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ENGINEERING MATHEMATICS - I
UNIT - I
INFINITE SERIES
1.
An Expression of the Form
x1 + x2 + x3 + ..... + xn +……
.......(1)
where each term is followed by another term following some definite rule is called
a series. When the series contains an unlimited number of terms it will be called an
infinite series. When the number of terms is finite, it is called a finite series.
The study of finite series presents no problem. We can find out all the terms and
find their sum if necessary. The behaviour of an infinite series is however a matter
of interesting study and the problem is to find whether the series is convergent and
possesses a finite sum or whether it is divergent. The sum of an infinite series is to
be defined first and then we have to find whether it possesses a sum according to
the accepted definition.
For the series (1) written above, we define a partial sum
Sn = x1 + x2 + … + xn
and take the infintie series to be the same as the sequence (Sn) of partial sums.
Definition 1:
If X = (xn) is a sequence in R then the infinite series (or simply the series) generated
by X is the sequence S = (Sn) defined by
S1 =
x1
S2 =
S1 + x2 = x1 + x2
S3 =
S2 + x3 = x1 + x2 + x3
Sn =
Sn-1 + xn = x1 + x2 +……+ xn
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Example :
Let
1
, then
2n
1
1
1
 2  ...  n  ...
2 2
2
is an infinite series.
The traditional symbols used to denote the series S = (Sn) are

x1 + x2 +.... + xn +.... or


(xn) or
n 1
The symbol
2.
x
n
x
n 1
n
is used to denote both the series and its sum.
CONVERGENT, DIVERGENT AND OSCILLATORY SERIES
Definition 1:
Let X = (xn) be a given series in R and S = (Sn) be a sequence of its partial sums. If
lim S = lim (Sn) exists and is finite, we say that the series
x
n
converges.
The limit is called the sum of the series. When the series converges to the limit s,
we simply write
x
n
= s.
s is called the sum to infinity i.e., lim Sn = s.
Definition 2 :
A series
x
n
is said to be divergent, if the sequence (Sn) of its partial sums is
divergent.
Definition 3:
A series
x
n
, is said to be oscillatory if the sequence S = (Sn) of its partial sums is
oscillatory.
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Example 1:
x
Consider the series
n
, where xn =
1
Here the corresponding sequence
n(n  1)
S = (Sn) of the partial sums can be obtained as follows.
n
Sn =
1
=
k (k  1)

k 1
n

k 1
1 
1
 

 k k 1 
1 
1 1 1
1

= 1   +    + … +  

2  2 3

 n n 1
=1-
1
n 1
1 

Clearly, lim S = lim Sn = lim 1 
= 1
n 1 


S = (Sn) converges to 1. Therefore the given
 series is convergent to the
sum 1.
Example 2:
Consider the series 1 + 3 + 5 + ......
Sn = 1 + 3 + 5 + ....+ (2n-l) = n2
We have
lim
Sn = lim n2 = 
 S = (Sn) is divergent
The given 1 + 3 + 5 + .... is divergent
Example 3:
Let X = (xn) , where xn = (-1)n
We observe that
S2n
= (-1)1 + (-l)2 +....+(-l)2n
= -1 + 1 - 1 + 1-…+1 = 0
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 lim S2n = 0
and
S2n+1
= S2n + x2n+1
= 0 + (-1)2n+1 = -1
 lim S2n+1 = - 1
Thus, the two subsequences (S2n) and (S2n+1) of (Sn) converge to two different
limits. Therefore. S = (Sn) oscillates.
3.
CONVERGENCE
OF
A
SUM
OR
A
DIFFERENCE
OF
TWO
CONVERGENT SERIES
Theorem 1: Let
(i) the series
(x
(ii)
n
(x
 yn ) =
the series
 (x
(iii)
x
n
n
y
n
be two convergent series. Then
 y n ) converges and
x
(x
 yn ) =
the series
and
n
y
n
+
n
 y n ) converges and
n
x -  y
n
 (x
n
n
) converges and
 (cx
n
) = c  ( xn ) , where is a real
number.
(iv)
If w is a fixed element of R , then the series
 (w.x
(v)
If
n
x
n
) = w.
x
n
converges to x and
y
n
converges to y, and p, q  R

(pxn
+ qyn) converges to px + qy.
Theorem 2:
The removal or addition of a finite number of terms of at the beginning of an
infinite series does not affect the convergence or divergence of a series.
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Theorem 3:
The multiplication all the terms of a series by fixed nonzero real number has no
effect on its convergence or divergence or oscillation.
Theorem 4:
If the terms of a convergent series are grouped in parenthesis, in any manner to
form new terms new terms, then the resulting series will convergence to the same
sum.
4.
A NECESSARY CONDITION FOR CONVERGENCE OF A SERIES
LEMMA
If
 (x
n
) converges in R, then lim (xn) = 0 .
Solution :
Suppose
x
n
converges to s. Then the sequence S = (Sn) of its partial sums
converges to s. Therefore it follows that Sn-1 also converges to s.
Thus lim xn = lim (Sn –Sn-1)
= lim Sn – lim Sn-1
=s–s=0
Note : If
x
n
is a given series such that lim xn  0 , then
x
Example:
Consider the series
x
n
=
1

4
2
n
+…..+
+…
2(n  1)
6
5
n
is not convergent
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We have
=


1
n
=
2(n  1)
2
xn =
1
2
1
 n 1 


 n 


1
1 
lim xn = lim
1/ 2
2 
1
 1  
n
 
x
n
1/ 2
n
n 1
=
1
2
1
1

1  
n

1/ 2


 = 1 0

2


is not convergent.
Theorem 3.5 : Let X = (xn) be a sequence of positive real numbers.
Then
x
n
converges if and only if the sequence S = (Sn) of partial sums is
bounded.
In this case
x
n
= lim (xn) = sup (Sn)
Proof: xn > 0  Sn – Sn-1 > 0  Sn > Sn-1
Hence S = (Sn) is a monotone increasing sequence.
According to monotone convergence theorem, the sequence S converges if and only
if it is bounded.
5.
ABSOLUTE CONVERGENT SERIES
Definition 3.5: Let X = (xn) be a sequence in R. We say that the series
absolutely convergent if the series

| xn | is convergent in R.
Example: The series
1 + nx +
n( n  1) 2
n(n  1)....( n  r  1) r
x +….+
x + ….
2!
r!
6
x
n
is
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is absolutely convergent when | x | < 1.
6.
CONDITIONALLY CONVERGENT SERIES
Definition 6 : A series is said to be conditionally convergent if it is convergent but
not absolutely convergent.

Example : The series

(-1)n+1
n 1
7.
n
in R is conditionally convergent.
n 1
CONVERGENCE OR DIVERGENCE OF GEOMETRIC SERIES

Theorem 3.6: The series 1 + a + a2 + … +an+1 +…  a n (a is common ratio) … (1)
n 0
(i)
Converges if | a | < 1
(ii)
diverges if a > 1
(iii)
oscillates finitely when a = -1
and (iv) oscillates infinitely when a < -1 .
Proof: (i) when | a | < 1 In this case we have Sn =
 Lim Sn
= lim
= lim
=
1 an
1 a
1 an
1 a
an
1
-lim
1 a
1 a
1
( | a | < 1  lim an = 0)
1 a
(in this case (Sn ) is convergent and therefore, the geometric series (1) is
convergent.
(ii) when a > 1
In this case Sn =
an 1
a 1
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
lim Sn = lim
an 1
=
a 1
( | a | > 1  lim an = 0)
 when a > 1, (Sn) is divergent. Therefore the geometric series is
divergent when a > 1 .
when a = 1
In this case : Sn = 1 + 1+.... to n terms

