20
Infinite Series
20.1 SEQUENCE
A sequence is a succession of numbers or terms formed according to some definite rule.
The nth term in a sequence is denoted by un.
For example, if un  2n  1.
By giving different values of n in un, we get different terms of the sequence.
Thus, u1  3, u2  5, u3  7 
A sequence having unlimited number of terms is known as an infinite sequence.
20.2 LIMIT
If a sequence tends to a limit l, then we write
lim un  l
n 
20.3 CONVERGENT SEQUENCE
If the limit of a sequence is finite, the sequence is convergent. If the limit of a sequence
does not tend to a finite number, the sequence is said to be divergent.
1
1 1 1
e.g.,
, ..., 2   is a convergent sequence.
1, , ,
4 9 16
n
3, 5, 7, ..., 2n  1,  is a divergent sequence.
20.4 BOUNDED SEQUENCE
u1, u2, u3, ..., un  is a bounded sequence if un  k for every n.
20.5 MONOTONIC SEQUENCE
The sequence is either increasing or decreasing, such sequences are called monotonic.
e.g.,
1, 4, 7, 10, ... is a monotonic sequence.
1 1 1
1, , , ,  is also a monotonic sequence.
2 3 4
1,  1, 1,  1, 1,  is not a monotonic sequence.
A sequence which is monotonic and bounded is a convergent sequence.
Exercise 20.1
Determine the general term of each of the following sequence. Prove that the following sequences are
convergent.
1, 1, 1, 1 , 
1
1.
Ans. n
2 4 8 16
2
1119
Infinite Series
1120
2.
3.
1, 2, 3, 4, 
2 3 4 5
1,  1, 1,  1, 
n
n1
Ans.  1n – 1
Ans.
12 , 22 , 32 , 42 , 52 , 
1 2 3 4 5
Which of the following sequences are convergent ?
n1
Ans. Convergent
5. un 
4.
Ans.
n
7.
un  n2
Ans. Divergent
6. un  3n
Ans. Divergent
1
8. un 
n
Ans. Convergent
n2
n
20.6 REMEMBER THE FOLLOWING LIMITS
(i)
lim xn  0 if x  1 and lim xn   if x  1
n 
(iii) lim
n 
log n
0
n
n
(iv) lim 1 
n  
n 
(ix) lim nh  
n 
n 
xn
 0 for all values of x
n!
n
1
e
n 
1n
(vi) lim [n !]1  n  
(ii) lim
 n ! 
(vii) lim 

n   n 
1
(x) lim h  0
n
n 

(v) lim n1  n  1
n 
1
e
(viii) lim n xn  0 if x  1
n 
sin x
tan x
 ax  1 
a1  n  1
(xi) lim 
 log a or lim
 log a (xii) lim
 1 (xiii) lim
1

x
x
x 0  x 
x 0
x 0
n 1n
20.7 SERIES
A series is the sum of a sequence.
Let u1, u2, u3,...............,un, ...............b e a given sequence. Then, the expression
u1  u2  u3  ............... un  ............... is called the series associated with the given sequence.
For example, 1  3  5  7   is a series.
If the number of terms of a series is limited, the series is called finite. When the number
of terms of a series are unlimited, it is called an infinite series.
u1  u2  u3  u4    un   

is called an infinite series and it is denoted by  un or  un. The sum of the first n terms of a series is denoted
by Sn.
n1
20.8 CONVERGENT, DIVERGENT AND OSCILLATORY SERIES
Consider the infinite series  un  u1  u2  u3    un   
S n  u 1  u 2  u3    u n
Three cases arise :
(i) If Sn tends to a finite number as n  , the series  un is said to be convergent.
(ii) If Sn tends to infinity as n  , the series  un is said to be divergent.
(iii) If Sn does not tend to a unique limit, finite or infinite, the series  un is called
oscillatory.
20.9 PROPERTIES OF INFINITE SERIES
1. The nature of an infinite series does not change :
Infinite Series
1121
(i) by multiplication of all terms by a constant k.
(ii) by addition or deletion of a finite number of terms.
2. If two series  un and  vn are convergent, then  un  vn is also convergent.
Example 1. Examine the nature of the series 1  2  3  4    n   .
n n  1
[Series in A.P.]
Solution. Sn  1  2  3  4    n 
2
n n  1
 
since
lim Sn  lim
2
n
n
Hence, this series is divergent.
Example 2. Test the convergence of the series 1 
Sn  1 
Solution.

1 1 1
    .
2 4 8
1 1 1
  
2 4 8
1
1
1
2
[Series in G.P.]
S  a 
 n 1  r


2
lim Sn  2
n
Hence, the series is convergent.
Example 3. Discuss the nature of the series 2  2  2  2  2   .
Sn  2  2  2  2  2   
Solution.
 0 if n is even.
 2 if n is odd.
Hence, Sn does not tend to a unique limit, and, therefore, the given series is oscillatory.
Exercise 20.2
Discuss the nature of the following series :
1.
1  4  7  10   
3.
651651651
4.
6.
3 3
3  2 
2 2
1 1 1
1   
2 4 8
1
3


4
n
10.  log
n1
1
12. 
n n  2
8.
14.

2

2
3

n
Ans. Divergent
2. 1 
5 6 7
  
4 4 4
Ans. Oscillatory
Ans. Convergent
5. 12  22  32  42   
Ans. Convergent
7.
Ans. Convergent
9. log 3  log
Ans. Divergent
11.  
n 1  
n 

Ans. Convergent
13.
 n n  1 n1 2 n  3
15.
 n2 n  12
 n  1 n  2 n  3
Ans. Convergent
20.10 PROPERTIES OF GEOMETRIC SERIES
The series 1  r  r2  r3    is
Ans. Divergent
1
13

1
1


35 57
Ans. Divergent
Ans. Convergent
4
5
 log    Ans. Divergent
3
4
2n  1
Ans. Divergent
Ans. Convergent
Ans. Convergent
Infinite Series
1122
(iii) oscillatory if r   1.
(i) convergent if | r |  1
(ii) divergent if r  1
1  rn
Proof. Sn  1  r  r2    rn – 1 
1r
(i) When | r |  1, lim rn  0
n 
lim Sn  lim
n
n
1  rn 1  0
1


1r 1r 1r
Hence, the series is convergent.
(ii) (a) When r  1, lim rn  
n
 lim Sn  lim
n
n
rn  1
 
r1
Hence, the series is divergent.
(b) When r  1, the series becomes 1  1  1  1   
Sn  1  1  1  1    n
lim Sn  lim n  
n
n
Hence, the series is divergent.
(iii) (a) When r   1, the series becomes 1  1  1  1  1   
Sn  0 if n is even
 1 if n is odd
Hence, the series is oscillatory.
(b) When r   1, let r   k where k  1.
rn   kn   1n kn
1  rn
1   1n kn
 lim
lim Sn  lim
1   k
n 
n
n 1  r
   if n is odd
   if n is even
Hence, the series is oscillatory.
Exercise 20.3
1.
3.
Test the nature of the following series :
1 1 1
1   2  3    Ans. Convergent
2 2 2
1 1 1
   Ans. Convergent
1  
3 9 27
2. 1 
3 9 27
 

