SEQUENCE AND SERIES OF REAL NUMBERS

Dr. Vimlesh Assistant Professor, Mathematics Sri Ram Swaroop Memorial University Lucknow-deva Road, U.P. .

Dr. Pragya Mishra Assistant Professor, Mathematics Pt. Deen Dayal Upadhaya Govt. Girl’s P. G. College, Lucknow, U.P. . What is a Sequence?

 In mathematics a sequence is an ordered list , like a set, it contains members called elements or terms of sequence. Most precisely, a sequence of real numbers is defined as a function S: N R, then for each n∈N, S(n) or 푆푛 is a real number. The real numbers 푆1 푆2 ,푆3,……. ,푆푛 are called terms of sequence. A sequence may be written as *푆1 푆2 푆3……. 푆푛} or {푆푛}.  For example 푡ℎ  The 푛 term of the sequence {-5n} is 푆푛 = −5푛 then the sequence becomes {-5,-10, -15, ……..-5n…….}. 푛  The 푛푡ℎ term of the sequence {푆 } is 푆 = , the sequence is 푛 푛 푛+1 2 3 {1, , , …}. 3 4

 A sequence is also given by its recursion formula where 푆1 = 1and 푆푛 = 3푆푛 , then the sequence is {1, 3, 3√3, … …}. INFINITE SERIES  A series is , roughly speaking , a description of the operation of adding infinitely many quantities one after the other, to a given starting quantity.

 An expression of the form 푎1 + 푎2 + ⋯ +푎푛 + ⋯, where each 푎푛 is real numbers , in which each term is followed by another term is known as infinite series of real ∞ 푡ℎ numbers. It is denoted by 푛=1 푎푛 표푟 푎푛 , here 푎푛 is 푛 term of the series.  The sum of n terms of series is denoted by 푆푛 , thus 푛  푆푛 = 푎1 + 푎2 + ⋯ + 푎푛 = 푛=1 푎푛.

SEQUENCE OF PARTIAL SUM

 Suppose 푎푛 is infinite series. We define a sequence *푆푛} as follows:

 푆1 = 푎1,  푆2 = 푎1 + 푎2,  푆3 = 푎1 + 푎2 + 푎3,  -  -

 푆푛 = 푎1 + 푎2 + 푎3 + ⋯ + 푎푛 and so on  The sequence *푆푛} is called a sequence of partial sums of series 푎푛.

CONVERGENCE, DIVERGENCE AND OSCILLATION OF A SERIES

1. CONVERGENT : A series 푎푛 is said to be convergent if the sequence *푆푛} of partial sums of series converges to a real number S. i.e. lim 푆푛 = 푆, where S is finite and unique. 푛→∞ 2. DIVERGENT: A series is said to be divergent if the sequence *푆푛} of partial sum diverges to +∞ 표푟 − ∞. i.e. i.e. lim 푆푛 = ±∞. 푛→∞ 3. OSCILLATORY: A series is oscillatory if sequence *푆푛} of partial sum of series oscillates i.e. lim 푆푛 does not tend to unique limit. 푛→∞ For example: consider the series 1 1 1 + + + ⋯ 1.2 2.3 3.4 1 1 1 1 Here 푆 = + + + ⋯ + 푛 1.2 2.3 3.4 푛(푛+1) 1 1 1 1 1 1 1 = 1 − + − + − + ⋯ + − 2 2 3 3 4 푛 푛+1 1 1 푆푛 = 1 − lim 푆푛 = lim 1 − = 1 − 0 = 1 푓푖푛푖푡푒 . 푛+1 푛→∞ 푛→∞ 푛+1 Hence the series converges to 1.

 Consider the series 3푛 = 3 + 32 + 33 + ⋯

푛 푛 Here 푆푛 = 3 − 1 , lim 푆푛 = lim 3 − 1 = ∞. 푛→∞ 푛→∞ Hence the series is divergent.

Example: Consider the series (−ퟏ)풏−ퟏ . 푛−1 • Here 푎푛 = (−1) 푛표푤 푆1 = 푎1 = 1 • 푆2 = 푎1 + 푎2 = 1 − 1 = 0, • 푆3 = 푎1 + 푎2 + 푎3 = 1 − 1 + 1 = 1, 푆4 = 0, 푆5 = 1 … . 푠표 표푛. • Therefore *푆푛+ = 1,0,1,0, … Which Oscillates between 0 and 1.So the series is oscillatory. Elementary Properties of series The alteration of a finite number of terms of a series has no effect on convergence and divergence. 1. If a series converges or has an infinite sums, their sum is unique. 2. Multiplication of the terms of a series by a nonzero constant K doesnot effect the convergence or divergence of a series. ∞ ∞

퐾푎푛 = 푘 푎푛 푛=1 푛=1

Necessary Condition For Convergence:

Theorem: If a series converges, its general term tends toward zero as n becomes infinity i.e. if the series 푎푛 converges , then lim푛→∞푎푛 = 0. But the converse is not true.