Sn = n
 lim Sn = lim n = 
 (Sn) is divergent when a = 1 and the geometric divergent a = 1.
Therefore, the geometric series is divergent when a > 1
(iii)
When a = -l; We have Sn =1-1 + 1-1 + .... to n terms
 1 when n is odd
 Sn = 
 0 when n is
even
 lim Sn = 1 or 0
 (Sn) oscillates finitely.
 the geometric series is divergent when a=-l.
(iv) when a < -1 . Let r = -1 -a then a < -l  -a > 1  x > 1
 Sn = 1 + a + a2 + … + an-1
=
1  a n 1  ( r ) n
( a = -r)

1 a
1  (r )



 Sn = 


1 rn
when n is even
1 r
1 rn
when n is odd
1 r
Now r > 1  lim rn = 
 -  when n is even
  when n is odd
 lim Sn = 
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 the Geometric series given by (1) oscillates
Thus (Sn) oscillates infinitely,
infinitely when a < -1.
Example 1:
 1
  2
Consider the series
common ratio

1
2

1
3
1 
 the series
3n 
1
2
n
is a geometric series with
1
< 1.
2
1
3
n
is also a geometric series with
1
< 1.
3
is convergent
n
 1
  2

is convergent and the series
n
common ratio

n
n

1 
 is convergent
3n 
1 1
1
Example 2: The geometric series 1   2  3  ... is convergent.
3 3
3
Since the common ratio a = -
1
1
 |a|=
<1
3
3

Example 3: Test the convergent of the series

n 1
1
3 1
n
Solution:
We have

1
3
n
xn =
1
1
1
where n
< n for n
3 1
3 1
3
n
is a Geometric series with common ratio
9
1
1
such that
<1
3
3
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
8.
1
3
is convergent.
Convergence or divergence of Harmonic series 
Theorem 3.7: The Harmonic series
x
n
=
1
n
1
:
n
is divergent.
1
n
and lim xn = 0, but we show that the series is not convergent. We construct a
subsequence (Skn) ) as follows.
If k1= 2, then Sk1 = 1 +
1
2
If k2= 22, then Sk2 = 1 +
1
1
1
+
+
2
3
4
and if k3 = 23 then Sk3 = 1 +
Now
Sk1 =
1 1 1 1 1 1 1
     
2 3 4 5 6 7 8
1 1

1 2
Sk2 = = 1 +
1 1 1
 
2 3 4
 Sk2 > Sk1 +
1 1

4 4
 Sk2 > Sk1 + 2.
i.e. Sk2 > Sk1 +
Similarly, Sk3 > Sk2 +
1
2
1
2
10
1
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… Skr > Skr-1 +


1
r
>1+
2
2
r
2
Skr > 1 +
the subsequence (Skn) is not bounded and the Harmonic series does not
converge.
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P-Series (Auxiliary Series).
Theorem 8 : The series
1
n
(i)
converges, if p > 1
(ii)
diverges, if p < 1
p

1
1
1
1
 p  p  .....  p  ...
p
1
2
3
n
Proof : Case (i) when p > 1
We combine the terms of the series into the groups and write.
1
n
p

1  1
1 
 p  p 
p
1
3 
2
1
1
1 
 1
+  p  p  p  p   ....
5
6
7 
4
(the groups consisting of 2, 4, 8,... terms etc.)
1
1
1
1
4
 p  p  p  p ….
p
4
5
6
7
4

1
n
p

1  1
1
 p  p
p
1
3
2
1
n
p

1
2
4

 +… < p  p  p  ....
1
2
4

1
1
1
 p 1  p 1 2  ...
p
1
2
(2 )
… (1)
the series on R.H.S of (i) is an infinite Geometric series with common
ratio =
1
2 p 1
<1
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Therefore, the series on R.H. S of (i) is convergent consequently
1
n
p
is
convergent.
Case II: when p = 1
1 1 1 1
    .....
n 1 2 3
We have  
… (2)
1
 n is divergent (refer 3. 9)
The Harmonic series
Case III : When p < 1 :
1
1
1
1 1
1

i.e.,  p  , p  ...
p
n
2 3
3
n
2
In this case np < n 

1
1 1
1
> 1 +   .....   ...
p
2 3
n
n
...(iii)
Thus in the present case, each term of the series exceeds the responding term of
the series on R.H.S of (iii)
1
 n is divergent.
But

1
is divergent in this case.
np
Example 1:
Examine the converge of the series
1
1
4
2/3

1
9

2/3
1
 ..... to 
16 2 / 3
Solution :
We have
x
n
 1
= 1+
=
1
4
1
(2 )
1
4/3
1

2/3
2 2/3

1
92 / 3

1
2
4/3
 ...
1
(3 )

2 2/3
1
3
4/3


1
(4 )
2 2/3
1
42 / 3
12
 ...
 ...
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
=
n
1
4/3
1
is of die form
1
n
where
p
p=
4
.
3
 Since P > 1 , the given series is
convergent.
Example 2:
Test the convergence of the series

1
n
Solution :
We have

1


n
1
=
1
n
p
where p =
1
<1
2
is divergent.
n
Theorem 3.9 : Let X = (xn) be a sequence of non-negative real numbers.
Then
x
n
converges if and only if the sequences S = (Sk) of partial sums is
bounded. In this case


= lim (Sk)
n 1
= sup sup {Sk. : k  N}
Proof: Since xn > 0 for all n  N , the sequence S = (Sk) of partial us is monotone
increasing. Therefore, we have
S1 < S2 < S3 < ..... < Sk <....
By monotone convergence theorem the sequence S = (Sn) converges if and only
if it is bounded, in which case the of lim (Sk) = sup {Sk. : k  N}.
Series positive terms (non-negative terms)
When every term of the series
 x
n
is positive we have a series of positive
terms: We discuss the following in connection with such series.
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1.
If
x
n
be a series of positive terms it must be convergent or
divergent to  : If the sequence of partial sums (Sn) is bounded, then
the series will be convergent. If it is unbounded then it is divergent to
+.
2.
(Comparision test I): If
x
n
and
x
terms, such that every term of
y
corresponding term of
x
convergent,
n
n
y
be both series of positive
n
is less than or equal to the
n
y
and if
is convergent, then is
n
is also convergent.
Let Sn = x1 + x2 +...+ xn and Sn = y1 +y2 +....+ yn
then Sn < Sn

lim Sn < lim Sn
 lim Sn < s’ where lim Sn' = s'
Thus
3.
x
n
is convergent to its sum where s = s'.
(Comparision test II): If
x
terms and if every term of
corresponding terms of
y
n
n
and
y
x
, then if
n
be both series of positive
n
is greater than or equal lo
y
n
is divergent.
x
n
is also
divergent.
Let Sn = x1 + x2 +.....+ xn and Sn' = y1 + y2 +....+ yn then ,Sn > Sn' so
that lim Sn > lim Sn'
y
n
Hence
is divergent  lim Sn'    lim Sn  
x
n
is divergent.