4 16 64
4. 1  2  4  8   
2
5.
7.
The series 1  1  1  1   is
(i) Convergent with its sum equal to 0.
(iii) Divergent.
Ans. Oscillatory
3
4
4
4
9 27
6. 1          
   Ans. Divergent

3 3 3
2 4
State, which one of the alternatives in the following, is correct :
23
Ans. Convergent
Ans. Divergent
(ii) Convergent with its sum equal to 1.
Ans. Oscillatory series
(iv) Oscillatory.
20.11 POSITIVE TERM SERIES
If all terms after few negative terms in an infinite series are positive, such a series is a
positive term series.
Infinite Series
1123
 10  6  1  5  12  20   is a positive term series.
By omitting the negative terms, the nature of a positive term series remains unchanged.
20.12 NECESSARY CONDITIONS FOR CONVERGENT SERIES
For every convergent series  un,
e.g.,
lim un  0
n
Solution. Let Sn  u1  u2  u3    un
lim Sn  k
(a finite quantity)
lim Sn – 1  k
(a finite quantity)
n
Also
n 
Sn  Sn – 1  un
un  Sn  Sn – 1
lim un  lim [Sn  Sn – 1]  0
n
n
lim un  0
n
Corollary. Converse of the above theorem is not true.
1
1
1
1


 
   is divergent.
e.g.,
1
4
2 

n

3 
1
1
1
1
1
1
1
1
1








Sn  1 
2
3
n
n
n
n
n
4








n





n

 
n
n

n  
lim Sn  lim 
n
n
Thus, the series is divergent, although lim un  lim
n
n
1
0
n

So lim un  0 is a necessary condition but not a sufficient condition for convergence.
n
Note : 1. Test for divergence
If lim un  0, the series  un must be divergent.
n
2. To determine the nature of a series we have to find Sn. Since it is not possible to find
Sn for every series, we have to devise tests for convergence without involving Sn.
20.13 CAUCHY’S FUNDAMENTAL TEST FOR DIVERGENCE
If lim un  0, the series is divergent.
n
n
2 3 4
  

3 4 5
n1
n
1
 lim
10
lim un  lim
Solution.
n 
n n  1
n 1 1
n
Hence, by Cauchy’s Fundamental Test for divergence the series is divergent.
Example 4. Test for convergence of the series 1 
Infinite Series
1124
3 8 15
2n  1
 n



5 10 17
2 1
1
1 n
2
2n  1
 lim
10
lim un  lim n
Solution.
n
n  2  1
n 1 1
2n
Hence, by Cauchy’s Fundamental Test for divergence the series is divergent.
Question. Examine the series for convergence
3
1
2



 lim un  0
 n

1  2– 1 1  2– 2 1  2– 3


[Try Yourself]
Ans. Divergent
Example 5. Test for convergence the series 1 
Exercise 20.4
Examine for convergence :
1.
1
2


4
2n
2

 n
 
4 1
17



5 

Ans. Divergent

nn 1

n
n1
Ans. Divergent
4.
 cos 1n
Ans. Divergent
5. 1 
6.
 6  n2
Ans. Divergent
7.   2n
8.
n1
2.

n1
3
3.

n1
1
2
2
Ans. Divergent
1
1
 3   4   Ans. Divergent
3
4
Ans. Divergent
Ans. Divergent
20.14 p-SERIES
1 1 1
The series p  p  p    is
(ii) divergent if p  1
(i) convergent if p  1
1 2 3
Solution. Case 1 : p  1
The given series can be grouped as
1 1 1 1
1 1
1
1
1
1
1
1
1
1
1
  p  p    p  p  p  p    p  p  p  p  p  p  p  p   
p
1 2 3  4 5 6
7  8 9
10 11 12 13 14 15 
1
1
1p
1 1 1 1 2
   
2p 3p 2p 2p 2p
1 1 1 1 1 1 1 1 4
       
4p 5p 6p 7p 4p 4p 4p 4p 4p
1 1
1
1 1
1 8
  p p p p p
8p 8p
15 8
8
8 8
On adding (1), (2), (3) and (4), we get :
1 1 1 1 1 1 1 1 1
1 

p   p  p    p  p  p  p   p  p   
1 2
3   4 5 6 7  8
8
15p 
Now

1 2 4 8
   
1p 2p 4p 8p
...(1)
...(2)
...(3)
...(4)
Infinite Series
1125
p–1
1
 1   
2
 
2p – 2
1
  
 2
3p – 3
1
  
 2

p–1

1
p–1

 1
G.P., r   

 2
1
1   
 2
 Finite number if p  1
Hence, the given series is convergent when p  1.
Case 2 : p  1
, S
When p  1, the given series becomes
1 1 1
1 1
1
1 1 1 1
1              
     
2  3 4   5 6 7 8  9 10
16 
1
1
1 1
2
2
1 1 1 1 1
   
3 4 4 4 2
1 1 1 1 1 1 1 1 4 1
        
5 6 7 8 8 8 8 8 8 2
1 1
1
1
1
1
8 1
   



9 10
16 16 16
16 16 2
On adding (1), (2), (3) and (4), we get
1 1
1
1 1 1 1
1
1 1
1              
     
16 
2  3 4   5 6 7 8   9 10
1
1 1 1 1
n
    1
2 2 2 2
2
1 

1  r
...(1)
...(2)
...(3)
...(4)
n  

Hence, the given series is divergent when p  1.
Case 3 : p  1
1 1, 1 1, 1 1


 and so on
2p 2 3p 3 4p 4
1 1 1 1
1 1 1
   1   
therefore,
2 3 4
1p 2p 3p 4p
 diver gent series p  1
As the series on R.H.S. 1  1  1  1   is divergent




2 3 4




Hence, the given series is divergent when p  1.
20.15 COMPARISON TEST
If two positive terms  un and  vn be such that
un
 k (finite number), then both series converge or diverge together.
lim
n   vn
Proof. By definition of limit there exists a positive number , however small, such that
Infinite Series
1126

 un
 v  k   for n  m

 n
un
k     k   for n  m
vn
i.e.,   
un
k
vn
Ignoring the first m terms of both series, we have
u
k    n  k   for all n.
vn
...(1)
Case 1.  vn is convergent, then
lim v1  v2    vn  h say
where h is a finite number.
n
From (1), un  k   vn for all n.
lim u1  u2    un  k   lim v1  v2    vn  k   h
n 
n
Hence,  un is also convergent.
Case 2.  vn is divergent, then
lim v1  v2    vn  
...(2)
n
k
Now from (1)
un
vn
un  k   vn for all n
lim u1  u2    un  k   lim v1  v2   vn
n 
n
lim u1  u2    un  
From (2)
n
Hence,  un is also divergent.
Note. For testing the convergence of a series, this Comparison Test is very useful. We
choose  vn (p-series) in such a way that
un
 finite number.
lim
n   vn
Then the nature of both the series is the same. The nature of  vn (p-series) is already
known, so the nature of  un is also known.

