SENGAMALA THAYAAR EDUCATIONAL TRUST WOMEN’S COLLEGE

(Accredited with ‘A ’grade by NAAC & An ISO 9001: 2015 Certified Institution)

SUNDARAKKOTTAI, MANNARGUDI-614 016

Sequences and Series – Unit – I Study Material for II B.Sc., Mathematics Students

Prepared by

Dr.S.Suganthi, M.Sc., M.Phil., Ph.d, SET.

Assistant Professor of Mathematics

STET Women’s College

CORE COURSE V

SEQUENCES AND SERIES

16SCCMM5

OBJECTIVES :

1. To lay a good foundation for classical analysis

2. To study the behavior of sequences and series.

Unit I

Sequences – Bounded Sequences – Monotonic Sequences – Convergent Sequence – Divergent Sequences – Oscillating sequences

Unit II

Algebra of Limits – Behavior of Monotonic functions

Unit III

Some theorems on limits – sub sequences – limit points : Cauchy sequences

Unit IV

Series – infinite series – Cauchy’s general principal of convergence – Comparison – test theorem and test of convergence using comparison test (comparison test statement only, no proof)

Unit V

Test of convergence using D Alembert’s ratio test – Cauchy’s root test – Alternating Series – Absolute Convergence (Statement only for all tests)

Book for Study

Dr. S.Arumugam & Mr.A.Thangapandi Isaac Sequences and Series – New Gamma Publishing House – 2002 Edition.

Unit I : Chapter 3 : Sec. 3.0 – 3.5 Page No : 39-55

Unit II : Chapter 3 : Sec. 3.6, 3.7 Page No:56 – 82

Unit III : Chapter 3 : Sec. 3.8-3.11, Page No:82-102

Unit IV : Chapter 4 : Sec. (4.1 & 4.2) Page No : 112-128.

Unit V : Relevant part of Chapter 4 and Chapter 5: Sec. 5.1 & 5.2 Page No:157-167.

Book for Reference

1. Algebra – Prof. S.Surya Narayan Iyer

2. Algebra – Prof. M.I.Francis Raj

Sequences and Series

UNIT - I

Set

Definition:

A set is a collection of well - defined and distinct objects.

Example

The elements or members of a set can be anything: numbers, people, letters of the alphabet, other sets, and so on

Note

Sets are conventionally denoted with capital letters. Sets A and B are equal if and only if they have precisely the same elements.

Function

Definition:

A function is more formally defined given a set of inputs X (domain) and a set of possible outputs Y (codomain) as a set of ordered pairs (x,y) where x∈X and y∈Y, subject to the restriction that there can be only one ordered pair with the same value of x. We can write the statement that f is a function from X to Y using the function notation f : X →Y.

3.1 Sequences

Definition:

Let f : N →R be a function and let f(n) = an . Then a1, a2, a3, a4,……… an,…….is called the Sequence in R determined by the function f and is denoted by (an). an is called the nth term of the sequence.

Note : The range of the function f, which is a subset of R, is called the range of the sequence.

Examples

1. The function f : N →R given by f(n) = n determines the sequence 1,2,3,……..n,…

(ie) f(1) = 1, f(2) = 2, f(3) = 3,….. f(4) = 4,…..

2. The function f: N →R given by f(n) = n2 determines the sequence 1,4,9,….n2,…

(ie) f(1) = 1, f(2) = 4, f(3) = 9,….. f(4) = 16,….. 3. The function f: N →R given by f(n) = (-1)n determines the sequence −1,1, −1,1……..…Thus the terms of a sequence need not be distinct. The range of this sequence is {ퟏ, −ퟏ}.

4. The Sequence ( (-1)n+1 ) is given by 1, -1, 1, -1…..The range of this sequence is also {ퟏ, −ퟏ}. Note The sequence ( (-1)n ) and ( (-1)n+1 ) are different . The first sequence starts with -1 and the second sequence starts with 1.