Proof: let us consider the series 푎푛 . Consider *푆푛} be the sequence of partial sums of series 푎푛 . 푆푛 = 푎1 + 푎2 + ⋯ + 푎푛−1 + 푎푛 푆푛−1 = 푎1 + 푎2 + ⋯ + 푎푛−2 + 푎푛−1 푆푛 − 푆푛−1 = 푎푛 Since the series 푎푛 converges , so *푆푛} converges Let lim 푆푛 = 푠 푡푕푒푛 lim 푆푛−1 = 푠 푛→∞ 푛→∞ Therefore lim 푎푛 = lim 푆푛 − lim 푆푛−1 = 푠 − 푠 = 0 푛→∞ 푛→∞ 푛→∞ Hence lim 푎푛=0 푛→∞ Thus the condition for convergence is necessary but not sufficient. Consider the harmonic series 1 1 1 1 = 1 + + + ⋯ + + ⋯ 푛 2 3 푛 1 1 Here 푎푛= lim푎푛 = lim = 0 푛 푛→∞ 푛→∞ 푛 1 But the series is divergent . (by p-test) 푛

RESULT FOR GEOMETRIC SERIES The geometric series 푟푛 = 1 + 푟 + 푟2 + 푟3 + ⋯ (푟 > 0) is convergent if 푟 < 1 and divergent if 푟 ≥ 1. Examples: 1 1 1 1  The series = + + ⋯ + + ⋯ is convergent. 2푛 2 22 2푛 Here the series is G.P. series where common ratio 푟 = 1/2 < 1. so the above series converges.  The series 3푛 = 3 + 32 + 33 + ⋯ is divergent, since 푟 = 3 > 1.

POSITIVE TERM SERIES An infinite series whose terms are positive or more generally, a non negative series (a series whose terms are nonnegative) are called positive term series. Theorem: (A test for a positive series)

A positive term series 푎푛 is convergent if and only if its sequence *푆푛} of partial sum bounded above.

Equivalently, a positive term series 푎푛 converges iff 푆푛 < 푘 ∀푛 ∈ 푁. Remark: Since monotonic sequence either converge or diverge but never oscillate. Therefore positive term series are either converge or diverge. Test for Divergence

Theorem: If 푎푛is a positive series such that lim 푎푛 ≠ 0, then the 푎푛 diverges. 푛→∞

휋 EXAMPLE(1): Test the convergence of series ∞ 푐표푠 . 푛=1 2푛 휋 휋 Here 푎푛 = 푐표푠 , Then lim 푎푛 = lim 푐표푠 = cos0 = 1 ≠ 0. 2푛 푛→∞ 푛→∞ 2푛

So by theorem lim 푎푛 ≠ 0 then the series is divergent. 푛→∞

1 2 푛 Example (2): consider the series + + ⋯ + + ⋯ 푛 6 2 푛+1

푛 푛 1/2 1 1/2 Here 푎푛 = lim 푎푛 = lim = lim 2 푛+1 푛→∞ 푛→∞ 2푛 1+1/푛 푛→∞ 2 1+1/푛

1 = ≠ 0 2 Hence the series is divergent.

SOME COMPARISON TEST

 FIRST COMPARISON TEST:

Theorem: let 푎푛 and 푏푛be two positive term series such that 푎푛≤ 푘 푏푛 ∀푛 ≥ 푚 (k being a fixed positive number and m a fixed positive integer. Then

1. If 푏푛 converges implies 푎푛converges. 2. If 푎푛diverges implies 푏푛diverges.

1 CONVERGENCE OF p-SERIES 푛푝 ퟏ ퟏ ퟏ ퟏ The series ∞ = + + ⋯ + + ⋯ (퐩 > ퟎ) 풏=ퟏ 풏풑 ퟏ풑 ퟐ풑 풏풑 Converges if (p >1) and diverges if 풑 ≤ ퟏ .

2 EXAMPLE consider the series 푒−푛 푒푥 > 푥 ∀푥 > 0\ 2 푒푛 > 푛2 ∀푛 1 1 −푛2 1 2 < 푒 < ∀푛 푒푛 푛2 푛2 1 Since ∞ converges (by p-series test here p=2 >1) 푛=1 푛2

2 Hence by first comparison test 푒−푛 is convergent.

1 EXAMPLE: Consider series ∞ 푛=2 푛2 log 푛 1 1 Since we know that > ∀푛 ≥ 2 푛2 log 푛 푛2 1 Here series ∞ is convergent since p = 2 > 1 푛=1 푛2 So by first comparison test 1 ∞ is convergent. 푛=2 푛2 log 푛 LIMIT FORM TEST

Theorem Let 풂풏 and 풃풏 be two positive term series such that 풂 퐥퐢퐦 풏 = 풍 (풍 풊풔 풏풐풏풛풆풓풐 풂풏풅 풇풊풏풊풕풆) 풏→∞ 풃풏

Then 풂풏 and 풃풏 converges and diverges together. i.e. 풃풏converges implies 풂풏 converges.

풃풏 diverges implies 풂풏 diverges. REMARKS: 1. If l = 0 표푟 푙 = ∞ then above test may not hold good.

2. To apply limit test on the series 푎푛, we have to select series 푏푛 called auxillary 푡ℎ series (which is usually p-series) in which the 푛 term of 푏푛 behaves as 푎푛, for large values of n written as 푎푛~푏푛. For large values of n we have 1 1 ~ 푛2 + 1 푛2

1 1 1 ~ = . 푛+ 푛+1 푛+ 푛 2 푛

We also take 푏푛 as 1 푏 = where 푎 푎푛푑 푏 are higher indices of n in denominator and numerator. 푛 푛(푎−푏) 푛 1 1 For example 푎 = then 푏 = = 푛 푛3+ 푛 푛 푛3−1 푛2 푎 And usually lim 푛 = 1. 푛→∞ 푏푛 1  The series converges if 푝 > 1 and diverges if 푝 ≤ 1. 푛푝 EXAMPLES 1 3 5 consider the series + + + ⋯ 1.2.3 2.3.4 3.4.5 The 푛푡ℎ term of the series is 2푛 − 1 푎 = 푛 푛(푛 + 1)(푛 + 2) 1 2 − 푎 = 푛 푛 1 2 푛2(1 + )(1 + ) 푛 푛