Example 6. Test the series 
n1
Solution.
Let
1
for convergence or divergence.
n  10
1
n  10
1
vn 
n
un
n
1
lim
 lim
 lim
 1  finite number.
v
n n
n   n  10
n   1  10
n
un 
According to Comparison Test both series converge or diverge together, but  vn is divergent
as p  1.
Infinite Series
1127
  un is also divergent.
1
3
5



123 234 345
1
2
n
1
2n  1

un 
Solution.
n n  1 n  2 n2 1  1  1  2 

n  
n 

1
2
un
n
1
 2  finite number.
 lim
Let vn  2
lim
v
1
n
n n
n   1   1  2 

n 
n  

According to Comparison Test both series converge or diverge together, but  vn is convergent as p  2.
  un is also convergent.
Example 7. Test for convergence the series

3n2 1
3
1 4 4n3  2n  7


Example 8. Test the following for convergence : 
13
Solution. un 
3n2  11  3
4n3  2n  71  4
1
31  3 n2  3 1  2 
3n



Let
14
2 7
n3  4 4  2  3 
n
n 

1
vn  1  12
n
13

1
31  3 1  2 
3n


14
2 7
n1  12 4  2  3 
n
n 

13
lim
un
 lim
vn n  
1
31  3 1  2 
3n


14

31  3
 finite number.
2

4  2  7 
 n2 n3 


According to Comparison Test both series converge or diverge together; but  vn is divergent
1
 1.
as p 
12
  un is also divergent.
Example 9. Test for convergence the series whose nth term is nlog x.
Solution. The series is 1log x  2log x  3log x  4log x   
1
1
1
1
  log x   log x   log x   log x   
1
2
3
4
Put  log x  p
1 1 1 1
 p p p p
1 2 3 4
(a) which is convergent if p  1, i.e.,  log x  1
1
1
1
 e or x 

loge  loge e 
x
x
e
n
Infinite Series
1128
(b) which is divergent if p  1   log x  1
1
1
1
e 
 x.

loge  loge e 
x
x
e
Example 10. Test the following series for convergence
Solution. Given series is
2 3 4 5
   
1p 2p 3p 4p
2 3 4 5
   
1p 2p 3p 4p
1
1
n
n1
un  p  p – 1
Here
n
n
1
un 1  n np – 1
1
1

1
vn  p – 1 

Let
vn np – 1
n
1
n
un
1
lim
n   vn
Therefore, both the series are either convergent or divergent.
But  vn is convergent if p  1  1, i.e., if p  2
and is divergent if p  1  1, i.e., if p  2
 The given series is convergent if p  2 and divergent if p  2.
1 2 2 33
Example 11. Test the series for convergence 1  2  3  4  
2 3 4
2
3
3
1 2
Solution. Let us consider 2  3  4  .....
2 3 4
nn
nn
nn
1


Here un 
n 
n
n
n1
1
n  1 n  1
1
n  1
1  1  1  1 
n 1    nn 1  
n

n
n


n  
n 

un
1
1

 Let vn 
n
vn 
1 n
1 
1  n  1  n 



un
1
1
  finite quantity.
 lim
lim
1 
1 n e
n   vn
n 
1  n  1  n 



Hence, either both the series are convergent or both the series are divergent. But  vn is
divergent as p  1.
 The given series is divergent.
Exercise 20.5
Examine the convergence or divergence of the following series :
3 1 4 1 5 1
1. 2     2   3   
2 4 3 4 4 4
12 123 1234



13 135 1357
3
5
7
3. 1      



2
3


4
2. 1 
Ans. Convergent
Ans. Convergent
Ans. Convergent
Infinite Series
1129
3
1
2



12 34 56
3
1
2
5.



13 35 57
4.
Ans. Divergent
Ans. Divergent
22 32 42
  
 3 4
2
1
2
3
7.



1  2 1  22 1  23
6. 1 
8. 1 

10.

n1

12.

n1

13.

n1
Ans. Convergent
Ans. Convergent
1 22 33
     Ans. Divergent
22 33 44
Ans. Divergent
an
xn  na
n1

19.

n1

20.

n1

21.

n1
Ans. Convergent
14.  

n2  1  n
Ans. Divergent
n1

4
Ans. Convergent

n

n2  1
4
16.
n

5n
Ans. Divergent
nn
n
Ans. Convergent

n1

2n  1
3n  n
Ans. Convergent
n2
en
Ans. Convergent
18.

n1
xn
x  0
1  x2n
Ans. If x  1, convergent; if x  1, divergent.
1
xn  x– n
Ans. If x  1, convergent; divergent if x  1

22.
2n3  5
4n5  1

Ans. Convergent
Ans. If x  a, convergent; if x  a, divergent
n1

11.
n1
15.  [

n  1  

n  1] Ans. Convergent

 3

1 2
  
3 32 33

1
n 1

n  


17.
9.
 sin 1n
Ans. Divergent
n1
23.
n2  n  11  2
n5  21  2
Ans. Convergent
20.16 D’ALEMBERT’S RATIO TEST
Statement. If  un is a positive term series such that
(i) the series is convergent if k  1
Solution.
un  1
k1
Case 1. When lim
un
n
lim
n
un  1
 k then
un
(ii) the series is divergent if k  1
By definition of a limit, we can find a number r  1 such that
Infinite Series
1130
u4
u3
 u2

 u  r, u  r, u  r 
2
3
 1

un  1
 r for all n  m
un
Omitting the first m terms, let the series be
u1  u2  u3  u4   
u u
u u u u u u
u




 u1 1  2  3  4    u1 1  2  3  2  4  3  2   
u
u
u
u
u
u
u
u
u
1
1
1
1
2
1
3
2
1




 u1 1  r  r2  r3   
u1
, which is a finite quantity.