5. The Constant function f: N →R given by f(n) = 1 determines the sequence 1,1,1,1… such a sequence is called a constant sequence 1 푛 푖푓 푛푖푠 푒푣푒푛 2 6. The function f: N →R given by f(n) = { 1 determines the (1 − 푛) 푖푓 푛푖푠 표푑푑 2 sequence 0, 1, -1, 2, -2,…….,n, -n,……. (ie) f(1) = 0, f(2) = 1, f(3) = -1, f(4) = 2, f(5) = -2,………… The range of this sequence is Z (integers). 푛 1 2 푛 7. The function f: N →R given by f(n) = determines the sequence , , …., ,… 푛+1 2 3 푛+1 8. The function f: N →R given by f(n) =2n+3 determines the sequence 5, 7, 9, 11, … 1 1 1 1 9. The function f: N →R given by f(n) = determines the sequence1, , , …., ,… 푛 2 3 푛 10. Let xєR. The function f: N →R given by f(n) = xn-1 determines the geometric sequence 1, x, x2, ………xn,….. 11. The sequence (-n) is given by -1, -2, -3,….-n,…The range of this sequence is the set of all negative integers.

Exercises

1.Write the first five terms of each of the following sequences.

(−1)푛 2 1 푐표푠푛푥 (−1)푛+1 (a) ( ) (b) ( (1 − )) (c) ( ) (d) ( ) 푛 3 10푛 푛2+푥2 푛!

1−(−1)푛 2푛2+1 (e) ( ) (f) ( ) (g) (n!) 푛3 2푛2−1 푛 푖푓 푛 푖푠 표푑푑 (h) f(n) = {1 푖푓 푛 푖푠 푒푣푒푛 푛

2. Determine the range of the following sequences.

(a) (n) (b) (2n) (c) (2n-1) (d) (1+(-1)n)

(e) The constant sequence a, a, a,……. 3.2 Bounded Sequences

Definition:

A sequence (an) is said to be Bounded Above if there exists a real number k such that an≤k for every n∈N. k is called an upper bound of the sequence (an).

A sequence (an) is said to be Bounded Below if there exists a real number k such that an≥k for every n∈N. k is called an lowerr bound of the sequence (an).

A sequence (an) is said to be Bounded if it is both bounded above and bounded below.

Note

1. A sequence (an) is bounded iff there exists a real number k≥ 0 Such that ⃒an⃒≤ k for all n. (ie) -k ≤ an ≤ k

Least upper bound:

An upper bound for a set A which belongs to A is a greatest element of A (or a maximum element). Notice that A can have only one greatest element. The least upper bound of A is also called the supremum of A. It can be written sup(A) or lub(A).

Greatest lower bound:

An lower bound for a set A which belongs to A is a least element of A (or a minimum element). Notice that A can have only one least element. The greatest lower bound of A is also called the infimum of A. It can be written inf(A) or glb(A).

`Examples

1 1 1 1. Consider the sequences 1, , , …….., ,………..Here 1 is l.u.b and 0 is the g .l.b. 2 3 푛 It is bounded sequence. 2. The sequence 1,2,3,….,n,…. is bounded below but not bounded above. 1 is the g.l.b of the sequence. 3. The sequence -1,-2,-3,….,-n,…. is bounded above but not bounded below. -1 is the l.u.b of the sequence. 4. 1,-1,1,-1,…..is a bounded sequences. Here 1 is the l.u.b., -1 is the g.l.b. of the sequence. 5. Any constant sequence is a bounded sequence. (eg) The sequence 2,2,2,2,….here l.u.b and g.l.b is 2 (ie) l.u.b = g.l.b = 2

Exercises

1. Give examples of sequences (an) such that (a). (an) is bounded below but not bounded above. (b). (an) is bounded above but not bounded below. (c). (an) is a bounded sequence. (d). (an) is neither bounded above nor bounded below.

2. Determine the l.u.b and g.l.b of the following sequences if they exist

(a). 2, -2, 1, -1, 1, -1, …………….

