Sarvajanik College of

Engineering and Technology

Topic: CONVERGENCE OF AN INFINITE SERISE

AND
APPLICATIONS
Subject : Calculus
Faculty : Prof Jigisha pandya
Submitted by: Roll no
31 ) Mr Tandlekar Alaukik
32 ) Mr Mulchandani Nikhil
33 ) Mr Roy Parth
34 ) Mr Shekhaliya Parth
35 ) Mr Navadiya Keval
Department: Computer Engineering (Shift 1) 1st Year
Department Code: 07
 INFINITE SERIES:-
Given a set of number {a1,a2,a3,a4,………..},the
sum
A1+a2+a3+a4+.........

Is called the infinite series

CONVERGENCE OF SERIES
If we can find the sum of given series then the
series is converge otherwise diverge.
TEST OF CONVERGENcE
 Nth term test
 P-Series test
 Integral test
 Root test
 Ratio test
 Alternating series test
 Geometric series
 Telescoping series
 Comparison test
 Limit comparison test
 P-series test



1 Is call p-siries

n 1 n p

 n  1  2  3  4  5  .... n  ....

1 1 1 1 1 1
n 1
 Is called the harmonic series and it diverges since p =1.
Geometric Series



1 1 1 1 1
     ...
n 1 2 n 2 4 8 16



n 1
ar Where a = first term and r=common ratio
n 1
Each term is obtained from the preceding number by multiplying
by the same number r.

To know series converge or diverge we have to find it’s sum.in

geometric series we can find the sum of series in this way.
 The sum of a geometric series

sn  a  ar  ar  ar  ...ar
n1
2 3 sum of n term

rsn  ar  ar  ar  ...ar
2 3 n maltiply each term by r

sn  rsn  a  a r n

a  ar n
a(1  r ) n
sn   ,r 1
1 r 1 r
if | r | 1, r  0 as n  .
n

So we can find that

a
sn  , | r | 1
1  r
Then we can find the sum of series so series converge.
If r>1 the geometric series diverges.

So we can write that

Geometric Series:



ar n 1

convg a
r 1
 1 r
n 1  divg r 1
Comparison Test
 b is converge if series  c
n n is converge
where The terms of an are less than or
equal to term of bn

 b is diverge if series  a
n n is diverge
where The terms of an are greter than or
equal to term of cn
It’s like that

 bn c n

a n

the set of all series

lim a n  0
 Limit comparison test
If two serise are with positive term and
an
lim x    c, 0c
bn
 an and  bn
 
Then the following series both are
Converge or diverge. n 1 n 1

  an
 
an bn
limx  0 then if converge then
bn n 1 n 1

converge

 bn diverge then  an
 
a
lim x  n   then if
bn
n 1 n 1
diverge.
 Nth term test
Nth term test is the test for diverges
Because sequence converge only in one condition

THEOREM:

a


i 1
i
lim an  0
n 
Convergent
And series diverge in other all condition.

THEOREM:THE TEST FOR DIVERGENCE

a

lim a n  0
or
i n
i 1
lim a n DNE
Divergent
n
 Integral Test
If f (n ) = an for all n and f (n ) is 1)positive.

2)continues.

3)decresing.

for[1,∞)
 f ( x ).dx is converge then  a ( n ) is converge.


 f ( x ).dx is diverge then  a ( n ) is diverge.

1


1
Ratio test
 an be a series with positive terms

Let
and n N
an 1
lim n  
an

Then
The series converges if ρ < 1
The series diverges if ρ > 1
The test is inconclusive if ρ = 1
Root test
 an be a series with non-zero term

Let
and n N
lim n  n | an |  L

Then
The series converges if L< 1
The series diverges if L > 1 or is
infinite
The test is inconclusive if L= 1
Power Series (infinite polynomial inx)


c n x n  c 0  c1 x  c 2 x 2  .....c n x n ...
n0

Is a power series centered at x = 0.

A
N
d


c n ( x  a ) n  c 0  c1 ( x  a )  c 2 ( x  a ) 2  .....c n ( x  a ) n ...
n0

Is a power series centered at x = a.

Geometric Power Series



x n
1 x  x 2
 x 3
 x 4
 ... x n

n0
We can find when

a  1 r  x
a n d
a 1
S   , x  1
1  r 1  x
Convergence of a Power Series
There are three possibilities
1)There is a positive number R such that the series diverges for
|x-a|>R but converges for |x-a|<R. The series may or may
not converge at the endpoints, x = a - R and x = a + R.

2)The series converges for every x. (R =  )

3)The series converges at x = a and diverges elsewhere. (R = 0)

Telescoping series
Telescoping Series:

 (b


n  bn  1 )
n 1

Telescoping Series:
If we can find the sum of the
 (b b { bn }


n n 1 ) telescoping series the we can

said that the series is
n 1
Convergent Convergent

s n  b1  bn 1
telescoping series

 (b  b

s n n 1 )  b1  lim bn 1
n
n 1
Alternating Series

Two versions

• When odd-indexed terms are negative



( 1) k
 ak  a1  a2  a3  ...
k 1

• When even-indexed terms are negative

 (1)

k 1
 ak  a1  a2  a3  ...
k 1
Test for the convergence
1) { ak } is a decreasing sequence
( that is ak + 1 ≤ ak for all k )
2) lim a  0
k
k 

3) All term are positive

If this three condition is satisfied then we can say

that the sequence is convergent.
But if
1) { ak } is a decreasing sequence
( that is ak + 1 ≤ ak for all k )
2) lim a  0
k
k 

3) All term are positive

If any one of them is not satisfied then we can say that
the sequence is diverge by Leibniz’s test.
Absolute convergent
First forget bout sing of term If series converge in for

a
this sequence is converges
k

then we can said that the sequence is converge

absolutely.
conditionally convergent
If any sequence is converge absolutely but not converge by
Leibniz’s test then we can said that the sequence is converge
conditionally.