ∈ N, n ≥ n₀, consequently, the sequence (u_n) does not have the limit 0. Then, according to the theorem 1.1.9, the series

∞

P

n=1

u_n is divergent.

Theorem 1.2.17 (the consequence of the root criterion) Let

∞

P

n=1

u_n be a series with positive terms, for which there exists lim

n→∞

√n

u_n. 1⁰ If

n→∞lim

√n

u_n<1, atunci seria

∞

P

n=1

u_n este convergent˘a.

2⁰ If

n→∞lim

√n

u_n>1, then the series

∞

P

n=1

u_n is divergent.

Proof. Leta:= lim

n→∞

√n

u_n. It holds a≥0.

1⁰Due to the fact thata∈[0,1[ we conclude that there exists the real numberq∈]a,1[.

Then, from a ∈]a−1, q[ it follows that there exists the natural number n₀ such that

√n

u_n ∈]a−1, q[, for all n∈N, n≥n₀. It follows that

√n

u_n≤q, for all n ∈N,n ≥n₀. By applying the root criterion we obtain that the series

∞

P

n=1

u_n is convergent.

2⁰ If 1< a, then there exists a natural number n0 such that

√n

u_n ≥1, for all n ∈N, n≥n₀. By applying the root criterion we obtain that the series

∞

P

n=1

u_n is divergent.

∞

X

n=1

√3

n³ + 3n²+ 1−√³

n³−n²+ 1n

. is convergent.

Solution. It holds

n→∞lim

√n

u_n= lim

n→∞

√3

n³+ 3n² + 1−√³

n³−n²+ 1

= 4 3 >1.

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and thus, according tot the consequence of the root criterion, the series (1.2.15) is divergent.

Remark 1.2.19 Having a series with positive terms

∞

P

n=1

u_n for which the limit lim

n→∞

√n

u_n exists and is equal to 1,the consequence of the root criterion does not decide weather the series

∞

P

n=1

u_nis convergent or divergent. There are series, either convergent, or divergent for which lim

n→∞

√n

u_n = 1.For exameple, given the series

∞

P

n=1

n⁻¹ and

∞

P

n=1

n⁻² it holds, ˆın ambele cazuri, lim

n→∞

√n

u_n = 1. The first one is divergent (see example 1.1.4) and the second one is convergent (see example 1.2.8).♦

Theorem 1.2.20 (Kummer’s criterion) Let

∞

P

n=1

1⁰ If there exists a sequence of real positive numbers (a_n)_n∈

N, de numere reale pozitive, there exists a real number r >0 and there exists a natural number n₀ such that

(1.2.16) a_n u_n

un+1

−a_n+1 ≥r, for all n∈N, n≥n₀, then the series

∞

P

n=1

u_n is convergent.

2⁰ If there exists a sequence of real positive numbers (a_n), such that the series

∞

P

n=1 1 an

is divergent and there exists a natural number n₀ such that

(1.2.17) a_n u_n

un+1

−a_n+1 ≤0, for all n ∈N, n ≥n₀, then the series

∞

P

n=1

u_n is divergent.

Proof. For each natural number n, we denote by s_n:=u₁+u₂+...+u_n the partial sum of degree n of the series

∞

P

n=1

u_n.

1⁰ Assume that there exists the sequence of real positive numbers (an), there exists a real number r >0 and there exists a natural number n₀ such that (1.2.16). We notice that the relation (1.2.16) is equivalent to

(1.2.18) a_nu_n−a_n+1u_n+1 ≥ru_n+1, for all n ∈N,n ≥n₀. Let n∈N, n≥n₀+ 1; then , from(1.2.18) we obtain:

a_n₀u_n₀ −a_n₀₊₁u_n₀₊₁ ≥ru_n₀₊₁,

· · ·

an−1un−1−anun ≥run,

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thus, by adding up, we obtain

a_n₀u_n₀−a_nu_n ≥r(u_n₀₊₁+· · ·+u_n).

This leads us to the conclusion that for all natural n ≥n₀ it holds s_n=

n

X

k=1

u_k =

n0

X

k=1

u_k+

n

X

k=n0+1

u_k ≤s_n₀ +1 r a_n₀_u_n

0 −a_nu_n

≤

≤s_n₀ +1

ra_n₀u_n₀, therefore the sequence (s_n) of the partial sums of the series

∞

P

n=1

u_n is bounded. According to theorem 1.2.4, the series

∞

P

n=1

un is convergent.

2⁰Assume that there exists a sequence of positive real numbers (a_n),with the property that the series

∞

P

n=1 1

an is divergent and there exists a natural numbern₀ such that (1.2.17) holds. Obviously (1.2.17) is equivalent to

1 an+1

1 an

≤ u_n+1

u_n , for all n ∈N,n ≥n₀. Since the series

∞

P

n=1 1

an is divergent, according to the third comparison criterion, the series

∞

P

n=1

u_n is divergent.