1 1 1 1 (b). 1, , , , …….., ,…… √2 √3 √4 √푛

1 2 3 푛 (c). , , , ….., ,…………. 2 3 4 푛+1 (d). 1, -1, 2, -2, 3,-3, ……….n, -n,………

1 1 1 1 (e) 1, , 3, , 5, , ………………(2n-1), ,……….. 2 4 6 2푛

1 1 1 (f). 1, ,1, , 1, ,…………… 2 3 4 (g). (1+n+n2)

(h). (-n2)

3.3 Monotonic Sequences

Definitions

A sequence (an) is said to be monotonic increasing if an≤ an+1 for all n.

A sequence (an) is said to be monotonic decreasing if an≥ an+1 for all n.

A sequence (an) is said to be strictly monotonic increasing if an< an+1 for all n

. A sequence (an) is said to be strictly monotonic decreasing if an> an+1 for all n.

A sequence (an) is said to be monotonic if it is either monotonic increasing or monotonic decreasing.

Examples

1. 1, 2,2, 3, 3, 3, 4, 4, 4, 4,,……….is a monotonic increasing sequence.(ie) an≤ an+1

2. 1,2,3,4,………,n, …….is a strictly monotonic increasing sequence. (ie) an< an+1 1 1 1 3. 1, , , …….., ,………..is a strictly monotonic decreasing sequence. (ie) an> an+1 2 3 푛 4. The sequence (an) given by 1, -1, 1, -1, 1,….is neither monotonic increasing sequence nor monotonic decreasing sequence. Hence (an) is not a monotonic sequence. ퟐ풏−ퟕ 5. Prove ( ) is a monotonic increasing sequence. ퟑ풏+ퟐ Proof: 2푛−7 2(푛+1)−7 Ler an = , an+1 = 3푛+2 3(푛+1)+2 2푛−7 2(푛+1)−7 Now an− an+1 = - 3푛+2 3(푛+1)+2 (2푛−7)(3푛+5)−(2푛−5)(3푛+2) = (3푛+2)(3푛+5)

−25 = < 0. (3푛+2)(3푛+5)

∴ an< an+1

Hence the sequence is monotonic increasing.

ퟏ ퟏ ퟏ 6. Consider the sequence (an) where an = 1+ + +……. + , Prove (an) is a ퟏ! ퟐ! 풏! monotonic increasing sequence. Proof 1 1 1 Given an = 1+ + +……. + 1! 2! 푛! 1 1 1 1 an+1 = 1+ + +……. + + 1! 2! 푛! (푛+1)!

∴ an< an+1

Hence the sequence is monotonic increasing.

Note

A monotonic increasing sequence (an) is bounded below and a1 is the g.l.b. of the sequence. A monotonic decreasing sequence (an) is bounded above and a1 is the l.u.b. of the sequence.

Solved Problem

푎1+푎2+⋯…..+푎푛 Show that (an) is a monotonic sequence then is also a monotonic 푛 sequence.

Solution

Let (an) be a monotonic increasing sequence.

∴ a1≤ a2≤ a3≤ a4≤……… ≤an≤ ……………….(1)

푎1+푎2+⋯…..+푎푛 Let bn = 푛 푎1+푎2+⋯…..+푎푛+푎푛+1 푎1+푎2+⋯…..+푎푛 Now, bn+1- bn = - 푛+1 푛

푛(푎 +푎 +⋯…..+푎 +푎 )−(푛+1)(푎 +푎 +⋯…..+푎 ) = 1 2 푛 푛+1 1 2 푛 (푛+1)푛

푛푎 −(푎 +푎 +⋯…..+푎 ) = 푛+1 1 2 푛 푛(푛+1)

푛푎 −(푎 +푎 +⋯…..+푎 ) ≥ 푛+1 푛 푛 푛 by (1) 푛(푛+1)

푛(푎 −푎 ) = 푛+1 푛 푛(푛+1)

