Tests for the Convergence

8/12/2019 Tests for the Convergence

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Tests for the Convergence of Infinite SeriesDC-I

Semester-IILesson: Tests for the Convergence of

Infinite Series

Course Developer: Dr. Chaitanya KumarDepartment/College:

Department of Mathematics,Delhi College of Arts and Commerce (D.U.)

University of Delhi

Institute of Lifelong Learning, University of Delhi pg. 1


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Tests for the Convergence of Infinite Series

Table of Contents:

Chapter : Tests for the Convergence of Infinite Series

1. Learning Outcomes 2. Introduction 3. D'Alember's Ratio Test 4. Cauchy's nth Root test 5. Convergence of the infinite integral ( )

1

u x dx

o 5.1. Cauchy's Integral Test Exercises-1 6. Alternating series

o 6.1. Leibnitz testo 6.2. Absolute convergenceo 6.3. Conditional Convergence

Exercise 2 Summary References

1. Learning Outcomes

After you have read this chapter, you should be able to

Define and understand the following tests for the convergenceof an infinite sereis.

o D'Alember's Ratio Testo Cauchy's nth Root testo Cauchy's Integral Test

Define the alternating series and convergence of thealternating series

Absolute convergence Conditional Convergence



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2. IntroductionThe purpose of this chapter is to discuss the following tests

for convergence, conditional convergence and absolute convergence

of an infinite series.

3. D'Alember's Ratio TestLet

nu be a positive term series such that

n

nn 1

ulim

u += (1)

then

(i) unconverges if > 1.(ii) undiverges if < 1.(iii) Test fails if = 1.

Proof: Case (i)let > 1, we can choose > 0 such that - > 1

or > 1, = -

Using (1), there exists a positive integer m1, such that

n1

n 1

u, n m

u

+

<

n1

n 1

u, n m

u

+

< < +

Consider

( )n 1n 1

u, n m

u

+

> =



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Tests for the Convergence of Infinite Seriesn

nn 1

ulim 1

u +=

The series n1

un

= is convergent, but

n

n n nn 1

u n 1 1lim lim lim 1 1

u n n +

+ = = + =

The series n 21

un

= is convergent, but

( )2 2

n

2

n n nn 1

n 1u 1lim lim lim 1 1

u n n +

+ = = + =

Value Addition: Note

1. The test fails for = 1 in the sense that it fails to give anydefine information.

2. Ifn

u is a positive term series such that nn

n 1

ulim ,

u += then

nu is convergent.

Example 1: Test for convergence the seriesn 1

n 1n 1

2

3

+=

.

Solution: We have

n 1 n

n n 1n n 1

2 2u , u

3 1 3 1

+ += =

+ +

n 1 n

n

nn n n

n 1n

13

u 1 3 1 1 3lim lim lim

1u 2 3 1 21

3

+

+

+ +

= = + +



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Tests for the Convergence of Infinite Series n

nn 1

u 3lim 1

u 2 += >

Hence by D'Alembert's Ratio Test the given series converges.

Example 2: Test for convergence the series2

n

2

n 1x

n 1

+

.

Solution: Here

( )

( )

22

n n 1

n n 1 22

n 1 1n 1u x , u x

n 1 n 1 1

++

+ = =

+ + +

( ) ( )

( )

22 n

n

22 n 1n nn 1

n 1 n 1 1u x 1lim lim .

u n 1 x xn 1 1 +

+

+ += =

+ +

Hence by D'Alembert's Ratio test the given series converges if

11 x 1

x> < and diverges if

11 x 1

x< > .

The test fails to give any information when x = 1,

when x = 1,2

n 2

n 1u

n 1

=

+.

2

n 2n n

n 1lim u lim 1 0

n 1

= =

+

The given series is divergent

Hence the given series converges if x < 1 and diverges if x > 1.

Example 3: Test for convergence the series2 3 4

1 1 1 1...

2 2.2 3.2 4.2+ + + +

Solution: Here,



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Tests for the Convergence of Infinite Series( )n n 12 n 1

1 1u ,u

n.2 n 1 2+ +

= =+

( ) ( )n 1

n

nn n n

n 1

n 1 2 n 1 .2ulim lim lim

u n.2 n

+

+

+ += =

=n

1lim 2 1 2 1

n

+ = >

By D'Alembert's Ratio test the given series converges.

Example 4: Test for convergence the series

( )

2 3 4x x x x

... x 01.3 2.4 3.5 4.6+ + + + > .

Solution: Here

( ) ( ) ( )

n n 1

n n 1

x xu ,u

n n 2 n 1 x 3

+

+= =+ + +

( ) ( )

( )

n

n

n 1n nn 1

x n 1 x 3ulim lim

u n n 2 .x +

+

+ +=

+

( ) ( )( )n

n 1 n 3 1lim .

n n 2 x

+ +=

+

2

n2

1 3n 1 1

1 1n nlim .

2 x xn 1

n

+ +

= =

+

By D'Alembert's Ratio test,n

u converges if1

1x

> i.e. x < 1 and

diverges if1

1x

< i.e. x > 1 and test fails for x = 1.