1 Let us consider the auxiliary series 푏 = 푛 푛2 푎 (2푛−1)푛2 Then 푛 = 푏푛 푛(푛+1)(푛+2) 풂 푛(2푛 − 1) 풏 = 풃풏 (푛 + 1)(푛 + 2) 풂 풏 ퟐ풏 − ퟏ lim 풏 = lim 푛→∞ 풃풏 푛→∞ ퟏ + 풏 풏 + ퟐ

푎 1 2 − 1\푛 푙푖푚 푛 = 푙푖푚 1 푛→∞ 푏푛 푛→∞ 1 + 푛 1 + 2\푛

풂 ퟏ ퟐ − ퟎ lim 풏 = 푛→∞ 풃풏 ퟏ + ퟎ ퟏ + ퟎ = 2 ≠ 0 푓푖푛푖푡푒 1 So 푎 and 푏 converges or diverges together since the series 푏 = converges 푛 푛 푛 푛2 (because p=2>1). Hence 푎푛 converges. ∞ 4 4 Example: Test the convergence of the series 푛=1 푛 + 1 − 푛 − 1 Sol: Here 푛푡ℎ term of the series will be 푛4 + 1 + 푛4 − 1 4 4 푎푛 = 푛 + 1 − 푛 − 1 × 푛4 + 1 + 푛4 − 1

푛4 + 1 − 푛4 − 1 2 푎푛 = ~ 푛4 + 1 + 푛4 − 1 1 1 n2 1 + + n2 1 − n4 n4 1 Consider 푏 = then 푛 푛2 푎 2 lim 푛 = lim 푛→∞ 푏푛 푛→∞ 1 1 1 + + 1 − 푛4 푛4 2 = = 1 ≠ 0(푓푖푛푖푡푒) 2 1 So 푎 and 푏 converges or diverges together . Since the series 푏 = is convergent. Hence 푛 푛 푛 푛2 the given series is convergent. Cauchy’s 풏풕풉 Root Test: ퟏ 풏 If 풂풏 be a positive term series such that 퐥퐢퐦 풂풏 = 풍 then 퐧→∞ a. 푎푛 is convergent , if 풍 < ퟏ b. 푎푛 is divergent, if 풍 > ퟏ c. Test fail if 풍 = ퟏ. Example: Test the convergence of the series ∞ 1\n 푛 (i) 푛=1 푛 − 1 Solution: The 푛푡ℎterm of the series be 1\n 푛 푎푛 = 푛 − 1

Now by Cauchy’s 푛푡ℎ root test 1\n 1\n 푎푛 = 푛 − 1 1\n 1\n lim 푎푛 = lim 푛 − 1 = 1 − 1 = 0 < 1 푛→∞ 푛→∞ Hence by Cauchy’s 푛푡ℎ root test given series is convergent. Example: Test the convergence of the series

1 2 3 2 4 3 + 푥 + 푥2 + 푥3 + ⋯ (x>0) 2 3 4 5 푛+1 푛 Solution: Here 푎 = 푥푛 (Neglecting the first term) 푛 푛+2 By Cauchy’s 푛푡ℎ root test

1\n 푛 + 1 lim 푎푛 = lim x = 푥 푛→∞ 푛→∞ 푛 + 2 푡ℎ Hence by Cauchy’s 푛 root test 푎푛 is convergent if x<1, 푎푛 is divergent if x>1 and test fail if x=1. Now when x=1 푛 푛 + 1 푎 = 1푛 푛 푛 + 2 푛 1 + 1\n 푒 1 lim 푎푛 = lim = = ≠ 0 푛→∞ 푛→∞ 1 + 2\n 푒2 푒

So the series 푎푛 is divergent at x=1. Finally the series converges if x<1and diverges if 푥 ≥ 1.

Comparison of Ratio test Or Second Ratio Test:

풂풏 풃풏 If 풂풏 and 풃풏 are two series of positive terms such that ≥ ∀ 풏 ≥ 풎 풂퐧+ퟏ 풃퐧+ퟏ

Then (i) 풃풏 converges implies that 풂풏 converges.

(ii) 풂풏 diverges implies that 풃풏 퐝퐢퐯퐞퐫퐠퐞퐬. D’ Alemberts Ratio Test:

Suppose 풂풏 be series of positive term such that 풂 lim 퐧 = 풍, then the series is 풏→∞ 풂퐧+ퟏ (i) Convergent if 풍 > ퟏ. (ii) Divergent if 풍 < ퟏ. (iii) The test fail to describe the nature of the series if 풍 = ퟏ. Remarks: (i) This test is applied when 푛푡ℎ term of the series involves factorials, product of several factors or combinations of powers and factorial.

풂퐧+ퟏ (ii) The another equivalent form of ratio test is lim = 풍, if 풂풏 is series of positive term 풏→∞ 풂퐧 then a. 풂풏 converges if 풍 < ퟏ. b. 풂풏 diverges if 풍 > ퟏ. c. Test fail if 풍 = ퟏ.

