Convergence of Series

A series is an expression of the form u1 + u2 + u3 + …..+ un in which the successive terms follow some regular law.

For e.g. 1 + 3 + 5 + 7+……….+ 2n-1+……… 1+ x + x2 + x3 + ……..+ xn + …………., |x| < 1 1 + 2x + 3x2 + 4x3 +………

If a series terminates at some particular term, it is called a finite series.

If the number of terms is unlimited, it is called an infinite series.

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211

Let Sn be sum of first n terms of the series i.e. Sn = u1 + u2 + u3 + ……..un. The series is said to be convergent to a number S if

SSlim nn

i.e. An infinite series converges to a sum S if for any given small positive no. we can find a positive no N, st |Sn – S| < , n N, where sn is the sum of the first n terms of un

The series Sn is said to be divergent if

orSlim nn

The series Sn is said to be oscillatory if it does not tend to a unique limit, finite or infinite.

Example

Sn = 1+ 2 + 3 + ………+ n = n(n+1)/2

2

)1n(nlimSlimnnn

Hence, this series is divergent.

Sn = 2 – 2 + 2 – 2 + 2 –……. S1 = 2, S2 = 0, S3 = 2, S4 = 0

.yoscillatorisitHence.existnotdoesSlim nn

Geometric Series

1n

1n1n2 ar........ar.....arara1n2

n ar.....araraS

)r1(aS)r1( nn

n1n2n arar.....ararSr

r1ar

r1a

)r1()r1(aSlim

nn

nn

1|r|ICase 0rthan n

SeriesConvergentr1

aSlim nn

1|r|IICase nrthan

seriesDivergentSlim nn

Case III r = 1

a + a + a + ……..+ a = na Divergent Series

Case IV r = -1

a - a + a - ……..

Sn = a, if n is odd

Sn = 0, if n is even Oscillatory Series

.........8421

Case IV r < -1, let r = -k where k > 1nnn k)1(r

))k(1()k)1(1(a

)r1()r1(aSlim

nnn

nn

even isn if -odd isn if

Oscillatory Series

Oscillatory Series

.........4

272932 Divergent Series

.........165

165

45

Convergent Series to -1

Nth Term Test

0ulim then converges, u If nn1nn

not true. is above of converse The

diverge.may It .convergent benot may series then the0ulimife.i nn

e.convergencfor condition sufficient anot but condition necessary a isulim Hence nn

n21n u..........uuS:Pf

n1nn uSS

S. beinfinity upto sum its and ,convergent be series theLet

SSlim nn

SSlim 1nn

nn1nnnnulimSlimSlim

0SSulim nn

diverge. series theexist then tofailsor 0ulimIf nn

..........1n

n.......54

43

321

1nn.......

54

33

321Sn

1nnu n

1n/11

1ulim nn

Hence the series is divergent.

..........n

1.......4

13

12

11

n1.......

41

31

211Sn

n1u n 0

n1ulim nn

n1...........

n1

n1

n1.......

41

31

211

nn

n

nn

n

SlimnS

Hence the series diverges.

Ratio Test

Let un be a series of positive terms such that then ru

ulim

n

1n

(a) The series converges if r < 1.

(b) The series diverges if r > 1.

(c) The test fails for r = 1.

Example

Test the series ...........x45x

34x

232 32

1nn x

n)1n(u

n1n x

)1n()2n(u

x)1n(

)2n(nu

ulim 2

n

1n

x

Hence if x < 1, the series is convergent; and if x > 1, the series is divergent.

If x = 1 ...........45

34

232

01n

1nlimu n

Thus the series is divergent.

Root Test

r u limnn n

Let un be a series of positive terms such that then

(a) The series converges if r < 1.

(b) The series diverges if r > 1.

(c) The test fails for r = 1.

Example

Discuss the convergence of the series

.............34

34

23

23

12

12

3

4

42

3

31

2

2

n

1n

1n

n n1n

n)1n(u

1

1n

1nn/1

n n1n

n)1n(u

11n

n

n/1nn n

11n11limulim

1n

n n11

n11

n11lim

11e

11e 1

Thus the series is convergent.

Cauchy’s Integral Test

A positive term series f(1) + f(2) + f(3) +………+ f(n) +………. where f(n) decreases as n increases, converges or diverges according to the integral

infinite.or finite is dx)x(f1

Examine the convergence of

1n

n2

ne

2xxe)x(f

1

x

m1

x

2elim dxxe

22

2e

2elim

1m

m

2

finite. is e2

1

Hence the series is convergent by Integral Test .