Example 1: Test the convergence of the series
n 2
Solution :
14
1
 (log n)
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 xn =
We have
But
1
 (log n)
n 2
1
1
for all n > 2

log n n
1
we find that xn > yn  n  N .
n
taking yn =
But

y
(Since
n
=
1
n
p
1
 n divergent.
where p = 1 is divergent for P < 1)
 By comparision test

1
 (log n)
is also divergent
n 2

Example 2: Test for the convergence of the series
log n
3
 1)
 ( 2n
n 1
Solution :
We have xn =
log n
n
n
1
 3  2 Taking y n  2 : we find that
3
2n  1 n
n
n
xn < yn for all n > 2

y
n
=

1
1

where p > 1
2
n
np

y
n
=

1
is convergent
n2
 By comparison test,
 xn =

log n
is convergent.
3
 1)
 ( 2n
1
EXERCISE
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1
1
1
 2  ....  n 1 is convergent.
3 3
3
1.
Show that the series l +
2.
Show that the series does not convergent.
1
n
(a)

(c)
n
(d)
 cos n
(e)
1 3 7
2 n1
   ....  n
 ...
3 5 9
2 1
(f)
x
(b) l + 3 +32 +....+3n-1+....
1
n
where xn 
1
1
1
n
(g) l + 2 + 22 + 23+.....
3.
Using Cauchy's Criterion. Prove that the Harmonic series

1
1
1
1
 n  1  2  3  ...  n  ...
n 1
4.
Using Cauchy's criterion prove that the series

1
1
1
1
 n!  1  1!  2!  ...  n!  ... converges.
n 1
5.
Using Cauchy's general principle of convergence show that
1
 2n 1
diverges.
Theorem 3.10: (Comparision test III):
Let
x
n
and
y
n
be two series of positive terms such that xn < kyn where k is a
fixed positive number. If
y
n
converges then
x
n
also converges.
Proof: We have xl + x2 +....+xn < k(y1 +y2 + ...+yn)
Let
y
n
Sn = x1 +.x2 +....+ xn
is convergent  There exists a real number M such that
16
...(1)
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(y1 +y2 +….+yn) < M  n
from (1) and (2) we have
...(2)
x1 +x2 +….+xn < kM
 Sn< kM for all n
x
Thus, the sequence (Sn) of partial sums (Sn) is bounded. Hence
n
is convergent.
Theorem 3.11: (Comparision Test IV):
Let
x
n
y
and
n
be two series of positive terms such that xn > kyn for all n > m
where k is a fixed positive number and is a fixed positive integer. If
divergent, then
x
n
y
n
is
is divergent.
Theorem 3.12: (Comparision test VI):
If
x
n
y
and
n
x n 1
y
 n 1
xn
yn
(i)
be series of positive terms and if
for all n , then when
y
is convergent,
n
x
n
is also
convergent.
(ii)
x n 1
y
 n 1 for all n , then when
xn
yn
y
n
is divergent
x
n
is also divergent.
Proof : From the given in equality.
xn1
x
x1
 n  ...... 
 k (say)
y n1
yn
y1
 xn+1 < k yn+1 for all n

x
n
is convergent (by comparision test)
(iii) Here we arrive at the result xn+l > k yn+1 for all n

x
Cor: If
x
n
n
is divergent.
is a series of positive terms and
17
x
n
is convergent, then
Prepared by Dindigala Raju
xn
 1 x
xn > 0  l + xn > l 
Proof:

x
n
1
<1
1xn
xn
< xn for all n
1xn
 0<
Since
is convergent
n
xn
< xn for all n
1xn
converges, by comparision test
xn
 1 x
is also converges
n
SEQ
Example 1: Test for the convergence of the series
1
3
5
2n  1


 .... 
 ....
1.2.3 2.3.4 3.4.5
n(n  1)( n  2)
Solution :
We have
xn 
2n  1
n(n  1)( n  2)
1

n 2  
n

=
1 
2

n 3 1  1  
n 
n

1
n
=
1
2



n 2 1  1  
n 
n

2
18
Prepared by Dindigala Raju
Let
1
n2
yn =
Then
lim


1


2
1 

n
=
...
 2
1 
2  1 
 n 1  1    2  
n 
n n 
 
xn
yn
1
n
2
= lim
 lim

=2 0
xn
is a finite and non-zero quantity.
yn
 By comparision test
But
1 
2

1  1  
n 
n

y
y
n
=
n
1
n
2
x
n
y
and
is in the form
n
behave alike.
p
, where p = 2 > 1
1
n
is convergent Therefore,
Example 2: Test the convergence of
x
is also convergent
n
x
n
where xn =
Solution :
We have xn =
n2  n 1
n3  2
1
1
 2
n n
2
1 3
n
n2 1

n3
19
n2  n 1
n3  2
Prepared by Dindigala Raju
1

Let
yn =
then
lim
1
1
 2
n n
2
1 3
n
1
n
1
n

1
1
 1  2
n n
= lim 
2

1 3
n

xn
yn


 =1



A finite and non-zero quantity.
x
 By comparision
But
and
n
y
n
=
y
n
1

n
both behave alike.

1
is divergent.
n1 / 2

1
1
1
   p   1 / 2 is divergent for p = < 1
2
n
n


x
n
=
n2  n 1
is also divergent
n3  2


Example 3: Test the convergence of the series
3
n3  1  n
n 1
Solution :
Let x n  3
n 3  1  n  (n 3  1)1 / 3  n
 (n )
3 1/ 3
1

1  
n

1/ 3
1/ 3


1
 n 1    1
n


20
n




Prepared by Dindigala Raju

1
1
 
 n 1  3  6  ....  1
3n
9n
 

Let
x
n
1 1
1

 3  ...
2 
n  3 9n

1
n2
1
1

 3  9n 2  ... 1

  0
1
3


 2
n 
xn
is finile and non-zero.
yn
 lim
y

xn
= lim
yn
lim
But
n 1
1

 3  ...
3 
n  3 9n

1
then
n2
yn =


n
=
y
and
1
n
2
By comparison test.
both behave alike.
n
is convergent by p-series test.
x
n
is also convergent.
Example 4:

Test for the convegence of the series

n 1
n
n 1
2
Solution :
We have
xn 
n
n 1

n1 / 2
2
n 2 (1 
1
)
n2
1

n 3 / 2 (1 
21
1
)
n2
Prepared by Dindigala Raju
Taking yn =
1
we get
n3/ 2
xn
= lim
yn
lim

x
n
y
and
1
1

1
n2
y
behave alike but
n
1
n
where P =
p
1
 1  0 ( a finite number)
1 0
n
n
=
1
3/ 2
is of the form
3
>1
2
  y n is convergent
  x n is convergent
Example 5: Examine the convegence or divergence of the series
1
1 2
1

2 3
1

3 4
 ....
Solution :
x
Let
n
1
=
1 2
1
 xn =
Taking yn =
 lim
 lim
1
n
n  n 1
xn
=
yn
we get
xn
= lim
yn
1
1 1
1
n
1

2 3


1
 1
n 1  
 n
1
1 1
=
1
n
1
1
 0
11 2
xn
is a finite and non-zero number.
yn
22
1
3 4
 ....
Prepared by Dindigala Raju

x
y
n
n
y
and
1

=
n
be have alike but
n
n
from p -series test
1
1/ 2
is of the form
y
 The given series
n
1
n
p
where p =
1
<1
2
is divergent.
x
n
is divergent

Example 6: Test for the convergence of the series

tan
n 1
1
n
Solution :
1
then
n
Let xn = tan
 xn = tan
=
1
n
Let yn =
1
1 1 1
2 1
=
 . 3  . 5  ...
n 3 n
15 n
n
1
2


1  3n 2  15n 4  ...


1
then
n
xn
= lim
yn
Lim
1
n
1
2


1  3n 2  15n 4  ...


1
 
n


1
2
= lim 1  2 
 ... = 1  0
15n4
 3n

 lim

xn
is a finite, non-zero real number.
yn
x
n
and
y
n
both behave alike but
y
divergent.
 By comparison test
x
n
is also convergent
23
n
=
1
n
(Harmonic series) is
Prepared by Dindigala Raju
Example 7:
Test the convergence of the series
2
3
4
 p  p  ...
p
1
2
3
Solution :
1
1 


n1   n1  p 
n 1
n
n
  p    p 1 
xn =
p
n
n
n
Let
Taking
yn =
1
n p 1
1
1
xn
lim n = lim p n1 .np-1
n
y
We get
1

= lim 1   = 1  0
n

xn
 lim n is a non-zero finite number.
y

x

x
n
n
and
y
n
both behave alike
converges when p-1 > 1 i.e.. when p > 2 and diverges when p - 1 < 1 i.e.,
when p < 2 .
 The given series is convergent when p > 2 and divergent when p < 2.
Example 8:
Test for convergence
2
n
1
1
Solution :
Let
x
n
=
2
2n + 1 > 2n 
n
1
1
1
1
< n
2 1
2
n
24
Prepared by Dindigala Raju
If We take yn =
1
then we have xn < yn  n  N
2n
n
1
But the series  y n =    , is a geometric series with common
2
1
< 1, therefore
2
y
By the comparison test
x
ratio
n
n
is convergent.
=
2
n
1
is also convergent.
1
Grouping of Series
We can group the terms of a series, in Parenthesis in any manner to form new terms (the
order of the terms remaining unaltered).
n2
 3 4  5 6 
 2n  1 2n  2 
For example:       .....(1) n 1
 .....  


