Convergence Lec. 1

8/3/2019 Convergence Lec. 1

1/16

Ms. Deepali Gupta. JIIT, NOIDA

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.

++++ ..........8

1

4

1

2

11


2/16


Let Sn

be sum of first n terms of the infinite series un

i.e. Sn

= u1

+ u2+ u3 +.. + un. Then the infinite series is said to be convergent to a

number S if Sn tends to a finite limit S as n increases indefinitely.

SSlim nn=

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.t


3/16


Example

Sn

= 1+ 2 + 3 + + n

= n(n+1)/2

+

=

2

)1n(nlimSlimn

nn

Hence, this series is divergent.

Sn = 2 2 + 2 2 + 2 .

S1 = 2, S2 = 0, S3 = 2, S4 = 0

.yoscillatorisitHence.existnotdoesSlimnn


4/16


Geometric Series

=

=+++++1n

1n1n2 ar........ar.....arara

1n2

n ar.....araraS++++=

n1n2

n arar.....ararSr ++++=

)r1(aS)r1( nn =

r1

ar

r1

a

)r1(

)r1(aSlim

nn

nn

=

=

1|r|ICase < 0rthan n

SeriesConvergentr1

a

Slim nn =


5/16


1|r|IICase > nrthan

seriesDivergentSlim nn

=

Case III r = 1

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

r < -1, let r = -k where k > 1nnn

k)1(r =

))k(1(

)k)1(1(a

)r1(

)r1(aSlim

nnn

nn

=

=

evenisnif-oddisnif

== Oscillatory Series

r > 1


6/16


++ .........8421 Oscillatory Series

++++ .........4

27

2

932 Divergent Series

5 5 5 .........4 16 64 + +

Convergent Series to -1

Case IV r = -1

a - a + a - ..

Sn= a, if n is odd

Sn= 0, if n is even Oscillatory Series


7/16


N t h Term Test

0ulimthenconverges,uIfnn1n n

=

=

not true.isaboveofconverseThediverge.mayIt.convergentbenotmayseriesthen the0ulimife.i n

n

=

e.convergencforconditionsufficientanotbutconditionnecessaryaisulimHence nn

n21n u..........uuS:Pf +++=n1nn uSS +=

S.beinfinityuptosumitsand,convergentbeseriestheLet

SSlim nn

=

SSlim 1nn

=


8/16


9/16


++++++ ..........

n

1.......

4

1

3

1

2

11

n

1.......

4

1

3

1

2

11Sn +++++=

n1u n = 0

n1ulim n

n==

n

1...........

n

1

n

1

n

1.......

4

1

3

1

2

11 ++++++++

nn

n==

=

n

n

n

Slim

nS

Hence the series diverges.


10/16


Rat io Test

Let un be a series of positive terms such that thenruulim

n

1n =+

(a) The series converges if r < 1.

(b) The series diverges if r > 1.

(c) The test fails for r = 1.


11/16


Example

Test the series ...........x

4

5x

3

4x

2

32 32 ++++

1n

n xn

)1n(u

+=

n

1n x)1n(

)2n(u

+

+=+

x)1n(

)2n(n

u

ulim

2

n

1n

+

+=+ x=

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

If x = 1 ...........4

5

3

4

2

32 ++++

01n

1nlimu n =+

= Thus the series is divergent.


12/16


Root Test

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.

nn

nlim u r

=


13/16


Example

Discuss the convergence of the series

+

+

+

.............3

4

3

4

2

3

2

3

1

2

1

23

4

42

3

31

2

2

n

1n

1n

nn

1n

n

)1n(u

+

+

++= ( )1

1n

1nn/1

nn

1n

n

)1n(u

+

+

++=

( )

11n

n

n/1n

n n11

n11limulim

+

+

+=

1n

n n11

n11

n11lim

+

+

+=

( ) 11e1

1e1


14/16


Cauc hy s In t egra l 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.orfiniteisdx)x(f1

=

1n

n2ne

2xxe)x(f =

Examine the convergence of

=

1

x

m1

x

2

elimdxxe

2

2

+

2

e

2

elim

1m

m

2

finite.ise2

1=

Hence the series is convergent by Integral Test .


15/16


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.offunctiondecreasingpositiveaisx1f(x)then1,pIf

p=>

dxxdx

x

1

1

p

1 p

=m

1

1p

m

1p

xlim

+

=+

=

1

m

1lim

p1

11pm

finite.is1p

1

=

The series converges by Cauchys Integral Test for p > 1.


16/16


When p < 1

=

dx)x(f1

( )

1mlim

p1

1 p1m

The series diverges by Cauchys Integral Test for p < 1.

When p = 1

.........

n

1..........

3

1

2

1

1

1+++++

[ ] ==

111 xlogdxx1

dx)x(f

The series diverges by Cauchys Integral Test for p = 1.

Convergence Lec. 1

Documents

Transcript of Convergence Lec. 1