풂퐧 (iii) If lim = ∞, then 풂풏 is convergent. 풏→∞ 풂퐧+ퟏ 풂퐧+ퟏ (iv) If lim = ퟎ, then 풂풏 is convergent. 풏→∞ 풂퐧 Example: Test the convergence of the given series 1 1.2 1.2.3 1.2.3.4 + + + + ⋯ 3 3.5 3.5.7 3.5.7.9 1.2.3.4…n Solution: Here 풂 = [since 3.5.7.9… are in A.P. 푛푡ℎ term is 3+2(n- 풏 3.5.7.9…(2푛+1) 1)=2n+1] 1.2.3.4…n(푛+1) 푎 = 푛+1 3.5.7.9…(2푛+1)(2푛+3)

푎 2푛 + 3 2 + 3\푛 lim 푛 = lim = lim = 2 > 1 푛→∞ 푎푛+1 푛→∞ 푛 + 1 푛→∞ 1 + 1\푛

Therefore by ratio test 풂풏 is convergent. 푥푛 Example: Test the convergence of the series 푥+푛 푥푛 푥푛+1 Solution: Here 풂 = and 푎 = 풏 푥+푛 푛+1 푥+(푛+1) 푛 푎푛 푥+푛+1 푥 *1+ 1+푥 \n+ 1 (1+0) 1 1 Now lim = lim 푛+1 = lim = = 푛→∞ 푎푛+1 푛→∞ 푥+푛 푥 푛→∞ (1+푥\n) 푥 (1+0) 푥 푥 1 1 By Ratio test 풂 is convergent if > 1 푖. 푒. 푥 < 1and 풂 is divergent if < 1 푖. 푒. 푥 > 풏 푥 풏 푥 1. For x=1 test becomes fail and we have 1 풂 = 풏 1 + 푛 1 1 푎 = choose 푏 = 푛 푛(1+1\푛) 푛 푛 푎 1 푙푖푚 푛 = 푙푖푚 = 1 ≠ 0(푓푖푛푖푡푒) 푛→∞ 푏푛 푛→∞ (1 + 1\푛)

By Comparison test 푎푛 and 푏푛 are either both convergent or divergent. 1 Since auxillary series 푏 = is divergent so 푎 is also divergent. 푛 푛 푛

Hence the series 푎푛 is convergent if 푥 < 1 and divergent if 푥 ≥ 1. Raabe’s Test:

Suppose 풂풏 be series of positive term such that 풂 lim 풏 퐧 − ퟏ = 풍, then the series is 풏→∞ 풂퐧+ퟏ (i) Convergent if 풍 > ퟏ. (ii) Divergent if 풍 < ퟏ. (iii) The test fail to describe the nature of the series if 풍 = ퟏ. Remarks: (i) Raabe’s Test is Stronger than D’Alemberts Ratio Test. 풂 (ii) When Ratio test fail i.e. lim 퐧 = 풍 then Rabbe’s test may apply. 풏→∞ 풂퐧+ퟏ

풂퐧 (iii) If lim 풏 − ퟏ = +∞ then series 풂풏 is convergent. 풏→∞ 풂퐧+ퟏ

풂퐧 (iv) If lim 풏 − ퟏ = −∞, then series 풂풏 is divergent. 풏→∞ 풂퐧+ퟏ

Example: Test the convergence of the series 푎(푎+1) 푎(푎+1)(푎+2) 1 + 푎 + + + ⋯ 2! 3! 풂 풂+ퟏ 풂+ퟐ +⋯+(풂+풏−ퟏ) Solution: Here 풂 = 풏 풏! 풂 풂 풂+ퟏ 풂+ퟐ +⋯+(풂+풏) 풏+ퟏ= 풏+ퟏ! 푎 풂 풂 + ퟏ 풂 + ퟐ … (풂 + 풏 − ퟏ) (풏 + ퟏ)! 푙푖푚 푛 = 푙푖푚 × 푛→∞ 푎푛+1 푛→∞ 풏! 풂 풂 + ퟏ … . . (풂 + 풏)

푎 푛 + 1 1 + 1\n 푙푖푚 푛 = 푙푖푚 = 푙푖푚 = 1 푛→∞ 푎푛+1 푛→∞ 푎 + 푛 푛→∞ 1 + 푎\n Therefore D’Alembert’s ratio test fail it is not able to describe the nature of series. Thus we apply Rabbe’s test 풂 풏 + ퟏ 풏 ퟏ − 풂 lim 풏 퐧 − ퟏ = lim 풏 − ퟏ = lim 풏→∞ 풂퐧+ퟏ 풏→∞ 풂 + 풏 풏→∞ 풏 ퟏ + 풂\n

풂 lim 풏 퐧 − ퟏ = ퟏ − 풂 풏→∞ 풂퐧+ퟏ So the series is convergent if ퟏ − 풂 > ퟏ 풊. 풆. 풂 < ퟎ and divergent if ퟏ − 풂 < ퟏ 풊. 풆. 풂 > ퟎ and test fail if ퟏ − 풂 = ퟏ 풊. 풆. 풂 = ퟎ . Now if 풂 = ퟎ then the series contains only first term and therefore the convergent. Hence finally the series is convergent if 풂 ≤ ퟎ and divergent if 풂 > ퟎ. Example: Test the convergence of series 푥푛 log 푛 푝 푛 푝 푛+1 푝 Solution: Here 푎푛 = 푥 log 푥 푎푛+1 = 푥 log(푛 + 1)

푛 푝 풑 −ퟏ 푎푛 푥 log 푥 ퟏ log(풏 + ퟏ) 푙푖푚 = 푙푖푚 푛+1 푝 = 푙푖푚 풑 푛→∞ 푎푛+1 푛→∞ 푥 log(푛 + 1) 푛→∞ 풙 (log 풏)

푎 ퟏ log 풏(ퟏ + ퟏ\n) −풑 ퟏ log 풏 + log( 1 + 1\n −풑 푙푖푚 푛 = 푙푖푚 = 푙푖푚 푎푛+1 푛→∞ 풙 log 풏 푛→∞ 풙 log 풏 푛→∞