n 1
2n  1 
 2 3  4 5 
 2n
is obtained by grouping the terms of the series
3 4 5 6
n2
    .....  (1) n 1
 .....
2 3 4 5
n 1
Note: The grouping the terms of a series does not affect the convergence the series.
Theorem 13 : If a series
x
n
is convergent, then any series obtained obtained form it
by grouping the terms is also convergent to the same value.
Proof: Let
x
n
be a convergent series and
y1 = x1 + x2 +…..+ xk1
y2 = xk1 + xk1+2 +…..+ xk2
Let Sn denote the nth partial sura of the sequence
sum of
y
k
x
n
and the tk denote the kth partial
, then we have t1 = y1 = Sk1, t2 = y1 + y2 = Sk2, …..
25
Prepared by Dindigala Raju
Thus, the sequence (tk) of partial sums of the grouped. Series
the sequence (Sn) of partial sums of
y
k
x
. Since
n
x
n
y
n
is a subsequence of
convergent, the grouped series
is also convergent.
Remark :
A given series may converge after parren thesis are inserted, but the series may not
convege when they are removed for example : Consider the series
(1 – 1) + (1 – 1) + (1 –1 ) + …
which converges to zero. When the parenthesis is removed the series can be written as
1 – 1 + 1 - 1 + ....
The above series oscillates between 0 and 1. Therefore it does not converge.
Rearrangement of series : For a finite number of terms, say
x1 + x2 + x3 +….+ xn
the algebraic sum is not affected by writing the series in a different order changing the
order of terms of a sequence is called rearrangement of series.
Definition 7:
Let
x
n
be a series in R . A series
y
k
is a rearrangement of
x
n
. If there is a
bijection f of N onto N such that yk = xf(k) for all k  N.
Remark : The rearrangement of given series may affect the convergence of the series.
Theorem 14 : Rearrangement Theorem :
x
n
be an absolutely convergent series in R. Then any rearrangement
converges to the same value.
Proof : Let
x
n
be an absolutely convergent series in R and lim xn = x .
For  > 0, let N be such that if n, q > N and Sn = x1 + x2 +…..+ xn then
26
y
k
of
x
n
Prepared by Dindigala Raju
| x – Sn | <  and
q

| xk | < 
k  N 1
Let M be a natural number such that all of the terms x1, x2 ,....,xN are contained as
summands in tM = y1 +y2 +...+ym. It follows that, if m > M, then tm - sm is the sum of a
finite number of terms xk with index k > N.
q

Hence. for some q > N , we have | tm – sn |
| xk | < 
k  N 1
Therefore, if m > M then we have
| t m – x | < | t m – sn | + | s n – n | <  +  = 2 
Since  > 0 is arbitrary we infer that
y
k
converges to x.
EXERCISE
1.
Test for the convergence of the series
(a) 1 
2.
1 1
1
  ..... 
 ...
3 5
2n  1
(b)
1
1
1


 ...
3  7 4.9 5.11
(c )
1
1
1


 ...
1.3.5 2.4.6 4.6.8
Show that
x
k
where x n 
1
1

1  
n
n
is divergent.
3. Test for the convergence of the series

(i)

( n 1  n )
n 1
4.

(ii)

( n4 1  n4 1)
n 1
Test for the convergence of
27
Prepared by Dindigala Raju

(i)


1
n
sin
n 1
(ii)
1
2

Show that
6.
Show that the series (a)
1 2
1 2 3


n 1
7.

3
 … is divergent
1 3 4
(n  2)( n  3)
and (b)
n(n  1)


n 1
n
are divergent.
1  2 n
Discuss convegence of the series

(i)
n
2
n 1

i


(iii)
1
(ii)
(iv)
n  n
2
n
3
(n  1)( n  2)
1
n3 n

n 1  n
n2

1
 1  n3

1
n 1  n
np

(vi)

1

(vii)


n

(v)

n2
8.
tan
n 1
5.
1
n

1
n log n
2
Test for convergence
(i)
1
1
1


 ...
1.2.6 2.3.7 3.4.8
(iii)
2 1
3 1
4 1
 3
 3
 ...
3
3 1
4 1
5 1
(ii)
1
3
5


 ...
4.6 6.8 8.10
(iv)
1
2
n

 ..... 
 ...
5
7
2n  3
Theorem 15: Cauchy's Root Test
(a) If X = (xn) is a sequence in R and there exists a positive number r < 1 and natural
number K such that | xn |1/n < r for n > K
then the series
(b)
x
n
is absolutciv convergent.
If there exists a number r > 1 and a natural number K such that
xn
1/ n
> r for n > K then the series
x
28
n
is divergent.
Prepared by Dindigala Raju
1/ n
Proof: Let x n
< rn
then for 0 < r < 1. then series
test
(c)
x
If x n
n
1/ n
r
n
is convergent. Therefore, from the comparision
is absolutely convergent.
> r for n > K then x n
1/ n
> rn
x
but r > 1  lim lim (1 xn |)  0 
Working rule: If: lim x 1n / n < 1, the series
If : lim x 1n / n > 1, the series
and
x
x
n
is divergent
n
n
is convergent
is divergent
If lim x 1n / n = 1, the test fails.
In the last case cannot conclude that the series is divergent.
Corollary 1: If r satisfies 0 < r < l and if the sequence X =(xn)
Satisfies x n
1/ n
> r for n > K (K  ER). then the partial sums Sn, n > k.
approximate the sum S =
|S – Sn| <
x
n
according to the estimate.
r n 1
for n > k
1 r
Proof : If m > n > k.
we have |Sm – Sn| = |xm+1 – xn+2 +….+ xn|
= | xn+1 | + | xn+2 | +….+ | xm |
< rn+1 +….+ rm <
r n 1
1 r
Now. taking the limit with respect to m , we obtain
| S – Sn | <
r n 1
for n > k
1 r
Corollary 2: Let X - (xn) be a sequence in RP and let r = lim (|x|1/n)
29
Prepared by Dindigala Raju
x
Whenever this limit exists. Then
n
is absolutely convergent when r < 1 and is
divergent when r > 1.
Proof: If limit (|x|)1/n exists and less than 1, then there is a real number ri with r < ri < 1
and a natural number K, such that |x|1/n < ri for n > K.
In this case the series is absolutely convergent. If lim (|x|)1/n >1, then there exists a real
number r2, and a natural number K’ such that |x|1/n > r2 for n > K2 in which case
is divergent.
SEQ

Example 1: Discuss the convergence of the series

N 1
Solution :
 n 
In this case we have x n  

 n 1 
 x 1n / n
 n 
= 

 n 1 
=
n2
1
 n 1 


 n 
 lim |xn|1/n = lim
n
1/ n
 n 
=

 n 1 

n
1
1

1  
n

1
1

1  
n

 by Cauchy's root test
n2
n
x
n

n
1
1
e
is convergent.

Example 2: Test for convergence

n 1

1 
1 

n 

30
n 3 / 2
 n 


 n 1 
n2
x
n
Prepared by Dindigala Raju
Solution :

1 
Let xn = 1 

n 

x
1/ n
n
n 3 / 2
then xn > 0 for all n  N and
n3 / 2


1 
= 1 


n 



1 
= 1 

n 

1/ n
 n
1
=
n

1 
1 

n 

1
 lim |xn|1/n = lim

1 
1 

n 

Hence, by Cauchy's rool test

n
x
n
1
1
e
converges.