For x = 1,( )n 2

1 1u

2n n 2n 1

n

= =+

+



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Tests for the Convergence of Infinite SeriesLet n 2

1v

n= which is convergent since p = 2 > 1.

2

n

n n2n

u nlim lim 1

2v

n 1 1n

= =

+

So by comparison testn

u and nv both converge and diverge

together.

Since n 21

vn

= converges consequently nu converges for x=1.

Hence the given series converges for x < 1 and diverges for x > 1.

Example 5: Show that the series2 3 4

1 2! 3! 4!...

5 5 5 5+ + + + is divergent.

Solution: We have

n n

n!u

5= and

( )n 1 n 1

n 1 !u

5+ +

+=

( )

n 1

n

nn nn 1

5u n!lim lim

u n 1 !5

+

+

=+

=n

5lim 0 1

n 1= .

Solution: Here

p

n

nu

n!= and

( )( )

p

n 1

n 1u

n 1 !+

+=

+



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Tests for the Convergence of Infinite Series( )

( )( )

pp

n

pn n n

n 1

n n 1 !u nlim lim lim n 1

u n 1n! n 1 +

+ = = + + +

p

n

n 1lim

1nn

+=

+

Hence by Ratio test, the given series is convergent.

4. Cauchy's nth Root testIf

nu is a positive term series such that

( )1

nn

nlim u

=

Then (i)n

u converges if < 1 (ii) nu diverges if > 1 (iii) the

fails if = 1.

Proof Case I. < 1.

Let us select a positive number such that + < 1.

Let + = < 1

Since ( )1

nn

nlim u

= therefore there exists a positive integer m such

that

( )1

nnu , n m <

( )1

nnu , x m < < +

( ) ( )n n n

nu , n m < < + =

nnu , n m<



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Tests for the Convergence of Infinite SeriesBut since n is a convergent geometric series with common ratio

< 1, therefore by comparison test the seriesn

u converges.

Case II. > 1

Let us select a positive number such that

- > 1

Let - = > 1

Since ( )1

nn

nlim u ,

= therefore there exists a positive integer m, such

that

( )1

nn 1

u , n m < < +

( ) ( )n n

n 1u , n m < < +

( )n n

n 1u , n m > =

But since n is a divergent geometric series with common ratio

> 1, therefore by comparison test, the seriesn

u diverges.

Value addition:NoteThe test fails to give any definite information for = 1.

Consider the two series1

n and 2

1

n . The series

1

n diverges,

while

1

n

2n

1lim

n

=

and the series2

1

n , converges, while

1

n

2n

1lim

n

=

.



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Example 7: Test for convergence the series( )

nn 2

1

logn

= .

Solution: Let( )

n n

1u

logn=

( )( )

1

n1

nn n

n n

1lim u lim

logn

=

=n

1lim 0 1

logn= <

Hence by Cauchy's nth root test the given series is convergent.

Example 8: Test for convergence the series whose general term is

32n

11

n

+

.

Solution: Let3

2n

n

1u

11

n

=

+

, then

( ) 32

1

n

1

nn

n n n

1lim u lim

1

1 n

=

+

nn

1 1lim 1

e11

n

= = 1 and divergent if p 1

and divergent if 0 < p < 1.

Solution: Let ( )( )

( )p1

u x , p 0x logx

= >

Clearly, for x > 2, u(x) is non negative, monotonically decreasing

and integrable function. Also u(n) = un, n N

Consider ( )( )

t t

p

2 2

1u x dx dx

x logx=

( )

( )( )

t1 p

2

t

2

1log x , if p 1

1 p

log log x , if p 1

=

=

or ( ) ( ) ( )

( ) ( )

1 p 1 pt

2

1log t log 2 ,if p 1

1 pu x dx

log log t log log 2 , if p 1

= =



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( ) ( )t

1 p

t2

, if p 1

1lim u x dx log 2 , if p 1

p 1

, if p 1

=

Thus ( )2

u x dx

is convergent if p > 1 and divergent if 0 < p < 1.

Hence, by Cauchy's integral test the given series is convergent if p

> 1 and divergent if 0 < p 2 and

diverges if p



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Tests for the Convergence of Infinite Series6.1. Leibnitz test: If the alternating series

( )n 1

n

n 1

1 2 3 4 nu u u u1 , u , nu 0

=

+ + > = is such that

(i) un+1

< un,

v nand

(ii)n

nlim u 0

=

then the series converges.

Proof: Let Sn= u1 u2+ u3- u4+ + (-1)nun

Now for all n,

2n 2 2n 2n 1 2n 2S S u u 0+ + + =

2n 2 2nS S+

The sequence is a monotonic increasing sequence.

Again 2n 1 2 3 2n 1 2nS u u u ... u u= + +

( ) ( ) ( )1 2 3 4 5 2n 2 2n 1 2nu u u u u ... u u u =

But since n 1 nu u+ for all n, therefore each bracket on the right is

positive and hence

2n 1S u , v n<

Thus the monotonic increasing sequence is bounded above

and is consequently convergent.