1r
r  1
Hence,  un is convergent.
un  1
k1
Case 2. When lim
un
n
By definition of limit, we can find a number m such that
un  1
 1 for all n  m
un
u
u2
u
 1, 3  1, 4  1
u1
u2
u3
Ignoring the first m terms, let the series be
u1  u2  u3  u4   
u 2 u3 u4
u2 u3 u2 u4 u3 u2




 u1 1       u1 1         
u
u
u
u
u
u
u
u
u
1
1
1
1
2
1
3
2
1




 u1 1  1  1  1  to n terms  nu1
lim u1  u2    un  ur
n
lim Sn  lim nu1  
n
n
Hence,  un is divergent.
un  1
1
un
The ratio test fails.
Note. When
k  1,
For Example. Consider the series whose nth term 
1
n
1
un  1
1
n1
n
lim
 lim
 lim
 lim
1
1
u
n
n 
n n  1
n 1 1
n 
n
n
1
Consider the second series whose nth term is 2 .
n
1
2
un  1
n  12
n 
lim
 lim
 lim 
 1
1
un
n
n 
n   n  1 
n2
...(1)
...(2)
Infinite Series
1131
Thus, from (1) and (2) in both cases lim
n 
un  1
1
un
But we know that the first series is divergent as p  1.
The second series is convergent as p  2.
un  1
 1, the series may be convergent or divergent.
Hence, when lim
un
n
Thus, ratio test fails when k  1.
Example 12. Test for convergence the series whose nth term is
Solution. un 
n2
.
2n
n  12
n2 ,

u

1
n
2n
2n  1
2
un  1
1
n  12 2n
1
1
 2  lim 1     1
 lim
n1
2
u
n
n
2
n
n
n 
n
 2

Hence, the series is convergent by D’Alembert’s Ratio Test.
2n
Example 13. Test for convergence the series whose nth term is 3 .
n
lim
2n ,
2n  1
un  1 
3
n
n  13
By D’Alembert’s Ratio Test
un  1
2
2n  1 n3

 n
3
3
un
n  1 2 
1

1
 n


Hence, the series is divergent.
un 
Solution.
lim
n
un  1
 lim
un
n
2
3
1  1 
 n


21
Exercise 20.6
Test for convergence the series :
n
1.

n1
2
3.
4.
2.

n1
2
n!
nn
Ans. Convergent
2
Ans. Convergent
2  5  8 2  5  8  11


1  5  9 1  5  9  13
Ans. Convergent
2
1


n1
7.
Ans. Convergent
 1    1  2   1  2  3    
 3  3  5   3  5  7 
  
 

n
5.
n
n2
3n
n ! . 2n
nn

Ans. Convergent
Prove that, if un  1 
6. 
n1
xn  1
n3n
Ans. Convergent if x > 3 Deavergent if x < 3.
k
, where k  0, u1  0, then the series  un converges to the positive root of the
1  un
equation x2  x  k.
8.

k  un , where k  0, u1  0, then the series  un converges to the positive root of
Prove that, if un  1  
2


the equation x x k.
9.
Prove that, if un  1  un2  k  k2, where u1  0 and un tends to a finite limit then l must be either k or 1  k.
Infinite Series
1132
20.17 RAABE’S TEST (HIGHER RATIO TEST)
 un

lim n 
 1  k, then
u
n
1

n


(ii) the series is divergent if k  1.
If  un is a positive term series such that
(i) the series is convergent if k  1
Proof. Case I. k > 1
Let p be such that k  p  1 and compare the given series  un with 
1
which is convergent
np
as p  1.
un
un  1

n  1 p
np
p
or
p p p  1 1
 un  
1

 u   1  n   1  n 
n2


1
n
2


 
 un

p p  1 1
n
 1  p 

n ! n2
u
 n1


 un
lim n 
 1  p
u
n

1
n


and k  p which is true as k  p  1;  un is convergent when k  1
If
Case II. k  1
Same steps as in Case 1.
Notes :
1. Raabe’s Test fails if k  1.
2. Raabe’s Test is applied only when D’Alembert’s Ratio Test fails.
x3
x
x2
x4



 ....
Example 14. Test the convergence for the series
1.2 3.4 5.6 7.8
Solution.
un 
xn
xn  1
and un  1 
2n  1 2n
2n  1 2n  2
1 

x 1 
un  1
2n 
2n  12n
xn  1




,
un
xn
2n  1 2n  2
1  1  1  2 

2n 
2n  

un  1
 x
lim
n un
(i) If x  1,  un is convergent (ii) If x  1, un is divergent (iii) If x 1, Test fails
Let us apply Rabee’s Test when x = 1
2n 1 2n  2  2n 2n  1 
 un

  2 n  1 2n  2

 1  lim n 
 1  lim n 
lim n 

u
n


2
1
2n 2n  1
2n
n

1
n 
 n 
 n  

2
2  
 4n 
 8n  2 

  2
 lim n 
 lim

n    2n 2n  1 
n   1 1  1 

2n 

So the series is convergent.
Infinite Series
1133
Hence we can say that the given series is convergent if x  1 and divergent. if x  1.
Ans.
1
Example 15. Test the following series for convergence 
n 1  1


1
1
, un  1 
un 
Solution.
n 2  1
n 1  1




un


n 2  1

un  1 
1  1
n

lim
n
un
un  1
 lim
n
1  2n  n1  1
1  1n  n1

2
2
D’Alembert’s test fails. Let us apply Raabe’s test.


 un


n 2  1
 1
 1  lim n 
lim n 
u
1


n
1

1
n




n

 n 





n 2  

n 1 

n 1  1 
n 2  1  
 lim n 
  lim n 


n


n
1
1
1
1 








n
 n 



2
1 

1 
1 
n
n 01
 lim n 

n


1
1
1  2


n n


 




Hence,  un is divergent.
Example 16. Test for convergence the series for positive values of x :
14 3
2
6
2n  2 n – 1
1  x  x2 
x  n
x

5
17
9
2 1
2n  1  2 n
2n  2 n – 1
x
x
and un  1  n  1
n
2 1
1
2
By D’Alembert’s Ratio Test
un  1 2n  1  2 n  2n  1 1 
  n1
x 

un
 1   2n  2 xn – 1 
2
Solution. un 
1
n
1
2n  1 1  n  x  2 1  2n  
2 

 
un  1

 lim
lim


u
1
1
  2n 1 

n
n
n   2n  1 1 
 2n  1   
2n – 1  
 

un  1
x
lim
un
n
(i) If x  1,  un is convergent.
(iii) If x  1, test fails.
(ii) If x  1,  un is divergent.
Infinite Series
1134
Let us apply Raabe’s Test when x  1
 2n  2  2n  1  1 


 u
lim n  n  1  lim n  n
  n  1 2  1
u

2
1
 
n
 n1
 2
 n   

 lim n [1  1]  0  1
n
so the series is divergent.
Hence, the given series is divergent if x  1 and convergent if x  1.
Example 17. Test the following series for convergence
3
369 3
36 2
3  6  9  12 4
1 x
x 
x 
x 
7
7
 10  13  16
7  10
7  10  13
3  6  9  12  3 n  1 n – 1
un 
x
Solution.
7  10  13  16  3n  1
3  6  9  12  3 n  1  3n
xn
un  1 
7  10  13  16  3n  1 3n  4
un  1
un  1
3n
3
xx

x, lim
 lim
un
u
3n  4 n  
n
n 3  4
n
If x  1,  un is convergent. If x  1,  un is divergent. If x  1, the test fails.
 un