Example 3: Test for convergence

1
2n
n3
Solution :
 2n
2n
Let xn = 3 ; then xn > 0  n  N and x 1n / n =  3
n
n
 lim | xn |1/n = lim
lim
1
n 
1/ n 3
2
n
3/ n
= 2. lim
1
n3/ n
= 2 ( lim n1/n = 1)
Hence, by Cauchy's root test.  xn is convergent.
31



1/ n
=
2
n
3/ n
Prepared by Dindigala Raju


Example 4: Test for convergence
n2
1
(log n) n
Solution :
Let xn =
x
1
then clearly xn > 0 for all n and
(log x) n
1/ n
n
1
=
(log n) n
1/ n
=
1
| log n |

1
1
1
1
  0 <
< and lim
= 0  lim
=0
n
n
log n
log n

x
Therefore by Cauchy's root test
n



is convergent
 nx 
Example 5: Examine the convergence of the series 

 n 1 
n
Solution :
 nx 
Let yn = 

 n 1 
 then y 1n / n
n
 nx  n 
= 
 
 n  1  
 | yn |1/n = lim
1/ n
=
nx
n 1
nx
= | x | lim
n 1
1
1
1
n
 1 
= x. 
=x
 1 0 
By Cauchy’s root test
y
n
is convergent when x < 1
x > 1.
32
and
y
n
is divergent when
Prepared by Dindigala Raju


n
1
 n 
when x = 1 we have yn = 
 = 

 n 1 
1
 1  
n
 
 lim yn = lim

x
and
1

1  
n

n
=
1
0
e
is convergent when x < 1
n
x
1
n


1
 =
n

1


1  
n


n
is divergent when x > 1


Example 6: Discuss the convergence of
e
n
xn, x > 0.
n 1
Solution :
Let yn = e
n
xn then yn > 0 ,  n  N
 lim = y 1n / n = lim ( e
 By Cauchy's root test
n
xn)1/n = lim e 1
y
n
y
n
.x = x
converges when x < l and diverges when x > 1.
When x = l, we have lim yn = lim e

n
n
= .
divergent when x = 1.
 The given series is convergent when x < 1 and divergent when x > l.
EXERCISE
1.
Test for convergence

(i)

n 1
1

1  
n

n 2

(ii)

n 1
 n3
 n
3
33




(iii)

n 1
1

1  
n

n
Prepared by Dindigala Raju

(iv)
nn
2
(n  1) n
3
(vii)  n. 
4
2.
n
(vii)

n2
(vi)

 n 1 
 n 
 3 
n
n2
3n
Discuss the convergence of the series

(a)

1

©

1

xn
,x>0
n
(b)

1
xn xn
, (x > 0)
n! n!
3. Test for convergence .
4.

(v)
2
 n 


 n 1 

(d)

1
x 2n
,x>0
2n
 n 1  n

 x , (x > 0)
 n2
n
2
32 2
(n  1) n n
. x +… x > 0
.
x

.
n

...

12
23
n n 1
Discuss the convergence of

(a)

nk xn when k > 0 , and x > 0
n 1

(b)

2
p n xn (where p > 0 , x > 0)
1
D' Alembert's Ratio Test: Theorem
(a)
If X - (xn) is a sequence of non-zero elements of R and there is a positive integer
r<1 and natural number K, such that
X n 1
Xn
< r for n > k.
(b) If there exists a number r > 1 and natural number K such that
then the series

(xn) is divergent.
Proof:
34
X n 1
Xn
> r for n > k
Prepared by Dindigala Raju
(a)
X n 1
Xn
< r for
n > k hold. Then by induction we have |Xk+m | < rm | xk | for
m>l.
It follows that for n > K. the terms of
terms of the geometric series
r
n
x
n
are dominated by a fixed multiple of
with 0 < r < 1, from the comparison test,
converges.
(b)
If
X n 1
Xn
> r for n > K then by induction we have
|Xk+m | > rm | nk | for m > 1
Since r > 1, it is impossible to have lim | xn | 0 .
Hence, the series cannot converge.
Corollary 1. If r satisfies 0 < r < 1 and if the sequence X = (xn) satisfies.
X m 1
Xn
< r for n > K (where r < 1 and k  N)
then the partial sums approximate the sum
S =
x
n
according to the estimate
| S – Sn | <
Proof:
X m 1
Xn
r
xn for n > k
1 r
< r  | Xm+1 | < r | Xn |
 | Xm+1 | < rK | Xn | when n > K
 therefore, if m > n > K, we have
| Sm – Sn | = | xn+1 +....+ xm | < | xn+1 |+......+ | xn |
< (r + r2 +.... + rm-n) | xn | <
r
xn
1 r
Taking limits with respect to m, we obtain
35
x
n
Prepared by Dindigala Raju
| Sm – Sn | <
r
xn for n > K
1 r
| x |
Corollary 2: Let X = (xn) be a sequence in R and set r =  n 1 
 xn 
x
Whenever the limit exists. Then the series
n
is absolutely convergent when r < 1,
and divergent when r < 1.
Proof: Let us suppose that the limit exists and r < 1.
If r1 satisfies r < r1 < 1, then there exists a natural number K such that
| xn1 |
< r1 for n > K
xn
By, D' AIembert's ratio test, the series is absolutely convergent.
If r < 1 and if r2 satisfies 1 < r2 < r, then there exists a natural number K. Such
that
| xn1 |
> r2 for n > K. Therefore in this case the series
xn
x
SEQ
1 1 1
Example 1: Test for convergent the series 1     .....
1! 2! 3!
Solution:
We have
x
then
 lim
n
1 1 1
1     .....
1! 2! 3!
xn =
1
1
1
 xn+1 =

(n  1)!
n! n.(n  1)!
| xn1 |
1
(n  1)!
= lim
n.(n  1)!
xn
= lim
1
=0<1
n
36
n
is divergent.
Prepared by Dindigala Raju
Hence by D' Alemberts ratio test
x
n
converges.
Example 2: Test for congence the series 1 
1 1
1
 2  .....  n
2
2 3
n
Solution :
1
nn
We have xn =
 Xn+1 =
Now
lim
1
( n  1) n 1
| xn1 |
1
= lim
.x n
n 1
xn
(n  1)
= lim
1
nn
.
(n  1) (n  1) n
= lim
=
1
nn
.
n  1  n  1n


 n 
lim
1
n 1
lim
1
 n 1 


 n 
n
=0<1
 by D Alemberts ratio test the given series is convergent

Example 3: Test for the convergence of the series

n 1
Solution :
Let xn =
2 n 1 (n  1)!
2 n .n!
then
x
=
n+1
(n  1) n 1
nn
37
2n.n!
nn
Prepared by Dindigala Raju
 lim
x n 1
2.2n ( x  1)n! n n
= lim
.
(n  1)( n  1) n 2n.n!
xn
 n 
= 2. lim 

 n  1
n
= 2.lim
1
 1
1  
 n
=
n
 Hence by D' Alemberts ratio lest


n 1


Example 4: Test for convergence
n 1
2
<1
e
2n.n!
converges.
nn
n3  a
(a > 0)
2n  a
Solution :
We have xn =
lim
x3  a
2n  a
x n 1
(n  1)3  a 2n  a
= lim
.
2n 1  a n3  a
xn
 1 3 a  
a 
n 3 1    3 .2 n 1  n 
 n  n   2 
= lim
a  
a

2 n 1 1  n 1  n 3 1  3 
 2   n 
  1 3 a  
a 
 1    3  1  3  
1
  n  n   n  
= lim  
a 
a 
2

 1  n 1 1  3  
  2  n  


=
1
<1
2
 by D' Alemberts ratio test
x
n
converges.
38
Prepared by Dindigala Raju
Example 5 : Test for the convergence or divergence of the series
xn
 3 .n
n
2
.( x  0)
Solution :
Here we have y n 
Now
xn
therefore,
3n n 2
y n 1
x n 1
x n .x
 n 1