≥ 0 by (1)

∴ bn+1 ≥ bn

∴ (bn) is monotonic increasing

The proof is similar if (an) is monotonic decreasing

Hence (an) is monotonic

Exercises

1. Give an example of a sequence (an) such that (an) is (a) Monotonic increasing and bounded above. (b) Monotonic increasing and not bounded above. (c) Monotonic decreasing and bounded below. (d) Monotonic decreasing and not bounded below. 2. Determine which of the following sequences are monotonic. 1 1 (a) (logn) (b) ((−1)푛+1푛) (c) (2 + ) (d) ( ) 푛 2푛 −1푛 (e) ( ) (f) .6, .66, .666, …… (h) 2, 1.9,1.8, ……… 푛

3.4 Convergent Sequences

Definition

A sequence (an) is said to converge to a number l if given є >0 there exists positive integer m such that ⃒ for all n ≥ m. We say that l is the limit of the sequence and we write lim 푎푛= l or (an) → l. ∞ Note

(an) → l iff given є >0 there exists a natural number m such that an ∈ ( l –є, l +є ) for all n ≥ m (ie) a finite number of terms of the ⃒an - l⃒< є sequence lie within the interval ( l –є, l +є ).

Theorem 3.1

A sequence cannot converge to two different limits.

Proof

Let (an) be a convergent sequence.

If possible let l1 and l2 be two distinct limits of (an).

Let є > 0 be given,

1 Since (an) → l1, there exists a natural number n1 such that ⃒an - l1⃒ < є for all n ≥ n1 → (1) 2 1 Since (an) → l2, there exists a natural number n2 such that ⃒an – l2⃒ < є for all n ≥ n2 → (2) 2 Let m = max{푛1, 푛2}.

Then ⃒ l1- l2 ⃒ = ⃒ l1 –am+am- l2⃒

≤ ⃒l1 –am⃒ + ⃒am- l2⃒

1 1 < є + є 2 2 = є

∴ ⃒l1- l2 ⃒ < є and this is true for every є>0.

Clearly this is possible only l1 - l2 = 0

Hence l1 = l2

Examples.

ퟏ 1. Prove: 퐥퐢퐦 = 0 ∞ 풏 Proof: Let є > 0 be given. 1 1 Then ⃒ - 0⃒ = < є if n > є 푛 푛 1 Hence if we choose m to be any natural number such that m > є 1 Then ⃒ - 0⃒ < є for all n≥m. 푛 1 ∴ lim = 0 ∞ 푛 2. Prove that the constant sequence 1, 1, 1, …….converge to 1. Proof: Let є > 0 be given. Let the given sequence be denoted by (an). Then an = 1 for all n.

∴ ⃒an - 1⃒= ⃒1-1⃒ = 0< є for all n∈N.

∴ ⃒an - l⃒< є for all n≥m where m can be chosen to be any natural number. ∴ lim 푎푛= 1 ∞ 풏+ퟏ 3. Prove 퐥퐢퐦 = 1 ∞ 풏 Proof Let є > 0 be given. 푛+1 1 1 Now, ⃒ - 1⃒ = ⃒1+ - 1⃒ = ⃒ ⃒ 푛 푛 푛 1 ∴ If we choose m to be any natural number greater than є 푛+1 We have⃒ - 1⃒ < є for all n≥m. 푛 푛+1 ∴ lim = 1 ∞ 푛 ퟏ 4. Prove 퐥퐢퐦 = 0 ∞ ퟐ풏 Proof Let є > 0 be given