Theorem 1.2.21 (Raabe-Duhamel’s criterion)Let

∞

P

n=1

1⁰ If there exists a real number q >1 and a natural number n₀ such that

(1.2.19) n

u_n u_n+1 −1

≥q for all n∈N, n≥n₀, then the series

∞

P

n=1

u_n is convergent.

2⁰ If there exists a natural number n₀ such that

(1.2.20) n

un

u_n+1 −1

≤1 for all n∈N, n≥n₀, then the series

∞

P

n=1

un is divergent.

Proof. According to Kummer’s criterion (teorema 1.2.20) consider a_n :=n, oricare ar fi n ∈N; then we get

an

u_n

u_n+1 −an+1 =n u_n

u_n+1 −1

−1.

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1⁰ If we taker:=q−1>0,then, ˆıntrucˆat (1.2.16) is equivalent to (1.2.19),we deduce that

∞

P

n=1

u_n is convergent.

2⁰ Since the series

∞

P

n=1

n⁻¹ is divergent and (1.2.17) is equivalent to (1.2.20), we get that the series

∞

P

n=1

u_n is divergent.

Theorem 1.2.22 (The consequence of Raabe-Duhamel’s criterion) Let

∞

P

n=1

u_n be a series with positive terms, for which there exits the limit

n→∞lim n u_n

u_n+1 −1

. 1⁰ If

n→∞lim n u_n

u_n+1 −1

>1, then the series

∞

P

n=1

u_n is convergent.

2⁰ If

n→∞lim n u_n

u_n+1 −1

<1, then the series

∞

P

n=1

u_n is divergent.

Proof. Let

b:= lim

n→∞ n u_n

u_n+1 −1

.

1⁰ Fromb >1 we conclude that there exists a real number q∈]1, b[.Then b∈]q, b+ 1[

the existence of a natural number n0 such that n

u_n u_n+1 −1

∈]q, b+ 1[, for all n ∈N,n ≥n₀,

thus (1.2.19) holds. By applying the Raabe-Duhamel criterion we get that the series

∞

P

n=1

u_n is convergent.

2⁰ Ifb <1,then there exits a natural numbern₀ such that (1.2.20) holds. By applying the Raabe-Duhamel criterion we get that the series

∞

P

n=1

u_n is divergent.

∞

X

n=1

n!

a(a+ 1)· · ·(a+n−1), unde a >0, is convergent if and only if a >2.

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Solution. We have

n→∞lim un+1

u_n = 1

therefore, according to the consequence of the quotient criterion, we cannot state the nature of the series Due to the fact that

n→∞lim n un

u_n+1 −1

=a−1,

according to the consequence of the Raabe-Duhamel’ criterion, if a > 2, the series is convergent, and if a <2 the series is divergent. If a= 2,then the series becomes

∞

P

n=1 1 n+1

which is divergent. This means that the given series is convergent if and only if a >2.

3. Exercises to solve - series

Exercise 1.3.1 Compute the sums of the following geometric series:

a)X

n≥3

3

5ⁿ, b)X

n≥4

2ⁿ⁻³+ (−3)ⁿ⁺³

5ⁿ , c)X

n≥5

eⁿ, d)X

n≥2

− 1 π

n

e)X

n≥3

(−3)ⁿ.

Exercise 1.3.2 Compute the sums of the following telescopic series:

a)X

n≥1

1

4n²−1, b)X

n≥1

√ 1 n+√

n+ 1, c)X

n≥5

1

n(n+ 1)(n+ 2) d)X

n≥1

ln

1 + 1 n

, e)X

n≥2

ln 1 + _n¹ ln (n^ln(n+1)). Exercise 1.3.3 Determine the nature of the following series:

a)X

n≥1

n+ 7

√n²+ 7, b)X

n≥1

1

√n

n, c)X

n≥1

1

√n

n!, d)X

n≥1

1 + 1

n n

.

Exercise 1.3.4 Determine the nature of the following series:

a)X

n≥1

2ⁿ+ 3ⁿ

5ⁿ , b)X

n≥1

2ⁿ 3ⁿ+ 5ⁿ. Exercise 1.3.5 Determine the nature of the following series

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a)X

n≥1

1

2n−1, b)X

n≥1

1

(2n−1)², c)X

n≥1

√ 1

4n²−1, d)X

n≥1

√n²+n

√3

n⁵−n.

Exercise 1.3.6 Determine the nature of the following series:

a)X

n≥1

100ⁿ

n! , b)X

n≥1

2ⁿn!

nⁿ , c)X

n≥1

3ⁿn!

nⁿ , d)X

n≥1

(n!)²

2ⁿ² , e)X

n≥1

n² 2 + ¹_nn.

Exercise 1.3.7 Determine, depending on the values of the parameter a >0, the nature of the following series:

a)X

n≥1

aⁿ

nⁿ, b)X

n≥1

n²+n+ 1 n² a

n

, c)X

n≥1

3ⁿ 2ⁿ+aⁿ. Example 1.3.1 For each a, b > 0, study the nature of the series:

a)

∞

X

n=1

aⁿ

aⁿ+bⁿ; b)

∞

X

n=1

2ⁿ

aⁿ+bⁿ; c)

∞

X

n=1

aⁿbⁿ aⁿ+bⁿ; d)

∞

X

n=1

(2a+ 1) (3a+ 1)· · ·(na+ 1) (2b+ 1) (3b+ 1)· · ·(nb+ 1). Example 1.3.2 Determine the nature of the series:

a)

∞

X

n=1

(2n−1)!!