3 (n  1) 2 3 n .3(n  1) 2
lim
yn1
x n .x
3n.n 2
= lim n
.
3 .3(n  1) 2 x n
yn
x n 
= lim 

3  n 1
=
x
3
2
1
lim
 1
1  
 n
2
=
x
3
=
x
3
By D' Alemberts test
(1)
y
n
converges if
(2)
y
n
diverges when
(3) When
x
x
< 1 i.e.,
< 1 converges if x < 3.
3
3
x
> 1 i.e.
3
y
n
diverges when x > 3.
x
= 1 i.e. if x = 3 the test fails when x = 3 we have
3
3n
 yn =  3n.n2 =
is of the form
1
n
By P-series list
1
n
 The given series
where p = 2 > 1
p
1
n
2
p
is convergent.
y
n
is convergent when x < 3 and divergent when x > 3.
39
Prepared by Dindigala Raju
Remark : In some cases we apply the modified form and divergent when x > 3 which
states that if
x
n
is a series whose terms are positive, then
(a) lim
xn 1
< 1, implies that
xn
x
(b) lim
xn 1
> 1, implies that
xn
x
(c) if lim
n
converges
n
diverges
xn 1
= 1 the test fails.
xn
EXERCISE
1.
Test for the convergence of the series.
(a) 1 
(c)
3 5 7
   ...
2 22 3
(b)
2! 3! 4!


 ...
3 3 2 33
1
2
3


 ...
2
1  2 1  2 1  23
Answer : (a) convergent, (b) divergent, (c) convergent

2.
Test for convergence (a)

1
(c)
n!
nn
n 2 (n  1) 2
 n!
(b)
2n
 n3
(d)
2
1
n 1
1
Answer : (a) convergent, (b) divergent, (c) convergent, (d) convergent
3.
Examine the convergence of the series
(i) 1 
x x2 x3


 ... (x > 0)
1! 2! 3!
40
Prepared by Dindigala Raju
(ii)
2 x 3x 2 4 x 3
(n  1) x n


 ... +
 ...
n3
1
8
27
3
8
(n 2  1) n
(iii) x  x 2  x 3  ...  2
x  ...
5
10
(n  1)
Answer : (i) convergent, (ii) divergent if n < 1, divergent if x > 1, (iii) convergent if
x < 1, divergent if x > 1.
4.
Show that the series
1
(a  1) (a  1)( 2a  1) (a  1)( 2a  1)(3a  1)


 ...
(b  1) (b  1)( 2b  1) (b  1)( 2b  1)(3b  1)
Converges if b > a > 0 and diverges if b > a > 0.
5.
Show that the series,
x
a.12  b

2x2
nx n

...

 ...
a.22  b
a.n 2  b
Is convergent where x < 1 and divergent when x > 1.
6.
Test for the convergence of the series
(a)
1
 x n  x 1
(b)
(n  1) n n
 n n 1 .x
(c)
xn
 n2
(d)
x 2 n2
 (n  1) n
Answer : (a) Converges n < 1 or x > 1, diverges when x = 1.
(b) Converges when x < 1 and diverges when x > 1.
(c) Converges when x < 1 and diverges when x > 1.
(d) Converges when x2 < 1 and diverges when x2 > 1.
7.
n! x n
Let x > 0 shown that the series  n converges or diverges according as x < e.
n
or x > e.

8.
2n
Discuss the convergence of 
n 1 1.3.5...(2n  1)
41
Prepared by Dindigala Raju

9.
(n!) 3 n 1
Show that for | x | > 27, the series 
converges and for | x | < 27, it
x
n 1 (3n)!
diverges.
Raabe's Test: Theorem 17 : Let X = (xn) be a sequence of non zero real numbers.
(a)
If there exists a > 1 and K  N. such that
then
(b)
x
n
x n 1
a
 1  for n > K (1)
xn
n
is absolutely convergent.
If there exist real numbers a < 1 and K  N, such that
then
Proof : (a) Let
x
n
x n 1
a
 1  for n > K
xn
n
is not absolutely convergent.
x n 1
a
 1  hold for n > K. Then replacing n by R and multiplying.
xn
n
(1) we have | xk+1 | < (k – 1) | xk | - (a – 1) | xk | for k > K.
From the deduce that the sequence (k |xk+1 |) is a decreasing sequence for k > K.
Taking k + K,....,n in (3) and adding, we obtain
[K – 1] | xk | - n | xn+1 | > [| xk | +....+ | xn |]
which shows that the partial sums of

xn are bounded. Hence
x
n
is
absolutely convent.
(b)
Let the relation
x n 1
a
 1  hold for n > K. Since a < 1, we have
xn
n
n | xn+1 | > (n – a) | xn | > (n – 1) | xn | for n > k.
Therefore the sequence (n |xn+1|) is increasing for n > k and there exists number
c > 0. Such that x n 1 
c
n
therefore, we conclude that
for n > K. We know that the harmonic series diverges,

xn also converges.
42
Prepared by Dindigala Raju


x
Corollary : Let X - (xn) be a non zero sequence in R and let =  lim n.1  n 1

xn


x
whenever this limit exists. Then
n




is absolutely convergent when a > l and is
not absolutely convergent when a < l.

x
Proof : Let a = lim n 1  n 1
xn





Exist and a > 1. Let a1 be any number such that a > a1 > 1. Then there exists K  N

x
such that a1 < n 1  n 1
xn

Therefore,

 for n > K.


xn1
a
 1 1 for n = k and Raab'es test (a) applies
xn
n
x
n
is absolutely
convergent.
Similarly, we can prove that
x
n
is divergent when a < 1 .
Example 1: Determine the nature of the series
1.5.9...(4n  3)
 3.7.11....(4n  1)
Solution :
Let xn =
1.5.9...(4n  3)
3.7.11....(4n  1)
Then xn+1 =
1.5.9...(4n  3) (4n  1)
3.7.11....( 4n  1) (4n  3)
We note that lim
x n 1
xn
1
4n  1
n =1
= lim
= lim
3
4n  3
4
n
4
Therefore, the nature of the series is not determined by the D'alemberts ratio test.

x 
Now consider: a = lim n 1  n 1 
xn 

43
Prepared by Dindigala Raju
 4n  3 
= lim n 1 
4n  1 

4n  1  4n  3
4n  1
= lim n
= lim
=
2
= lim
4n  1
2
4
1
n
2
1
 <1
40 2
By Raabe's test the given series
x
n
is divergent.
Example 2 : Test the convergence of the series 1 
3
3.6 2
3.6.9
x
x 
 x 3  ...
7
7.10
7.10.13
Solution :
Neglecting the first term, the general term can be written as
yn =
3.6.....(3n)
.x n
7.10....(3n  4)
 yn+1 =

lim
3.6.....(3n) (3n  3)
x n 1
7.10....(3n  4) (3n  7)
y n 1
3.6.....(3n) (3n  3)
7.10.....(3n  4) x n 1
=
.
. n
7.10....(3n  4) (3n  7)
3.6....(3n)
yn
x
3
 3n  3 
n .x=x
= lim 
 . X = lim
7
 3n  7 
3
n
3
By D'Alemberts test.
y
n
is convergent if x < l and divergent if x > 1.
when x = 1 the test Fails
We apply : Raabe's test.
44
Prepared by Dindigala Raju

y
 lim n 1  n 1
yn


 = lim

 
3n  3 
n1  3n  7 

 
  3n  7  3n  3 
= lim n

3n  7

 
= lim
4n
= lim
3n  7
4
3
7
n

4
>1
3
by Raabe's test  y n is convergent.

y
n
convergent when x < 1 and divergent when x > 1
Example 3: Test for the convergence
1
1  a (1  a) (2  a)

 ...  ...
1  b (1  b) (2  b)
Solution :
Here, we have
so that

lim
xn =
xn+1 =
(1  a) (2  a)...( n  a)
(1  b) (2  b)...( n  b)
(1  a ) (2  a)...( n  a ) (n  1  a)
(1  b) (2  b)...( n  b) (n  1  b)
x n 1
n 1 a
= lim
=1
n 1 b
xn

u 
Therefore, D'Alemberts ratio test fails. Now lim n 1  n 1 
un 

 n 1 a 
= lim n 1 
 n  1  b 
 n 1 b  n 1 a 
= lim n 

n 1 b

= (b - a) lim
n
n  (1  b)
45
Prepared by Dindigala Raju
= (b - a)
It is given that b – a  0.
Therefore, by Raabes test, it follows that the series
x
n
is convergent if b – a > 1
and divergent b – a < 1.
EXERCISE
1.
Test the convergence
(a)

© 1
12.4 2.7 2.....(3n  2) 2
3 2.6 2.9 2...(3n) 2
(b)
2
2.4
2.4.6


 ...
3.5 3.5.7 3.5.7.9
1
1.2 2 1.2.3 3
x
x 
x  ...
2
2.5
2.5.8
(d) x 
1 x 3 1.3 x 5 1.3.5 x 7
.