 3n  4

4 4
 1
 1  lim n 
 1  lim n
lim n 
n
u
3
3n
3
n
 n
 n1
 n 
Hence, the series is convergent if x  1 and divergent if x  1.
Example 18. Test the following series for convergence
1
9
25 4
x  x2  x3  x4 
x 
8
2
32
Solution.
un 
n  12  xn  1
n2  x n ,
un  1 
n
2
2n  1
2
un  1 n  12 xn  1 2n
n  1  x
 2 n

un
n x  n  2
2n  1
un  1
 lim
un
n
2
1  1  x  x

n  2 2
n

x
x
(i) If  1 or x  2,  un is convergent. (ii) If  1 or x  2,  un is divergent.
2
2
x
(iii) If  1 or x  2 the test fails.
2
Let us apply Raabe’s test.
 un

 n2  n2  2n  1   2n2  n
 n2 2
n
 1  n 
 1  n 
  n  12
2
u
2
n  12
 n1



 n  1

1
2
n
 un

 1  lim
lim n 
2 21
u
n
1
 n1
 n 
1  n 


lim
Infinite Series
1135
Hence,  un is divergent if x  2, and convergent if x  2.
|2

|3
1




Example 19. Show that the series
x x x  1 x x  1 x  2
converges if x  2 and diverges if x  2.
|n
x x  1 x  2  x  n  1
|
n  1
un  1 
x x  1 x  2  x  n  1 x  n
un 
Solution.
By D’Alembert’s test
1
un  1 1  n
un  1 n  1 ,


1
lim
un
x
x  n n   un
1
n
Test fails. Let us apply Raabe’s Test.
x1


 un
x  n
x  1
 lim
 1  lim n 
 1  lim n 
x1
lim 

n    un  1
 n   n  1
 n   n  1  n   1  1
n
If x  1  1 or x  2,  un is convergent.
If x  1  1 or x  2,  un is divergent.
Example 20. Discuss the convergence of the series
Solution.
x3
x4
x2



2 log 2 3 log 3 4 log 4
x4
x3
x2



2 log 2 3 log 3 4 log 4
xn  1
xn  2
, un  1 
n  1 log n  1
n  2 log n  2
n

1
an  1
n  2 log n  2
x


un
n  1 log n  1
xn  2
2
 1  2  log 1  
n



1
n
1  n  2  log n  2
1
 
 lim




x  n  1  log n  1 n   x 
x
1
 1
1
n  log 1  n 



1
 1 i. e., x  1, the series is convergent
(i) when
x
(ii) x  1, the ser ies is divergent. (iii) when x  1, the test fails.
2
log n  log 1  
n
un

 n  2  log n  2  n  2 




un  1  n  1  log n  1  n  1 
1
log n  log 1  
n

2 1 4
2
log n    2  

1
n 2 n
n  2
 n  2
n log n





 n  1 log n  1  1  12    n  1  1  1  
n log n
n 2 n
un 
Infinite Series
1136
1  
 n  2 
1 
1 
1
1
1

 1 
1






n
n
n
n
n
n
log
log
log
n  1 
 

 n  1 
1

 un
1
 1  n 1 
 1 
01
n
n
n
u
n
log
log


 n1

the series converges if x  1 and diverges if x  1.
Example 21. Test the series for convergence

   1     1 2
   1   2     1   2 3
1
x
x 
x 
1
1  2  3     1   2
1  2      1
Solution.
   1   2  [  n  1]     1  [  n  1] n
un 
x
n !    1  [  n  1]
   1   2  [  n  1]   n     1 [  n  1]   n n  1
un  1 
x
n  1 !    1  [  n  1]   n
un  1
un
 


1  n  1  n 
  n   n


x

x
n  1   n
1  1 1   

n 
n  

un  1
x
lim
un
n
(i) If x  1, the series is convergent.
(ii) If x  1, the series is divergent.
(iii) If x  1, the test fails.
Let us apply Raabe’s Test.
 n  1   n
 un


 n   n2    n     n   n   n2 
n
 1  n 
 1  n 

  n   n
   n   n
 un  1





 un
 1  lim
lim n 
u
n
 n
 n1




1
n
n
1
 




1
1



n
 n


(i)   1      1 or     ,  un is convergent.
(ii)   1      1 or     ,  un is divergent.
Example 22. Test for convergence the series

4  7    3n  1 n
x
12n
4  7   3n  1 3n  4 n  1
x
1  2   n n  1
By D’Alembert’s Test
3  4  x
 n
un  1
u
3n  4
n1

 3x

x, lim
 lim 
un
un
1
n1
n
n
1
n
Solution. un  1  
Infinite Series
1137
1,
1
 un is convergent.
If 3x  1  x  ,  un is divergent.
3
3
1
If 3x  1  x  , the test fails.
3
By Raabe’s Test
 un

n  1

n  1  n  4  3 
13
1
 1  lim n 
 1  lim n 
 1
lim n 
  lim
u
4
4
3
3


n
4
n

1


n
n

 n   n 

 n  1 

n
3
3


Thus,  un is divergent.
1
1
The series is convergent if x  and divergent if x  
3
3
n1
3
4 3
x    3 x3   
Example 23. Test the series for convergence 2x  x2 
8
27
n
If 3x  1  x 
Solution.
un 
n 2 n1
n1 3
x , un  1 
x
n3
n  13
By D’Alembert’s Test
un  1
n  2 xn  1
n3
 lim

lim
3
n  1
n  1 xn
n   un
n
 lim
n
2
1  1  
n
n  2 x

 lim
xx
4
4
n  1
n 
1
1  n 


n3
(i) If x  1,  un is convergent. (ii) If x  1,  un is divergent. (iii) If x  1, the test fails.
Let us apply Raabe’s Test when x  1.
 un


 n  14
 1  lim n  3
 1
lim n 
u
n
 n1
 n    n n  2

 n4  4n3  6n2  4n  1  n4  2n3 
 lim n 

n4  2n3
n


6 4 1
2  2  3
n n n
2n3  6n2  4n  1
lim
 lim

21
3
2
2
n  2n
n
n
1
n
Thus, the series is convergent for x  1 and divergent if x  1.
Example 24. Test for convergence the series
1  3  5 x7
1 x3
1  3 x5
x  




 
3
2
24 5
246 7
Solution. The given series is :
1 x3 1  3 x5 1  3  5 x7
x  
 
 
2 3 24 5 246 7
nth term of x, x3, x5  is x  x2n – 1, i.e., x2n – 1
(G.P.)
[. Tn  arn – 1]
Infinite Series
1138

un 

un  1 
1  3  5    2n  3 x2n – 1

2  4  6    2n  2 2n  1
1  3  5    2n  3 2n  1 x2n  1

2n  1
2  4  6    2n  2  2n
2
2  1 

n 
2
un  1
2n  1

2
x 

x2
un
1

2n 2n  1
2 2  
n

 un  1  2
lim 
x .
n    un 


 The given series is convergent if x2  1 and divergent if x2  1.
If x2  1, then, D’Alembert’s Test fails. Now apply Raabe’s Test.
 un
 2n 2n  1


n
 1  n 
 1
Here
2
u
 n1
 2n  1


1
1
n2 6  
6
n
n
 4n  2n  4n  4n  1  n 6n  1

n

  2n  12 
2
2
2
2
1


n
1
1



2
n 2  
2  n 
n



2
2
 u
 6 3
lim n  n  1   , i.e.,  1
u
n

1
n

 4 2

 un is convergent.