ퟏ ퟏ −풑 log 풏 + − + ⋯ 푎푛 ퟏ 풏 ퟐ 푙푖푚 = 푙푖푚 ퟐ풏 푛→∞ 푎푛+1 푛→∞ 풙 log 풏

−풑 푎푛 ퟏ ퟏ ퟏ 푙푖푚 = 푙푖푚 ퟏ + − ퟐ + ⋯ 푛→∞ 푎푛+1 푛→∞ 풙 nlog 풏 ퟐ풏 log 풏

푎푛 ퟏ 풑 푷 ퟏ 푙푖푚 = 푙푖푚 ퟏ − + ퟐ − ⋯ = 푛→∞ 푎푛+1 푛→∞ 풙 nlog 풏 ퟐ풏 log 풏 풙 ퟏ Hence by ratio test the series is convergent if > ퟏ 풊. 풆. 풙 < ퟏ and divergent if 풙 ퟏ < ퟏ 풊. 풆. 풙 > ퟏ. Test fail when x=1. 풙 For x=1, Applying Rabbe’s test

풂퐧 풑 풑 lim 풏 − ퟏ = lim 풏 ퟏ − + ퟐ − ⋯ − ퟏ 풏→∞ 풂퐧+ퟏ 풏→∞ nlog 풏 ퟐ풏 log 풏 −풏 풑 풑 = lim − − ⋯ = 0 < 1 풏→∞ 풏 log 풏 2nlog 풏 So by Raabe’s test the series diverges if x=1. Hence finally series converges if x<1 and diverges if 푥 ≥ 1. Example: Test the convergence of the series 푝 푝(푝 + 1) 푝(푝 + 1)(푝 + 2) 1 + + + + ⋯ 푃 > 0 푞 > 0 푞 푞(푞 + 1) 푞 푞 + 1 (푞 + 2) Solution: Neglecting First term 푝 푝+1 푝+2 …(푝+푛−1) 푝 푝+1 푝+2 …(푝+푛−1)(푝+푛) 푎 = , 푎 = 푛 푞 푞+1 푞+2 …(푞+푛−1) 푛+1 푞 푞+1 푞+2 …(푞+푛−1)(푞+푛)

푎 푞 + 푛 1 + 푞\n 푙푖푚 푛 = 푙푖푚 = 푙푖푚 = 1 푛→∞ 푎푛+1 푛→∞ 푝 + 푛 푛→∞ 1 + 푝\n So D’Alembert’s test fail Now applying Raabe’s test 풂 푞 + 푛 푞 − 푝 lim 풏 퐧 − ퟏ = lim n − 1 = lim n = 푞 − 푝 풏→∞ 풂퐧+ퟏ 풏→∞ 푝 + 푛 풏→∞ 푝 + 푛 By Raabe’s test the series is convergent if 푞 − 푝 > 1 i. e. q > 1 + p and divergent if 푞 − 푝 < 1 푖. 푒. 푞 > 1 + 푝. The test fail if 푞 − 푝 = 1 푖. 푒. 푞 = 1 + 푝.

푝 푝 + 1 푝 + 1 … (푝 + 푛 − 1) 푎 = 푛 푝 + 1 푝 + 2 … (푝 + 푛 − 1)(푝 + 푛)

푝 1 푎 = Let 푏 = 푛 (푝+푛) 푛 푛 푎 푝푛 푝푛 푝 Then 푙푖푚 푛 = 푙푖푚 = 푙푖푚 = 푙푖푚 = 푝 > 0 푛→∞ 푏푛 푛→∞ 푝+푛 푛→∞ 푛(1+푝\n) 푛→∞ 1+푝\n ퟏ So by comparison test 풂 converges or diverges. Here series 풃 = divergent. 풏 풏 풏

Hence 풂풏 diverges when q=1+p. Therefore finally the series is convergent if q>1+p and diverges if 푞 ≤ 1 + 푝. Logarithmic Test:

Suppose 풂풏 be series of positive term such that 풂 풍풊풎 풏 풍풐품 풏 = 풍, then the series is 풏→∞ 풂풏+ퟏ (i) Convergent if 풍 > ퟏ. (ii) Divergent if 풍 < ퟏ. (iii) The test fail to describe the nature of the series if 풍 = ퟏ. Remarks: (i) Logarithmic test applied only when ratio test fails and it involves the exponential ‘e’. (ii) Logarithmic test is alternative form of Raabe’s test.

De Morgans and Bertrand’s Test (D&B Test):

Suppose 풂풏 be series of positive term such that 풂 lim 풏 퐧 − ퟏ − ퟏ 풍풐품풏 = 풍, then the series is 풏→∞ 풂퐧+ퟏ (i) Convergent if 풍 > ퟏ. (ii) Divergent if 풍 < ퟏ. Remark: This test is applied only when D’Alembert’s Ratio test and Raab’s test fail to describe nature of the series. Higher(Second) Logarithmic Ratio Test: OR Altenative To Bertrand’s Test:

Suppose 풂풏 be series of positive term such that 풂 lim 풏 풍풐품 퐧 − ퟏ 풍풐품풏 = 풍, then the series is 풏→∞ 풂퐧+ퟏ (i) Convergent if 풍 > ퟏ. (ii) Divergent if 풍 < ퟏ. Remark: This test applied when Logarithmic test failed.