1 1 1 Now,⃒ - 0⃒ = < ( since 2n > n ∀ n∈N) 2푛 2푛 푛

1 1 ∴ ⃒ - 0⃒ < є for all n≥m where m is any natural number greater than 2푛 є

1 ∴ lim = 0 ∞ 2푛

5. Prove the sequence ((−ퟏ)풏) is not convergent Proof Suppose the sequence ((−1)푛) converges to l Then, given є>0, there exists a natural number m such that ⃒(−1)푛- l ⃒< є for all n≥m. ∴ ⃒(-1)m - (-1)m+1⃒ = ⃒ (-1)m - l + l - (-1)m+1⃒ ≤ ⃒ (-1)m - l ⃒ +⃒(-1)m+1 -l⃒ < є + є = 2є. But ⃒(-1)m - (-1)m+1⃒ = 2. ∴ 2 < 2є (ie) 1<є which is contradiction since є > 0 is arbitrary. ∴ The sequence ((−1)푛) is not convergent

Theorem 3.2

Any convergent sequence is a bounded sequence

Proof

Let (an) be a convergent sequence.

lim 푎푛= l ∞

Let є > 0 be given. Then there exists m∈N such that ⃒an - l⃒< є for all n≥m..

∴ ⃒an⃒ < ⃒ l⃒ + є for all n≥m Now, let k = max { ⃒a1⃒, ⃒a2⃒, ⃒a3⃒, …. ⃒am-1⃒, ⃒ l⃒ + є }

Then ⃒an⃒≤ k forall n

∴ (an) is a bounded sequence

Note

The converse of the above theorem is not true. For example, the sequence ((−1)푛) is a bounded sequence. However it is not a convergent sequence.

Exercises

1 1. Prove that lim = 0 ∞ 푛2 1 2. Prove that lim (1 + ) = 1 ∞ 푛! 2푛+1 3. Prove that lim = 1 ∞ 2푛 4. Prove that the following sequences are not convergent (a) ((−1)푛 푛) (b) (n2)

3.5 Divergent and Oscillating sequences

We now proceed to classify sequences which are not convergent as follows.

1. Sequences diverging to ∞. 2. Sequences diverging to -∞. 3. Finitely oscillating sequences. 4. Infinitely oscillating sequences.

Definition

A sequences (an) is said to diverge to ∞ if given any real number k>0,there exists m∈N such that an>k for all n≥m.In symbols we write (an) → ∞ or lim 푎푛= ∞ ∞

Note

(an) → ∞ iff given any real number k>0 there exists m∈N such that an ∈ ( k, ∞) for all n≥m.

Examples

1. Prove (n) → ∞ Proof Let k>0 be any given real number. Choose m to be any natural number such that m >k. Then n >k for all n≥m ∴ (n) → ∞

2. Prove (n2) → ∞

Proof

Let k>0 be any given real number.

Choose m to be any natural number such that m >√푘.

Then n2 >k for all n≥m ∴ (n2) → ∞ 3. Prove (ퟐ풏) → ∞. Proof Let k>0 be any given real number Then 2n > k ⟺ n log 2 > log k. log 푘 ⟺ n > log 2 log 푘 Hence if we choose m to be any natural such that m > , log 2 then 2n > k for all n≥m ∴ (2푛) → ∞

Definition

A sequences (an) is said to diverge to - ∞ if given any real number k<0,there exists m∈N such that an

Note

(an) → −∞ iff given any real number k<0 there exists m∈N such that an ∈ ( -∞, k) for all n≥m.

Theorem 3.3

(an) → ∞ iff (-an) → − ∞.

Proof

Let (an) → ∞

Let k<0 be any given real number. Since (an) → ∞ there exists m ∈ N such that an > -k for all n≥m. (since k < 0 ⇒ - k > 0)

∴ - an < k for all n≥m.

(-an) → − ∞.

Conversely, Let (-an) → − ∞.

Let k>0 be any given real number. Since (-an) → − ∞ there exists m ∈ N such that

-an< -k for all n≥m. (since k > 0 ⇒ - k <0)

∴ an > k for all n≥m.

(an) → ∞.

Theorem 3.4

If (an) → ∞ and an ≠0 for all n∈N then (1/an) → 0

Proof

Let є > 0 be given.