Hence,  un is convergent if x2  1 and divergent if x2  1.
Exercise 20.7
Determine the nature of the following series :
|2
 |3

 |4
1. 1  2  2  2   
2 3 4
2.
1 13 135 1357




1 1  4 1  4  7 1  4  7  10
3. 1 

4.
Ans. Divergent

n1
1   1   2  

 
1   1   2  
Ans. Convergent
Ans. If     1, convergent. If     1, divergent.
n3
Ans. Convergent
en
2x2 3x3 4x4



2! 2! 4!
1 3
1
1
6. 1  x  x2 
x 
5
10
2
5. x 
7. 1 
8. x 
x 13 2 135 3
x 

x 
2 24
246
2x2
2!

3x3 4x4 5x5



3!
4!
5!
Ans. Convergent
Ans. Convergent if – 1  x  1 and divergent if | x |  1.
Ans. If x  1, convergent; divergent if x  1.
Ans. If x 
1
1
, convergent; divergent if x  .
e
e
Infinite Series
1139
Ans. If x2  4, convergent; divergent if x2  4.
 1! 2 2 2 !2 4  3 !2 6
x 
x 
x    x  0
2!
4!
6!
Find the values of x for which the following series converges :
9.
1
10. x2 log 2q  x3 log 3q  x4 log 4q  

11.

n1


 1n n ! xn
10n
13.

xn
2n 2n  1
14.

15.

12.
n0
Ans. If x  1, convergent; divergent if x  1.
 1 x  1
2n  n2
n
12n
n
Ans. If  3  x, convergent; if x  – 3, divergent
Ans. x  1, convergent; x  1, divergent.
xn
4  7  3n  1
12n
xn
7  10  3n  4
Ans. If 0  x  3, convergent and divergent if x  3.
Ans. If 0  x  3, convergent and divergent if x  3.
20.18 GAUSS’S TEST
Statement. If  un is a positive term series such that
un
 
   2
un  1
n n
where
(i)
(ii)
0
if   1, convergent
if   1, divergent, whatever  may be
  1, convergent
if   1 and 
divergent
  1,
22  42 22  42  62 22  42  62  82



32  52 32  52  72 32  52 72  92
22  42 22  42  62 22  42  62  82
Solution. The given series is 2 2  2 2 2  2 2 2 2  
3 5
3 5 7
3  5 7  9
2 42 62
2    82  2n  22
 2 2 2 2

3  5  7  9  2n  32
22  42  62  82  2n  22 2n  42
un  1  2 2 2 2
3  5  7  9  2n  32 2n  52
16 16
4

un  1 2n  42 4n2  16n  16
n n2



un
2n  52 4n2  20n  25 4  20  25
n n2
16 16
4

un  1
n n2
lim
 lim
1
un
n
n   4  20  25
n
n
D’Alembert’s Test fails. Let us apply Raabe’s Test.
Example 25. Test for convergence the series
Infinite Series
1140
4n2  20n  25

 u

 1
lim n  n  1  lim n  2
u
n
 n1
 n   4n  16n  16

 49 

 4n2  9n 
n 
 lim 
 lim  2
 1, Raabe’s Test fails

16
16 
n    4n  16n  16 
n   4 
 2
n n 

Let us apply Gauss’s Test
2
un
un  1
1  5 

2n 

–2
2n  52
2
5 25  

 1  
1 
2n  42  2 2  n 4n2   n 
1  n 


5 25  
4 23 4  
5 25  
4 12


 1   2  1  
2   1  n 
2  1  n  2  
n
n
4
4
n
n
n
n
2









 un
 
4 12 5 20 25
1 7
1  2   2  2 1  2
    2
u
n n
n n 
n n
n n
4n
 n1

Hence,   1,   1. Thus, the series is divergent.
20.19 CAUCHY’S INTEGRAL TEST
Statement. 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

 f x dx
1
is finite or infinite.
Proof. In the figure, the area under the curve from x  1 to x  n  1 lies between the sum
of the areas of small rectangles and sum of the areas of large rectangles.
n1
 f 1  f 2    f n 

f x dx  f 2  f 3    f n  1
1
n1
Sn 

f x dx  Sn  1  f 1
1
As n  , from the second inequality that if the integral
has a finite value then lim Sn  1 is also finite, so  f n is
n
convergent.
Similarly, if the integral is infinite, then from the first
inequality that lim Sn  , so the series is divergent.
n
Example 26. Apply the integral test to determine the
convergence of the p-series
1
1 1 1
   p 
n
1p 2p 3p
Infinite Series
1141

1
xp
f x 
Solution. (i) When p > 1
m
m
1
1
 x1 – p 
1 – p  1]
f x dx  lim  p dx  lim 
  lim 1  p [m
m  1 x
m    1  p
m

1
1
 lim
m
1
1  1
 1 
, which is finite.

p

1
1  p m
 p1
By Cauchy’s Integral Test, the series is convergent for p  1.
(ii) When p  1


f x dx 
1
Thus, the series is divergent, if p  1.
(iii) When p  1
1
 lim m1 – p  1  

1  p m  



 1x dx

 [log x]  
1
1
Thus, the series is divergent.
1
Thus,  p is convergent if p  1 and divergent if p  1.
n

Example 27. Examine the convergence of

n2
1

n log n
1
Solution. Here f x 
x log x

1
 x log x dx
m
 lim [log log x]  lim [log log m  log log 2]  
2
2
m
m 
By Cauchy’s Integral Test the series is divergent.

2
Example 28. Examine the convergence of  ne – n
n1
2
f x  xe – x
Solution. Here
m


Now
1
 e– x 
 e– m e– 1  e– 1 1
dx  lim 

 , which is finite.

  lim 
2 2e
2 
m   2
m   2
2
2
xe – x
2
1
Hence, the given series is convergent.
Exercise 20.8
Examine the convergence :
1.
2.
1
2
2
1
x
2

x
x2 x2
    x  0
32 43
n  1n
32 2 43 3
x  4 x    n  1 xn  
3
2
3
n
Ans. Convergent
Ans. x  1, convergent; x  1, divergent.
Infinite Series
1142
3.
1
4.


n1

6.

n1
2 2 2 2  4 2 22  4 2  6 2



32 32  52 32  52  72

1
n

Ans. Divergent
1
nn
Ans. Convergent

8.
1
n2  1


7.
3
 n2e– n
Ans. Convergent
Ans. Convergent
n1
n log n2
n1
5.
n1
1

Ans. Divergent
Ans. Convergent
20.20 CAUCHY’S ROOT TEST
Statement. If  un is a positive term series such that lim un1  n  k, then
n
(i) if k  1, the series converges.
(ii) if k  1, the series diverges.
Proof. By definition of limit
un1  n  k   for n  m


k    un1  n  k   for n  m
(i)
k1
kr1
un1  n  k

un  kn
u1  u2     k  k2    kn   
1

a finite quantity
1k
 The series is convergent.
(ii)
k1
k–1
un1  n  k    1
un  1
Sn  u1  u2    un  n
lim Sn  
n 
 The series is divergent.
(iii) k  1
If lim un1  n  1, the test fails.
n
For example,  un  
1
np
1n
1
lim un1  n  lim  p 
n
n
n  
1
 lim  1  n 
n
n 