Ratio Test

Comparison Test Test For Divergence

Raabe’s Test Logarithmic Test

Higher Logarithmic Test D & B Test

Example: Test the convergence of following series 푥 22 33 (i) 1 + + 푥2 + 푥3 + ⋯ 1! 2! 3! 1 푝 1.3 푝 (ii) (1)푝+ + + ⋯ 2 2.4 푛푛푥푛 Solution: Here 푎 = 푛푒푔푙푒푐푡푖푛푔 퐹푖푟푠푡 푡푒푟푚 푛 푛! (푛 + 1)푛+1푥푛+1 푎 = 푛+1 푛 + 1 ! 풂 풏 풏 ퟏ ퟏ ퟏ lim 퐧 = lim = lim ퟏ + ퟏ\n −풏 = 풏→∞ 풂퐧+ퟏ 풏→∞ 풏 + ퟏ 풙 풏→∞ 풙 풆풙 ퟏ By D’Alembert’s Ratio Test the series is convergent if > ퟏ 풊. 풆. 풙 < ퟏ\e and divergent if 풆풙 ퟏ ퟏ < ퟏ 풊. 풆. 풙 > and the test fail if x=1\e. 풆풙 풆 1 Now when 푥 = 푒 풂 퐧 = 풆 ퟏ + ퟏ\n −풏 풂퐧+ퟏ

풂 풏풍풐품 퐧 = nlog 풆 − 풏ퟐ log(ퟏ + ퟏ\n) 풂퐧+ퟏ 풂 풏 log 퐧 = 풏 − 풏ퟐ ퟏ\n − ퟏ\ퟐ풏ퟐ + ퟏ\ퟑ풏ퟑ − ⋯ 풂퐧+ퟏ 풂 ퟏ ퟏ ퟏ lim 풏 log 퐧 = lim − + ⋯ = < ퟏ 풏→∞ 풂퐧+ퟏ 풏→∞ ퟐ ퟑ풏 ퟐ So by logarithmic test, given series is divergent if x=1\e ퟏ ퟏ Finally given series is convergent if 풙 < and diverges if 풙 ≥ . 풆 풆 1.3.5….(2푛−1) 푝 1.3.5….(2푛+1) 푝 (II). Here 푎 = 푎 = 푛 2.4.6….2푛 푛+1 2.4.6….2푛+2 푎 (2푛 + 2) 푝 푙푖푚 푛 = 푙푖푚 = 1 푛→∞ 푎푛+1 푛→∞ 2푛 + 1 So D’Alembert’s test fail to describe nature of series. Now we applying Logarithmic test: 푎 (2푛 + 2) 푝 (1 + 1\푛) 푛 log 푛 = 푛 log = 푛푝 log 푎푛+1 2푛 + 1 (1 + 1\2푛) 푎푛 1 1 1 1 1 1 푙푖푚 푛 푙표푔 = 푙푖푚 푛푝 − 2 + 3 − ⋯ − − 2 2 + 3 3 − ⋯ 푛→∞ 푎푛+1 푛→∞ 푛 2푛 3푛 2푛 22 푛 32 푛 1 3 7 푝 = 푙푖푚 p − + − ⋯ = 푛→∞ 2 8푛 24푛2 2 푝 푝 So by Logarithmic test, the series is convergent if > 1 i. e. p > 2 and if < 2 2 1 푖. 푒. 푝 < 2 and test fail at p=2. Now applying Higher Logarithmic Test

푎푛 1 3 7 푛 푙표푔 − 1 = 2 − + 2 − ⋯ − 1 푎푛+1 2 8푛 24푛 푎 3 7 log 푛 푙푖푚 푛 푙표푔 푛 − 1 log 푛 = 푙푖푚 − + − ⋯ = 0 < 1 푛→∞ 푎푛+1 푛→∞ 4 12푛 푛 Therefore by Higher Logarithmic Test the series is divergent if p=2. Thus finally given series converges if p>2 and diverges if p<=2. Example: Test the convergence of the series: 12 12.32 12.32.52 (i) + 푥 + 푥2 + ⋯ 22 2242 2242.62 (ii) 푥 + 푥1+1\2 + 푥1+1\2+1\3 + 푥1+1\2+1\3+1\4 + ⋯ 12.32.…(2푛−1)2 12.32.…(2푛+1)2 Solution: Here 푎 = 푥푛−1 푎 = 푥푛 푛 2242….(2푛)2 푛+1 2242….(2푛+2)2 ퟐ 풂 2푛 + 2 ퟏ ퟏ lim 퐧 = lim = 풏→∞ 풂퐧+ퟏ 풏→∞ 2푛 + 1 풙 풙 ퟏ Therefore by D’Alembert’s Test the given series is convergent if > ퟏ 풊. 풆. 풙 < ퟏ and 풙 ퟏ divergent if < ퟏ 풊. 풆. 풙 > ퟏ and the test fail if x=1 풙 Now when x=1 ퟐ 풂 ퟐ풏 + ퟐ 퐧 − ퟏ = − ퟏ 풂퐧+ퟏ ퟐ풏 + ퟏ 풂퐧 ퟒ풏 + ퟑ lim 풏 − ퟏ = lim 풏 × ퟐ = ퟏ 풏→∞ 풂퐧+ퟏ 풏→∞ ퟐ풏 + ퟏ Therefore Raabe’s test fail to describe nature of series, now applying DeMorgans and Bertrand test 풂퐧 ퟒ풏 + ퟑ 풏 − ퟏ − ퟏ = 풏 × ퟐ − 1 풂퐧+ퟏ ퟐ풏 + ퟏ 풂퐧 −풏 − ퟏ −ퟏ − ퟏ\n log 풏 lim 풏 − ퟏ − ퟏ log 풏 = lim ퟐ log 풏 = lim ퟐ = ퟎ < ퟏ 풏→∞ 풂퐧+ퟏ 풏→∞ ퟐ풏 + ퟏ 풏→∞ ퟐ + ퟏ\n 풏 Thus by D & B Test given series is divergent when x=1. hence finally series is convergent if x<1 and divergent if x>=1. Cauchy’s Condensation Test: If f(a) be a positive function of positive integral value of n and f(n) is a monotonically decreasing function of n ∀푛 ∈ 푁 then the two infinite series 푓(푛) and 푎푛푓(푎푛) converge or diverge together, here a being a positive integer greater than unity. OR If f(n) be a positive monotonically decreasing function of n ∀푛 ∈ 푁, then the two series f(1)+f(2)+f(3)+ …..…+f(n) and 푎푓 푎 + 푎2푓 푎2 + 푎3푓 푎3 + ⋯ + 푎푛푓(푎푛) converge or diverge together, here a being a positive integer greater than unity. Remark:

This test generally applied when 푎푛 contains log 푛. Theorem: 1 The auxiliary series ∞ is convergent if 푝 > 1 and divergent if 푝 ≤ 1. 푛=1 푛 log 푛 푝 Proof: Case I if 푝 ≤ 0 1 1 Then ≥ for 푛 ≥ 2 푛 log 푛 푝 푛 1 1 So by comparison test since is divergent so the series is also divergent. 푛 푛 log 푛 푝 Case II If 푝 > 0 the consider 1 푓 푛 = 푛 log 푛 푝

1 Here f(n) is positive for all 푛 ≥ 2 and since 푛 log 푛 푝 is increasing sequence so 푛 log 푛 푝 i.e. f(n) is decreasing sequence. Hence by Cauchy’s condensation test should apply. By CCT the series 푓 푛 is convergent or divergent according as 푎푛푓 푎푛 is convergent or divergent. 푎푛 1 1 1 Now 푎푛푓 푎푛 = = = 푎푛(log 푎푛)푝 (푛 log 푎)푝 푛푝 (log 푎)푝 1 1 Since is constant so the series 푎푛푓 푎푛 is converges or diverges as is (log 푎)푝 푛푝 convergent or divergent. 1 Since is convergent if p>1 and divergent if 푝 ≤ 1. hence by CCT the given series 푛푝 1 is convergent if p>1 and diverges 푝 ≤ 1. 푛 log 푛 푝 Example: test the convergence of the series (log 푛)2 (i) 푛2 (log 푛)2 Solution: The 푛푡ℎterm of the series 푎 = 푓 푛 = 푛 푛2 Which is positive for n≥ 2and monotonically decreasing as n increases. (log 푎푛)2 푛2(log 푎)2 Now 푎푛푓 푎푛 = 푎푛 = 푎2푛 푎푛 Where a being a positive integer greater than one . (log 푎푛)2 Consider 푎푛푓 푎푛 = 푎푛 = 푏 (푠푎푦) 푎2푛 푛

푛2(푙표𝑔 푎)2 (푛+1)2(푙표𝑔 푎)2 푏 = 푏 = 푛 푎푛 푛+1 푎푛+1 2 푏푛 푎푛 푙푖푚 = 푙푖푚 2 2 = 푎 > 1 푛→∞ 푏푛+1 푛→∞ 푛 (1 + 1\푛 ) 푛 푛 So by D’Alemberts Ratio test the series 푏푛 푖. 푒. 푎 푓 푎 is convergent. (푙표𝑔 푛)2 Hence by CCT the series is convergent. 푛2 Convergence Of Infinite Integrals: ∞ The infinite integrals ퟏ 풇 풙 풅풙 is said to be convergent (divergent) if 풕 퐥퐢퐦 풇 풙 풅풙 풊풔 풇풊풏풊풕풆 풐풓 풊풏풇풊풏풊풕풆 . 풕→∞ ퟏ Definition: Let f(x) be real valued function with domain ,ퟏ, ∞) the function f(x) is said to be non negative if 풇 풙 ≥ ퟎ ∀풙 ≥ ퟏ and f(x) is said to be monotonically decreasing if 풙 ≤ 풚 풊풎풑풍풊풆풔 풇 풙 ≥ 풇 풚 풙, 풚 ∈ ퟏ, ∞ . Cauchy’s Integral Test: If f(x) is nonnegative monotonically decreasing and integrable function such that 퐟 퐧 = 풂풏 ∀풏 ∈ 푵 ∞ ∞ Then the series 풏=ퟏ 풂풏 is convergent and the integral ퟏ 풇 풙 풅풙 are either both convergent or both divergent.

Example: Test the convergence of the series by using Cauchy’s integral test 1 (i) ∞ 푛=1 푛(푛+1) 1 (ii) 푛 (log 푛)푝 1 Solution: Here 푓 푥 = so that 푓 푛 = 푎 푥2+푥 푛 For 푥 ≥ 1, 푓 푥 is nonnegative, decreasing and integrable function. Now

푡 푑푥 푡 1 1 푡 푥 푡 = − 푑푥 = log = log − 푙표푔2 1 푥(푥+1) 1 푥 푥+1 1 푥+1 푡+1 푡 푑푥 Therefore lim = 푙표푔1 + 푙표푔2 = 푙표푔2 (푓푖푛푖푡푒) 풕→∞ 1 푥(푥+1) ∞ 푑푥 Hence the integral is convergent and so by Cauchy’s integral test the given 1 푥(푥+1) 1 series ∞ is also convergent. 푛=1 푛(푛+1) Alternating Series 푛−1 A series of the form 푎1 − 푎2 +푎3 −푎4 + ⋯ + −1 푎푛 + ⋯ where 푎푛 > 0∀푛 ∈ 푁 is called an alternating series and denoted by ∞ 푛−1 −1 푎푛 푛=1 Leibnitz Test: ∞ 풏−ퟏ An alternating series 풏=ퟏ −ퟏ 풂풏 where 풂풏 > ퟎ∀풏 ∈ 푵 is convergent if