Let2n

nlimS S

=

We shall now show that the sequence also converges to

the same limit S.



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Tests for the Convergence of Infinite SeriesNow 2n 1 2n 2n 1S S u+ += +

2n 1 2n 2n 1

n n nlimS limS lim u+ +

= +

But by condition (ii)

2n 1nlim u 0+

=

2n 1 2n

n nlimS limS S+

= =

Thus the sequences and both converge to the same

limits.

We shall now show that the sequence also converges to S

Let > 0 be given.

Since the sequences and both converge to S,

therefore there exists positive integers m1, m2such that

2n 1S S , n m < (1)

and 2n 1 2S S , n m+ < (2)

Thus from (1) and (2), we have

( )n 1 2S S , n max m ,m <

converges to S

The series ( )n 1

n1 u

converges.

Example 14: Show that the seriesp p p p

1 1 1 1...

1 2 3 4 + + converges for

p > 0.



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Tests for the Convergence of Infinite SeriesSolution: Let n n 1p p 1

1 1u u

n n+ +

= =

Here n 1 nu u , v n+ <

and n pn n

1lim u lim 0

n = =

Hence by Leibnitg test the alternating series( )

n 1

p

1

n

converges.

6.2. Absolute convergence: A series unis said to be absolutely

convergent if the seriesn

u is convergent.

6.3. Conditional Convergence: A seriesn

u is said to be

conditionally convergent, if

(i) unis convergent and

(ii) unis not absolutely convergent.

Value Addition: Illustrations

1. The series n 2 31 1 1u 1 ...2 2 2

= + + is absolutely convergent,

since n 2 31 1 1

u 1 ...2 2 2

= + + + + , being a geometric series with

common ratio1

12

< , is convergent.

2. The series n 1 1 1u 1 ...2 3 4

= + + is not absolutely convergent,

since n1 1 1

u 1 ... ...2 3 n

= + + + + + is not convergent.

Theorem 2: Every absolutely convergent series is convergent.



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Tests for the Convergence of Infinite SeriesProof: Let

nu be absolutely convergent, so that nu is

convergent.

Hence for any > 0, by Cauchy's General principle of convergence,

there exists a positive number m such that

n 1 n 2 n pu u ... u , n m+ + ++ + + < and p > 1

Also for n and p > 1,

n 1 n 2 n p n 1 n 2 n pu u ... u u u ... u , n m+ + + + + ++ + + + + + < and p >1.

Hence by Cauchy's General principle of convergence the series un

converges.

Value Addition: Remark

The divergence of |un| does not imply the divergence of un.

For example, if( )

n 1

n

1u

n

= , we have seen above that |un| is

divergent, whereas unis convergent.

Example 15: Test for convergence and absolute convergence

the series.

1 1 1 1 ...1 3 5 7

+ +

Solution: We have

n

1u

2n 1=

n 1

1u

2n 1+ =

+



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Tests for the Convergence of Infinite Series n 1 nu u ; v n.+ <

Now,n

n n

1lim u lim 0

2n 1 = =

Hence, by leibnitg's test, the series unis convergent. Now we show

that unis not absolutely convergent.

We haven

1u

2n 1=

Letn

1v

n=

nn n

n

u n 1lim lim 0

v 2n 1 2 = =

Since vn is divergent, so |un| is divergent. Hence un is not

absolutely convergent.

Example 16: Show that for any fixed value of x the series

2n 1

sin nx

n

=

is convergent.

Solution: Let n 2sin nx

un

= so thatn 2

sin nxu

n=

Nown 2 2

sin nx 1u , v n

n n=

and2

1

n converges.

Hence by comparison test, the series2

sin nx

n converges.

2

sin nx

n is absolutely convergent.



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Tests for the Convergence of Infinite SeriesSince every absolutely convergent series is convergent, therefore

2

sin nx

n is convergent.

Example 17: Show that the series

2 3x xx ...2! 3!+ + + converges

absolutely for all values of x.

Solution: Letn

n

xu

n!= and

( )

n 1

n 1

xu

n 1 !

+

+ = +

Now nn n

n 1

u n 1lim lim

u x +

+= , except, when x = 0. Hence by Ratio

test the series converges absolutely for all x except possibly zero.

But for x = 0, the series evidently converges absolutely. Hence the

series converges absolutely for all values of x.

Note: Since for a convergent series un, nnlim u 0

=

n

n

xlim 0n!

= , is a useful result.

Example 18: Test for convergence and absolute convergence the

series( )

( )

n 1

n 1

1 1 1 1...

log n 1 log 2 log 3 log 4

+

=

= +

+

Solution: We have

( )n

1u , n N

log n 1

=

+

Clearly( )nn n1

lim u lim 0log n 1

= =+

Since log x is an increasing function for all x > 0.

( ) ( ) ( )log n 2 log n 1 n 2 n 1+ > + + > +



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( ) ( )1 1

, v nlog n 2 log n 1

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Tests for the Convergence