–p
 1 for all p, k  1
Infinite Series
But
1143
1
 np is convergent for p  1 and divergent for p  1.
Thus, we cannot say whether  un is convergent or divergent for k  1.
1
Example 29. Examine the convergence of the series 
n2
1  1 

n 

Solution.
un 
1
1 


n2
, un1  n
1
n
1
 
n2

 1  1 
n

lim un1  n  lim
n
n
1
1
n 
n

1  1 

 n




1
n
1  1 
 n



1
1
e
Hence, the given series is convergent.
Example 30. Discuss the convergence of the following series :
 22 2 
 12  1 


Solution.
–1
–2
–3
33 3 
 44 4 
 3    4   
2 2
3 3
–n
 n  1n  1 n  1 
un  

n1
n 
 n
–n
1
n  1n  1 n  1   n n  1n  1 n  1  – 1
 
[un]1  n  


n1
n1
n 
n 

 n
 n
–1
1
1 n1 
1
lim un1  n  lim 1  
 1     e  1 – 1 
1
n
n
e
1

n
n 





Hence, the given series is convergent.
2
Example 31. Discuss the convergence of the series
3
1 2
3
4
 x    x2    x3   
5
2 3
4
 
 
Solution. Ignoring the first term, we have :
n
n  1 n
un  
 x
n  2
n  1
un1  n  
x
n  2
1  1 

n
n
1

lim un  lim 
xx
n 
n   1  2 
n

Hence, the series converges if x  1 and diverges if x  1.
If x  1
[Cauchy Root Test]
Infinite Series
1144
n
1
n 1  
 n  1 
n
un  

 
2

n

 1  2 
n

n
1  1
 n
e 1


lim un  lim
 2  / 0
n 2
e
e
n   
2 2 
n
1  n  

 
 The series is divergent by Cauchy Fundamental theorem.
Exercise 20.9
Discuss the convergence of the following series :

1.

n1

3.

n1
5.
8.
10.
2.

n1
2
1  1 

n 


32
–n
Ans. Convergent
4.

n1
n

1
Ans. Convergent
Ans. If k  1, convergent
1
Ans. If k  1, convergent; divergent if k  1.
n log nk
 n log n– 1 log log n– k

Ans. Divergent
log n n
 n– k
n2
7.
Ans. Convergent
1  1 

n 



6.

1
nn
n
1
 1  n 
Ans. If k  1, convergent; divergent if k  1.
2
Ans. Convergent
9.
a  b  a2  b2  a3  b3  
xn
 nn
Ans. Convergent
Ans. convergent if a  1, b  1; divergent if a  1, b  1.
20.21 LOGARITHMIC TEST
If  un is a positive term series such that
un 

k
lim n log
un  1 
n 
(i) if k  1, then the series is convergent. (ii) if k  1, then the series is divergent.
Proof. (i) If k  1
1
Compare  un with  p , if k  p  1,  un converges.
n
if
if
p
n  1 p  1 
 1  
un  1
n
np

Taking logarithm of both sides of (1), we have :
un
1
 p log 1  
log
un  1
n

un
1
1
1
1
 p   2  3  4  
log
n 2n 3n 4n
un  1


un
1
1
1

 p 1 

 
n log
2n 3n2 4n3
un  1


un
p
lim n log
un  1
n
un

...(1)
Infinite Series
1145
un


 lim n log un1 k 
n  

i.e., k  p which is true as k  p 1.
Hence,  un is convergent.
When p  1
Similarly when p  1,  un is divergent.
When p  1, the test fails.
Example 32. Test the convergence of the series
Solution. x 
x
22  x2 33  x3 44  x4




2
3
4
nn  xn
22  x2 33  x3 44  x4





n
2
3
4

un 
n  1n  1  xn  1
nn  x n ,
un  1 

n  1
n

n
1
nn 1
1
nn  x n
1



n x
n x
n  1  xn  1
un  1
n



n
1

n
1

1  1
 n


un
1
1 1 1
 lim
 
lim
n
n   un  1
n 
1 x e x
1  n 


un

1
1
 1 or x  , the series is convergent.
e
ex
1
1
1
1
 1 or  x, the series is divergent. If
 1 or x  , the test fails.
If
ex
e
ex
e
If
log
un
1
n
1

n  e  log e  log 1  n 
un  1


1  1 
 n


1 1
1
1

 1  n log 1    1  n   2  3  
n
n
n
n
2
3




 log
11
1
1
1
1



2n 3n2
2n 3n2
un
 lim
lim n log
u
n1
n
n
Thus, the series is divergent.
 1  1   1  1.
 2 3n  2


Example 33. Test the convergence of the series
Solution. The given series is
2x 32x2 43x3



1
2! 3!
4!
n–1 n–1
x
n
un 
Here
n!
1
2x 32x2 43x3



2! 3!
4!
 un  1 
n  1n x
n  1 !
n
Infinite Series
1146

un  1 n  1n xn
n  1n
n!
 n–1 n–1 

x
un
x
n
n  1 !
n  1 nn – 1
 n  1
n  1n – 1
x


n–1
n
 n 
n–1
1
x  1  
n

n–1
n
1
1
x  1    1  
n
n
 

–1
x
n
. .

1


 . lim 1  n  e

 n  

1
By D’Alembert’s Ratio Test,  un is convergent if | ex |  1, i.e., | x |  and is divergent
e
1
if | ex |  1, i.e., | x |  .
e
1
ex  1, i.e., x  , this test fails.
If
e
1
x
If
e
un  1
 ex
lim
n   un
n–1
un  1 
1
1
 1  

un
n
e


un
e

n–1
un  1
1  1 

n

e
 un 

log 
  log 
u
 n  1

1 n – 1
 1  
n








1
1
1
1
 log e  n  1 log 1    log e  n  1   2  3  
n
n
n
2
3n




1
1
1 1
1
1
1
1


 1  1 
 2  2
 2     2    1  1 
n
2
n
n
2n
3n
2n
3n
2n


3
5


2n 6n2
By Logarithmic Test
u
3 5

n log n  
un  1 2 6n
un  3

 , i.e.,  1
lim n log
un  1  2
n 

1
,  un is convergent.
e
1
1
Hence, if x  ,  un is convergent and if x  ,  un is divergent.
e
e

for x 
Exercise 20.10
Examine the convergence for the following series :
1.
12 52 92 132

 