(i) 풂풏+ퟏ ≤ 풂풏 ∀풏 ∈ 푵

(ii) 퐥퐢퐦 풂풏 = ퟎ 풏→∞ Example: Test the convergence of the series 1 1 1 (i) 1 − + − + ⋯ p > 0 2푝 3푝 4푝 푙표𝑔2 푙표𝑔3 푙표𝑔4 (ii) − + − ⋯ 22 32 42 1 1 Solution: Here 푎 = 푎 = 푛 푛푝 푛+1 (푛+1)푝 1 1 푎 − 푎 = − > 0 , 푝 > 0 푛 푛+1 푛푝 푛+1 푝 푎푛 − 푎푛+1 > 0 푎푛 > 푎푛+1 ∀푛 1 And 푙푖푚 푎푛 = 푙푖푚 = 0 푠푖푛푐푒 푝 > 0 푛→∞ 푛→∞ 푛푝 Hence by Leibnitz test the given series is convergent. log (푛+1) (ii) Here 푎 = ∀풏 ∈ 푵 푛 (푛+1)2 log (푛+1) log 푛 푙푖푚 푎푛 = 푙푖푚 = 0 [since 푙푖푚 = 0 ] 푛→∞ 푛→∞ (푛+1)2 푛→∞ 푛2

Next to show that 풂풏+ퟏ ≤ 풂풏 ∀풏 ∈ 푵

푙표𝑔푥 푥21\푥−2푥푙표𝑔푥 Let 푓 푥 = 푓′ 푥 = < 0 ∀ 푥 > 푒1\2 푥2 푥4 (Since 푥 > 푒1\2 ↔ 푙표푔푥 > 1\2 ↔ 1 − 2푙표푔푥 < 0) Which implies that f(x) is decreasing function ∀ 푥 > 푒1\2 Thus f(n+2) ≤f(n+1) ∀풏 ∈ 푵 풏 + ퟐ > 풏 + ퟏ > 푒1\2 log (푛+2) log (푛+1) ≤ ∀풏 ∈ 푵 (푛+2)2 (푛+1)2 This shows that 풂풏+ퟏ ≤ 풂풏 ∀풏 ∈ 푵 Hence by Leibnitz’s test the given series is convergent. Absolute convergence:

A series 풂풏 퐢퐬 퐬퐚퐢퐝 퐭퐨 be absolutely convergent if the series 풂풏 퐢퐬 퐜퐨퐧퐯퐞퐫퐠퐞퐧퐭.

ퟐ ퟑ Example: Consider a series 풂풏 = ퟏ − ퟏ\ퟐ + ퟏ\ퟐ − ퟏ\ퟐ + ⋯ which is absolutely convergent ퟐ ퟑ Since 풂풏 = ퟏ + ퟏ\ퟐ + ퟏ\ퟐ + ퟏ\ퟐ + ⋯ Which is geometric series with common ratio r=1\2<1 then the series 풂풏 is convergent. Hence 풂풏 is absolutely convergent. Note: The given series is also convergent ퟐ ퟑ 풂풏 = ퟏ − ퟏ\ퟐ + ퟏ\ퟐ − ퟏ\ퟐ + ⋯

Clearly 풂풏+ퟏ ≤ 풂풏 ∀풏 ∈ 푵 1 푙푖푚 푎푛 = 푙푖푚 = 0 푛→∞ 푛→∞ 2푛 So by Leibnitz’s test the given series is convergent.

Remark: Every absolutely convergent series is always convergent but converse need not be true. Example: consider a series

푎푛 = 1 − 1\2 + 1\3 − 1\4 + ⋯

Clearly 푎푛+1 < 푎푛 ∀푛 ∈ 푁

And lim 푎푛 = lim 1\푛 = 0 푛→∞ 푛→∞ So by Leibnitz’s test the given series is convergent. 1 But 푎 = 1 + 1\2 + 1\3 + 1\4 + ⋯ = , which is divergent 푛 푛

Hence 푎푛 is not absolutely convergent. Conditional Convergence:

A series 푎푛 is said to be conditionally convergent, if

(i) 푎푛 is convergent.

(ii) 푎푛 is not absolutely convergent.

Example 푎푛 = 1 − 1\2 + 1\3 − 1\4 + ⋯ is conditionally convergent.

Since 푎푛 is convergent, but 푎푛 is not convergent.

Summery of tests:

Let us take a series 푎푛 of positive term, now to check the convergence of series we proceed as follows:

1. First find 푙푖푚 푎푛. 푛→∞ i. If 푙푖푚 푎푛 ≠ 0 then series is divergent. 푛→∞ ii. If 푙푖푚 푎푛 = 0 then series may or may not be convergent. 푛→∞ 2. In this case we apply comparison test if 푎푛 is algebraic function in n First comparison Test Limit for Comparison Test

3. If in 푎푛, n as an exponent form then Cauchy’s nth root test applied .

4. If above test fail and 푎푛 contains the term of logn then Cauchy’s Condensation tesr must be applied. 5. Next to find the nature of series D’Alembert’s Ratio Test should be applied. 6. If this test fails then apply Raabe’s test. 7. Again if Raabe’s test fails for l=1, then immediately DeMorgan’s and Bertrand test must be applied. 푎 푎 8. If 푛 − 1 cannot be easily calculated then evaluate log 푛 , and apply Logarithmic test 푎푛+1 푎푛+1 and after failure if this test we always use Higher Logarithmic Test.