(2n)!!

1

2n+ 1; b)

∞

X

n=1

(2n−1)!!

(2n)!! ; c)

∞

X

n=1

1 n!

n e

n

.

Example 1.3.3 For each a : 0 determine the nature of the series:

a)

∞

X

n=1

n!

a(a+ 1)...(a+n); b)

∞

X

n=1

a⁻(¹⁺¹2+...+_n¹); c)

∞

X

n=1

aⁿ·n!

nⁿ . Remark 1.3.4 For further detalis go to [5].

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CHAPTER 2

Taylor’s Formula

1. Taylor’s Polynomial: definition, properties

Taylor’s formula, mainly used in approxiamting functions by the means of the poly- nomials, is one of the most important formulae in mathematics..

Definition 2.1.1 Let D be a nonempty subest ofR, x₀ ∈D andf :D→R be a n times differentiable function at x₀. The (polynomial) function T_n;x₀f :R→R defined by

(T_n;x₀f) (x) =

n

X

k=0

f^(k)(x₀)

k! (x−x₀)^k (T_n;x₀f) (x) = f(x₀) + f⁰(x₀)

1! (x−x₀) + f⁰⁰(x₀)

2! (x−x₀)²+...

...+f⁽ⁿ⁾(x₀)

n! (x−x0)ⁿ, for all x∈R,

is called Taylor’s Polynomial of order n attached to the function f, centered at the point x₀. ♦

Remark 2.1.2 Taylor’s polynomial of ordern, has the degree at mostn.♦ Example 2.1.3 For the exponential function f :R→R defined by

f(x) = expx, for all x∈R, it holds

f^(k)(x) = expx, for all x∈R¸si k ∈N.

Taylor’s polynomial of ordernattached to the exponential function exp :R→R,centered at the point x₀ = 0 is

(T_n;0exp) (x) = 1 + 1 1!x+ 1

2!x²+· · ·+ 1 n!xⁿ, for all x∈R. ♦

Remark 2.1.4 We notice the foolowing:

• the domain of definition of Taylor’s polynomial is R, in contrast to the domain of definition for the function f, which sometimes is much smaller;

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• being a polynomial function T_n;x₀f is indefinite differentiable on R ¸si, for all x∈R, so it holds

(T_n;x₀f)⁰(x) =f⁽¹⁾(x₀) + f⁽²⁾(x0)

1! (x−x₀) +· · ·+ f⁽ⁿ⁾(x0)

(n−1)! (x−x₀)ⁿ⁻¹ =

= (Tn−1;x0f⁰) (x), (T_n;x₀f)⁰⁰(x) =f⁽²⁾(x₀) + f⁽³⁾(x₀)

1! (x−x₀) +· · ·+f⁽ⁿ⁾(x₀)

(n−2)! (x−x₀)ⁿ⁻² =

= (Tn−2;x₀f⁰⁰) (x),

· · · ·

(T_n;x₀f)⁽ⁿ⁻¹⁾(x) =f⁽ⁿ⁻¹⁾(x₀) + f⁽ⁿ⁾(x₀)

1! (x−x₀) =

= T_1;x₀f⁽ⁿ⁻¹⁾ (x),

(T_n;x₀f)⁽ⁿ⁾(x) =f⁽ⁿ⁾(x₀) = T_0;x₀f⁽ⁿ⁾ (x), (T_n;x₀f)^(k)(x) = 0, for all k ∈N, k≥n+ 1.

This leads to

(Tn;x0f)^(k)(x0) = f^(k)(x0), for all k∈ {0,1,· · ·, n}

¸si

(T_n;x₀f)^(k)(x₀) = 0, for all k ∈N, k≥n+ 1.

Remark 2.1.5 At the point x₀, not only the value of Taylors’ polynomial of order n attached to the functionf, centered atx₀, but also the values of its derivatives, up to the order n, coincide to the values of the function f, and its derivatives up to the order n, respectivley .

2. Taylor’s Formula

Definition 2.2.1 LetDbe a nonempty subset ofR, x₀ ∈Dandf :D→Rbe a function n time differentiable at the point x0. The function Rn;x0f :D →R defined by

(R_n;x₀f) (x) = f(x)−(T_n;x₀f) (x), for all x∈D

is called Taylor’s remainder of order n attached to the function f centered at x₀.

When the form of the reminder is given by a certain computational expression, the following

f =Tn;x0f+Rn;x0f,

Series of real numbers

CALCULUS ON R

Mathematics and Computer-Science, February 2020

Coordinators: Dorel I. DUCA and Anca GRAD

Contents

Series of real numbers

Taylor’s Formula