.

.
 ...
2 3
2.4 5
2.4.6 7
(e) 1 +
a.b a(a  1) (b  1) a(a  1) (a  2)b(b  1)(b  2)


1.c
1.2.c(c  1)
1.2.3c(c  1)(c  2)
where a b, c are positive constants such that a + b  c
Answer: (a) convergent, (b) convergent, (c) convergent if x<3 and divergent if x >3
(d) convergent if x2 < 1 and divergent if x2 > 1. .
(e) convergent if a + b < c and divergent a + b > c.
2.
Find the nature of the following series
(a) x 
x3 x5

 ....
3
5
(b)
x
x2
x3
(c)


 ....
1.2 2.3 3.4
46
1
1.2 2 1.2.3 3
x
x 
x  ...
3
3.5
3.5.7
Prepared by Dindigala Raju
(d)
x 1 x 2 1.3 x 3
 .

.
 ...
1 2 3
2.4 5
x 1 x 3 1.3 x 3
 .

.
 ...
1 2 3
2.4 5
(e)
(a) convergent if , x < 1, divergent if x > 1
(b) convergent if x < 2, divergent if x > 2
(c) convergent if x < 1, if divergent if x > 1
(d) Convergent if x < 1, divergent if x > 1
(e) convergent if x < 1, divergent if x > 1.
Integral Test: Let f be a positive decreasing function on {  : t > 1}. Then the series


1
f(k) converges if and only if the improper integral.


b

f(t)dt = lim
b
1
f(t) dt
1

Exists: In the case of convergence, the partial sum Sn =


f(k) and the sum s =
1

1
f(k) satisfy.


f(t)dt < s – Sn <
n 1


n
f(t) dt ….
(1)
Proof: f is a positive and decreasing function on [k–1, k]. Therefore. Taking integrals
between t – k-1 and t - k , we have
f(k) =

k
k 1
f(t) dt < f(k-1).....
If we add these inequalities for k = 2, 3, ....n we obtain
f(2) + f(3) +….+ f(n) <
 Sn – f(1) <

n
1

n
1
f(t) dt < f(1) + f(2) +….+ f(n-1)
f(t) dt < Sn-1
From which, it clear that either both or neither of the limits.
47
(2)
Prepared by Dindigala Raju
lim Sn and lim
n
n

n
f(t) dt, exist
1
If the limits exist, then on adding (2) for k = n + 1, n + 2,…., m we get
Sm – Sn <

m
f(t) dt < Sm-1 – Sn-1 …..
n
Now

m

m1
f(t) dt < Sm-1 – Sn-1
n

(3)
f(t) dt < Sm – Sn …..
n 1
(4)
From (3) and (4) we obtain

m1
n 1
f(t) dt < Sm – Sn <

m
f(t) dt …..
n
(5)
taking limits in (5) as m   We obtain


n 1
f(t) dt < S – Sn 


n
f(t) dt
Hence proved

Example 1 : Using integral test, show that the series

n 1
1
converges.
n( n  1)
Solution :
Let f(t) =
1
for t  [1,  ]
t (t  1)

Then


f(k) =
1
K ( K  1)

n 1
1
Clearly, f(t) > 0 and f(t) is monotonically decreasing

n
1
f(t) dt
=

n
1
dt
t (t  1)

n
1
1 
1
 
 dt
 t t 1 
= (log t – log (t + 1))n
= log n – log (n + 1) + log 2
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Prepared by Dindigala Raju
= log
 lim
n

n
n

= lim log
 log
n
n 1

f(t) dt
1
n
+ log 2
n 1
 
  1
= lim log 
n
 1  1
n
 




2



  log




2 = log 2


f(t) dt converges to log 2.
1
Hence

f(k) converges i.e., the given series converges.

Example 2 : Show that series
1
converges
n 1
2
Solution :
Let f(t) =
1
for t  [1,  ].
t 1
2
So that f(x) =

1
n 1
2
Clearly f(t) > 0 and monotone decreasing in (1,  )


n
1
f(t) dt =

n
dt
= [tan-1 t]1n = tan-1 n – tan-1 1
t 1
2
1
= tan–1 n 
lim
n

n
1


f(t) dt = lim  tan -1 n -  (tan-1 n)
n
4

=



1

4

2
f(t) dt converges to


4


4

4
49
Prepared by Dindigala Raju
Hence, by integral test

f(x) converges.
Example 3 : Discuss the convergence of the series

1
(p > 0) or Show that p –
np
series converge if P > 1 and diverge if 0 < p < 1.
Solution :
Let f(t) =
1
, p > 0 and [t : t > 1]. Then f(n)
tp

1
np
Clearly f9t) > 0 and monotine decreasing.
 By integral test
Now,

n

f(t) dt =
1
f(x) and

n
1


1
f(t) dt by converge or diverge together.
n1 P
1
1
dt =
when p  1

p
t
1 p 1 p
and log h when  = 1.

When p < 1 :


n
1
n1 P
1
f(t) dt =

1 p 1 p
n
1
 n1 P
1 
  
f(t) dt = lim 

n
 1 p 1 p 
when p > 1 :

lim
n



lim
1

n
lim
n
n
1

n

n
1
1
n
1
f(t) dt =
n
1 P
1
1

(1  p) 1  p

1
1 
1
 
f(t) dt = lim  1 P

n
 n (1  p) 1  p  1  p
n
When p = 1;

f(t) dt = log n
f(t) dt = lim log n = 
n
 when p  1

f(t) dt = 
1
 1  p when p  1

50
Prepared by Dindigala Raju





f(t) dt converges to
1
f(x) i.e.,

1
when p > 1 and diverges when p < 1
1 p
1
converges when p > 1 and diverges when P < 1
np
1 1
1


Example 4 : Show that lim 1    ...   log n  exists
n
2 3
n


Solution :
1
for [t : t > 1]
t
Let f(t) =
Clearly f(t) > 0 and f(t) is monotone decreasing
Let Sn = f(1) + f(2)+…+ f(n) = 1 
 In =

n
f(t) dt =
1

n
1
1
1
 ... 
2
n
1
dt = log n
t
Now
f(n) < (Sn – In) < f(1) for all n  N
 0<
1
< (Sn – In) < 1 for all n  N
n
Thus the sequence {Sn – In} is bounded and monotone decreasing sequence.
 {Sn – In} converges.
1
1


Hence lim 1   ...   log n  exists
n
2
n


EXERCISE
1.
Using integral test. Show that the series converges.
(a)
2.
 n( n
1
2
 1)
(b)
ne
n 2
Show that the series, diverges.
51
(c)
1
n
2
Prepared by Dindigala Raju
(a)
2n 3
4
3
n
(b)