Since (an) → ∞, there exists m∈N such that an > 1/є for all n≥m. 1 ∴ < є for all n≥m. 푎푛

1 ∴ ⃒ ⃒ < є for all n≥m. 푎푛

∴ (1/an) → 0

Note

The converge of the above theorem is not true.

(−1)푛 For example, consider the sequence (an) where an = clearly (an) → 0. 푛

푛 Now (1/an) = ( ) = -1, 2, -3, 4,…….. which neither converges nor diverges to ∞ or - ∞. (−1)푛

Thus if a sequence (an) → 0, then the sequence (1/an) need not converge or diverge.

Theorem 3.5

If (an) → 0 and an > 0 for all n ∈ N, then (1/an) → ∞.

Proof

Let k>0 be any given real number. Since (an) → 0 there exists m∈N such that

⃒an⃒ < 1/k for all n≥m.

∴ an < 1/k ( Since an > 0 )

∴ 1/an > k for all n≥m..

∴ (1/an) → ∞

Theorem 3.6 Any sequence (an ) diverging to ∞ is bounded below but not bounded above.

Proof

Let (an) → ∞. Then for any given real number k >0 there exists m∈N such that an > k for all n≥m. ……………….(1)

∴ k is not an upper bound of the sequence (an).

∴ (an) is not bounded above.

Now let l = min { a1, a2, a3,………..am, k}.

From(1) we see that an ≥ l for all n.

∴ (an) is bounded below.

Theorem 3.7

Any sequence (an ) diverging to −∞ is bounded above but not bounded below.

Proof

Let (an) → −∞. Then for any given real number k <0 there exists m∈N such that an < k for all n≥m. ……………….(1)

∴ k is not an lower bound of the sequence (an).

∴ (an) is not bounded below.

Now let l = max { a1, a2, a3,………..am, k}.

From(1) we see that an ≤ l for all n.

∴ (an) is bounded above.

Note

The converge of the above theorem is not true.

0 푖푓 푛 푖푠 표푑푑 For example, the function f : N → R defined by f(n) = {1 푛 푖푓 푛 푖푠 푒푣푒푛 2 Determines the sequence 0, 1, 0, 2, 0, 3,……. Which is bounded below and not bounded above. Also for any real number k>0, we cannot find a natural number m such tha an >k for all n≥m..

Hence this sequence does not diverge to ∞

0 푖푓 푛 푖푠 표푑푑 Similarly the function f : N → R defined by f(n) = { 1 − 푛 푖푓 푛 푖푠 푒푣푒푛 2 Determines the sequence 0, -1, 0, -2, 0, -3,……. Which is bounded above and not bounded below. Also for any real number k>0,we cannot find a natural number m such tha an >k for all n≥m. Hence this sequence does not diverge to - ∞.

Definition

A Sequence (an) which is neither convergent nor divergent to ∞ or -∞ is said to be an oscillating sequence. An oscillating sequence which is bounded is said to be finitely oscillating. An oscillating sequence which is unbounded is said to be infinitely oscillating.

Examples

1. Consider the sequence ((−1)푛). Since this sequence is bounded it cannot diverge to ∞ or -∞. Also this sequence is not convergent because this sequence is -1, 1, -1, 1….have a l.u.b is 1 and g.l.b is -1.( two finite limit) Hence ((−1)푛) is a finitely oscillating sequence 1 푛 푖푓 푛푖푠 푒푣푒푛 2 2. The function f: N →R defined by f(n) = { 1 determines the (1 − 푛) 푖푓 푛푖푠 표푑푑 2 sequence 0, 1, -1, 2, -2,…….,n, -n,……. The range of this sequence is Z. Hence it cannot converge or diverge to ± ∞. This sequence is infinitely oscillating.

Exercises.

1. Discuss the behaviour of each of the following sequences 1 1 1 (a) (n!) (b) 1, , 2, , ……., , n,…. (c) ((−1)푛5) 2 3 푛

(d) ((−1)푛 + 5) (e) ( -n2) (f) (√푛