Show that the p-series

.........n1..........

31

21

11

n1

pppp1n

p

( p a real constant ) converges if p > 1, and diverges if p 1

x.offunction decreasing positive a is x1 f(x) then 1,p If p

dxxdxx1

1

p

1 p

m

1

1p

m 1pxlim

1m

1limp1

11pm

finite. is1p

1

The series converges by Cauchy’s Integral Test for p > 1.

When p = 1

.........n1..........

31

21

11

111xlogdx

x1dx)x(f

The series diverges by Cauchy’s Integral Test for p = 1.

When 0 < p < 1

dx)x(f1

1mlim

p11 p1

m

The series diverges by Cauchy’s Integral Test for 0 < p < 1.

When p < 0

1n

pn

pn nu

p

nnnnLtuLt

The series diverges by Nth Test for p < 0.

Comparison Test

If two positive term series un and vn be such that

nonzero) and finite(rvu

limn

n

n

then both series converge or diverge together.

Here un is given and vn is chosen for which convergence and divergence is known.

Example

Test the convergence of series

........5.4.3

54.3.2

33.2.1

1

)2n)(1n(n1n2u n

n21

n11

n12

n1

2 2n n1v

2

n21

n11

n12

limvu

limn

n

n

n

vn is convergent as p = 2. So by comparison test un is convergent.

Higher Ratio Test

If un is a series of positive terms such that thenr1uu

nlim1n

n

n

(a) The series converges if r > 1.

(b) The series diverges if r < 1.

(c) The test fails for r = 1.

It is applied only when ratio test fails.

Test the convergence of series ...........8.7

x6.5

x4.3

x2.1

x 432

)n2)(1n2(xu

n

n

n221

n211

n211

xx

)n2)(1n2()2n2)(1n2(

xu

un

1n

n

1n

)2n2)(1n2(xu

1n

1n

xu

ulim

n

1n

n

Hence if x < 1, the series is convergent; and if x > 1, the series is divergent.

If x = 1 test fails

r1uu

nlim1n

n

n

1

)n2)(1n2()2n2)(1n2(nlim1

uu

nlimn

1n

n

n

2)n2)(1n2(

)2n8(nlimn

So the series is convergent.

)n2)(1n2(1u n

Logarithmic Ratio Test

If un is a series of positive terms such that thenruu

lognlim1n

n

n

(a) The series converges if r > 1.

(b) The series diverges if r < 1.

(c) The test fails for r = 1.

Test the convergence of series

0x.......,..........!4x4

!3x3

!2x21

3322

!nxnu

1n1n

n

)!1n(x)1n(u

nn

1n

1n1n

nn

n

1n

xn!n

)!1n(x)1n(

uu

1n

n

n)1n()1n(x

1n

n11x

xen11

n11xlim

uu

lim1n

nn

1n

n

Hence if x < 1/e, the series is convergent; and if x > 1/e, the series is divergent.

If x = 1/e, the test fails.

e1

n11

en!n

)!1n(e)1n(

uu 1n

n1n

1nn

n

1n

n11log)1n(elog

uu

log1n

n

1n1n

n

n11

euu

..........n31

n21

n1)1n(1

uu

log 321n

n

1n

1n

n e!nnu

..........n31

n21

n1..........

n31

n2111

uu

log 3221n

n

...............

n21

n31

n1

n21

22

..........n65

n23

uu

log 21n

n

....n65

23

uu

logn1n

n

123....

n65

23

uu

lognlim1n

n

n

The series is convergent.

Alternating Series

A series in which the terms are alternately positive and negative is an alternating series.

...............uuuu 4321

......n)1(.............

41

31

211

1n

Lebnitz’s Rule for convergence of an alternating series.

if converges..........uuuu u(-1) series The 43211n

n1n

0 u limn n

• un’s are all positive.

• un un+1 for all n N, for some integer N.

u1

s1

u2

s2 s3s4

u4

u3

Test the convergence of the series

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411

311

211)11(

011n

nlimulimnnn

.......56

45

34

232

.....

41

31

211.....)11111(

Series is oscillatory.

Absolute Convergence : A series un converges absolutely if the corresponding series of absolute values |un| converges.

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41

211

The geometric series converges absolutely because the corresponding series of absolute values converges.

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41

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211

The series is convergent by Lebnitz Test

.divergentis.........51

41

31

211

Conditional Convergence : A series that converges but does not converge absolutely.

.convergentabsolutelyisn

nsin1n

2