3.
Using integral test, show that

n2

1
n
1
converges if P > 1 and diverges if
n(log n) p
0 < P < 1.
Alternating Series :
In the preceding sections, we have discussed the test for absolute convergence of
the series. In the section we discuss tests for non absolute convergence of the series.
Definition : A sequence X = (xn) non zero real numbers is said to be alternating if the
terms (-1)n+1 xn, n  N are all positive (or all negative) real numbers. If the sequence
X = (xn) is alternating we say that the series
Thus a series of the form
x
n
generated by X is an alternating series.
x1 - x2 + x3-......+(-l)n+1 xn+...
where xn > 0 for all n , is an alternating series.
In the case of alternating series, we set xn = (-1)n+1zn, where zn > 0 for all n  N
Alternating series test (Leibnitz test): Let Z = (zn) be a decreasing sequence of strictly
positive numbers with lim (zn) = 0. Then the alternating series

(-1)n+1Zn is
convergent.
Proof: We have S2n = (z1 - z2) + (z3 - z4)+..... + (z2n-1 - z2n)
Z = (Zn) is a decreasing sequence
 zn – zn+1 > 0 for all n. Therefore, it follows that the subsequence (S2n) of partial
sums is increasing..
Now S2n = z1 - (z2 - z3) -…-(z2n-2 - z2n-l) - z2n
 S2n < z1 for all n  N.
Therefore, by montone convergence theorem, the subsequence (S2n) convergence to
some real number say s.
52
Prepared by Dindigala Raju
We now prove that the entire sequence (Sn) convergences to s.
for a given  > 0, let K  N, such that n > k  | S2n - s | <
1

2
1
.
2
and | z2n+1 | <
It follows that if n > K then
1
1
 +  =
2
2
|S2n+1 - s | = | S2n + z2n+1 - s | < | S2n - s | + |z2n+1 | <
for a sufficiently large n, every partial sum of an odd number of terms is also within
 of s.
But  > 0 is arbitrary. Therefore the sequence (Sn) converges to s. Hence
 (1)
n 1
z n converges.
Example 1: Test for convergence of the series 1 
1
2

1
3

1
4
 ....
Solution :
We have zn =
1
 zn+1 =
n
1
n 1
n  1  n for n  N.
Since
We have
1
n 1
<
1
n

zn+1 < zn

(zn) is a decreasing sequence.
Further lim zn = lim
1
n
=0
Therefore by Leibwitz’s test, the given series is convergent.
Example 2 : Test for convergence
1
1
1
 3 (1  2)  3 (1 + 2 + 3)+…
3
2
3
4
53
Prepared by Dindigala Raju
Solution :
1  2  3  .....  n
(n  1) 3
We have
zn =
i.e.,
1
 n(n  1) 
zn = 
.
2  (n  1) 3

=
and
n
2( n  1) 2
n 1
2(n  2) 2
zn+1 =
 zn+1 – zn =
(n  1)
n

2
2(n  2)
2(n  1) 2
(n  1) 3  2(n  2) 2
=
2(n  2) 2 (n  1) 2
n2  n 1
=
<0
2(n  2) 2 (n  1) 2
 zn+1 < zn = lim
n
=0
2( n  1) 2
Both the conditions of Leibnitz’s test are satisfied.
Therefore, the given series is convergent.

Example 3 : Test for convegence of the series

n 2
Solution :
Let
zn =
then
xn
n(n  1)
zn+1 =
x n 1
(n  1) n
54
(1) n x n
(0 < x < 1)
n(n  1)
Prepared by Dindigala Raju

zn+1 – zn =
xn
x n 1

n(n  1) ( x  1) n
=-
xn  1
x 


n  n  1 n  1 
 n(1  x)  (1  x) 
= - xn 

 n(n  1) (n  1) 
< 0 for n > 2 9since 0 < x < 1)
 Sn+1 < zn for n > 2, also, we have lim zn = lim
xn
=0
n(n  1)
( 0 < x < 1)
 by Leibnitz’s test, given series is convergent.
EXERCISE
Test for convergence of the series
1. 1 
3.
1
1
1
 2  2  ...
2
2
3
4
2.
12 12.32
12 32 5 2


. . ....
2 2 2 2.4 2 2 2 4 2 6 2
5. 1 
1
2 2

1
3 3

4. 2 
1
6. 1 
4 4
Answer : (1) Convergent
(2) Divergent
(4) Not convergent oscillatory
(5) Convergent
1 2 3 4
    ...
2 3 4 5
(6) Convergent
II. Find the nature of the following series
(a) 1 – 2x + 3x2 – 4x3 + …. (x < 1)
55
3 4 5
   ...
2 3 4
1 1 1
   ...
2 3 4
(3) Convergent
Prepared by Dindigala Raju
(b)
1
1
1
-… (x . 0, a > 0)


x x  1 x  2a
1
1
1
1



 ...x  0
x x 1 x  2 x  3
©
(d)
x
x2
x3


 ... (0 < x < 1)
2
1 x 1 x
1  x3
Answer : (a) Convergent
(b) Convergent
Conditional convergence :
In this section we discuss conditional convergence of the series.
 x contain both positive and negative terms. If | x | is
convergent then  x is absolutely convergent. If  u is convergent but
| x | is not convergence, then the series  x is said to conditionally
Let the series
n
n
n
n
n
n
convergent (or semi convergent) series.
Example 1:
Show that the series
1
2
1


3
1
4

1
5
+….. is conditionally convergent.
Solution :
We have x
1
n
n2 >
Now
1


1
= and xn+1 =
n2
n 1
n 1
>
1
n 1
xn+1 < xn  x > 1
further, lim xn = lim
1
n 1
=0
 by Leibnit'z that the given series is convergent now consider
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Prepared by Dindigala Raju
| x
n
| (1)
|=
1
Taking yn =


| x
| x
n
n

xn ) | =
y
1
n 1
behave alike but
n
.
| xn |
 1

= lim 
 n =1
yn
 n 1

, we get lim
n
| and
n1
y
n
=

1
n
isdivergent.
| is divergent.
 The given series is conditionally converent.
Example 2:
Examine the following series for absolute convergence.
1
1 1
3

1
x
2 1
3
1
3 1
3
n 2 ....
Solution :
The given series

| y
n
| 
 y  (1)
n
| x | n 1
n3  1
| y n 1 |
lim
= lim
yn
| y
n3  1
n3  1
|x|
(n  1) 3 1
n
x n 1
, so that


1
1 3


n

= lim 
1  1 

 1  n 3   n 3 


 The series
n 1
1/ 2
|x|=|x|
| is convergent if | x | < 1 and divergent if | x | > 1
If | x | = 1 we have | yn | =
1
n3  1
1

n3 / 2 1 
57
1
n3
Prepared by Dindigala Raju
Let vn = n3/2 we find that
| yn |
= lim
vn
Lim
But
V   n
n
Thus
| y
n
1
1
1
1
1 3
n
is convergent, therefore
3/ 2
| y
n
| is also convergent.
| is convergent if | x | < 1 and divergent if | x | > 1
 The given series is absolutely convergent for | x | < 1.
EXERCISE
x2 x3 x4
1. Show that the series x 


 .... is absolutely convergent.
2!
3!
4!
2. Show that the series 1 
1
1
1
 p  p  .....,(P >0) is absolutely convergent
p
2
3
4
for P > 1, and conditinoally convergent for 0 < p < 1.
3.
Determine the intervals in which the following series are convergent.
(a) 1 
x
3
x2

5

x3
3
…
(b) 1  2 x 
3 2 4
x  x3  ....
2
2
Answer:
(a) The given series is convergent for - 1 < x < 1
(b) Convergent for –l < x < l
4.
Test for absolute convergence and conditional convergence of the series.
(1) n 1 .n
(2)  2
n 1
1
1
1
(1) 1  2  2  2  …..
2
3
4
(3)
 (1)
n 1

n 1  1

Answer:
(a) Absolutely convegrent
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Prepared by Dindigala Raju
(b) Conditionally convergent
(c) Conditionally convergent
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