42 82 122 162
Ans. Convergent
Infinite Series
|1
x
1147
|2
 2 |3

x  3 x3   
32
4
2.
1
3.
12  32  52 2
12 12  32
x 
2  2
2 x 2
2 2 4
2  42  62
Ans. Convergent If x  1, and divergent if x  1
4.
a  x a  2x2 a  3x3



2!
3!
1!
Ans. Convergent if x 
2
Ans. If x  e, convergent and divergent if x  e
1
1
, divergent if x 
e
e
20.22 DE MORGAN’S AND BERTRAND’S TEST
A series  un of positive terms is convergent or divergent according as

 
  u
lim n  n  1  1 log n  1 or  1.
n
n     n  1

 
p
p
p
1  3  1  3  5 
1
Example 34. Test for convergence the series 1p     
 
 
2
 
2  4   2  4  6 
Solution. The given series is :
p
p
p
1
1  3   1  3  5 
1p     
 
 
2
 
2  4   2  4  6
 1  3  5    2n  3 
un  

 2  4  6    2n  2 
Here
p
 1  3  5    2n  3 2n  1 
un  1  

 2  4  6    2n  2 2n 
p
p
un  1  2n  1 
1


  1  2n 
un
 2n 




lim
n
p
un  1
1
un
 D’Alembert’s Test fails.
Now let us apply Raabe’s Test.
–p
Here



1
 un

 1
 1  n 1  
n
2n 


 un  1


 p p p  1
p p p  1

   1  

 n 1 
2
2n
8n
8n

 2
 p
 un
lim n 
 1 
u
1

n


n
 2

p
p
 1, i.e., p  2, the series is convergent and divergent if  1, i.e., p  2.
2
2
p
This test fails if  1, i.e., p  2.
2
If
Infinite Series
1148
Now let us apply De Morgan’s Test. When p  2

 un
3
 1  1 

n
n
u
4
n

1


3
  un


   1 log n
 1  1 log n  lim 1 
Now
lim n 
u
4n
n 
n   n1



 lim
n
  un is divergent when p  2.
3  log n
   0  1
4  n

 lim log n  0


n
 n 

20.23 CAUCHY’S CONDENSATION TEST
If  n is positive for all positive integral values of n and continually diminishes as n
increases and if a be a positive integer greater than 1, then the two series   n and
 an  an are either both convergent or both divergent.
Example 35. Test for convergence the series
1
1
1
1



2 log 2 p 3 log 3 p
n log n p
is convergent if p  1 and divergent if p  1 or  1.
Solution. We apply Cauchy’s Condensation Test.
1
 n 
Here
n log n p
 nth term of the second series  an  an is :
1
1
1
1
1

an  n
i.e.,
i.e.,
i.e.,
 p
n p
n p
p
p
n
log
log
a


n
a

a
a
a
log
log


 the given series will be convergent or divergent if

1
1
 log a p  np  is convergent or


1
divergent, i.e., if  p is convergent or divergent.
n
1
But we know that  p is convergent when p  1 and divergent when p  1 or  1.
n
Hence, the given series is convergent if p  1 and divergent if p  1 or  1.
20.24 ALTERNATING SERIES
A series in which the terms are alternately negative is called the alternating series.
e.g.,
u1  u2  u3  u4   
20.25 LEIBNITZ’S RULE FOR CONVERGENCE OF AN ALTERNATING SERIES
(i) Each term is numerically less than its preceeding term. (ii) lim un  0
n 
1 1 1 1 1
    
2 3 4 5 6
Solution. The terms of the given series are alternately positive and negative.
1
1
| un |  , | un  1 | 
n
n1
Example 36. Test the following series for convergence 1 
Infinite Series
1149
(i) | un  1 |  un as
1
1

n1 n
(ii) lim un  lim
n
n
1
0
n
As both the conditions for convergence are satisfied, the given series is convergent by
Leibnitz’s rule.

Example 37. Discuss the convergence of the series

n1
 1n
n

n2  1
Solution. The terms of the given series are alternately positive and negative; and each
term is numerically less than its preceeding term.
n
1
 lim
0
lim 2
n n  1
n n  1
n
Hence, by Leibnitz’s rule, the given series is convergent.
3
4
5
1 2





Example 38. Test the convergence of the series
6 11 16 21 26
Solution. The terms of the given series are alternately positive and negative.
n
un   1n – 1
5n  1
n
1
n1
5n2  6n  5n2  5n  n  1



| un |  | un  1 | 
5n  1 5n  6
5n  1 5n  6
5n  1 5 n  1  1

| un |  | un  1 |
n
1
1
 lim
 0
5n  1 n   5  1 5
n
Hence, the series is not convergent. It is oscillatory.
lim un  lim
n
n
Exercise 20.11
Discuss the convergence of the following series :
1
1
1
1. 1 



4
2 

3 
2.
3.
4.
1  2x  3x2  4x3    x  1
3
2

n1
 1n – 1
Ans. Convergent
4
x
x
x
x



   0  x  1
1  x 1  x2 1  x3 1  x4
1 1 1 1
   
1p 2p 3p 4p

5.
Ans. Convergent
Ans. Convergent
Ans. If p  0, convergent; oscillatory if p  0.
n
Ans. Oscillatory
2n  1
1
1
1
1




x1 x2 x3 x4
is convergent for all real values of x other than negative integers.
6.
Show that the series
7.
Prove that the series x 
x2 x3
   converges if  1  x  1.
2 3
20.26 ALTERNATING CONVERGENT SERIES
There are two types of alternating convergent series :
Infinite Series
1150
(1) Absolutely convergent series. (2) Conditionally convergent series.
Absolutely convergent series. If u1  u2  u3   be such that | u1 |  | u2 |  | u3 |   be
convergent then u1  u2  u3    is called absolutely convergent.
Conditionally convergent series. If | u1 |  | u2 |  | u3 |  ....... be divergent and
u1  u2  u3   be convergent then u1  u2  u3    is called conditionally convergent.
1 1 1 1 1
Example 39. Show that the series 1       
2 3 4 5 6
is convergent but not absolutely convergent.
1 1 1
Solution. 1     
2 3 4
The terms of the series are alternately positive and negative.
1
1

(ii) lim un  0
(i) | un  1 |  | un | as
n1 n
n
Thus, the given series is convergent.
1 1 1
But 1       is divergent. (p-series, p  1
2 3 4
Hence, the given series is conditionally convergent.
1
1 1
Example 40. What can you say about the series 1  2  2  2    ?
2 3 4
1 1
1
Solution. 1  2  2  2  
2 3 4
1
1
| un |  2 , | un  1 | 
n
n  12
1
(i) | un  1 |  | un |
(ii) lim un  lim 2  0
n
n  n
Thus, the given series is convergent by Leibnitz’s rule.
1
1 1
And 1  2  2  2   is also convergent. (p-series, p  2
2 3 4
Thus, the given series is absolutely convergent.
1 1 1 1 1 1 1
Example 41. Discuss the series for convergence 1    3  2  5  3  7  
3 2 3 2 3 2 3
Solution. The given series is rewritten as
1 1 1
1
1 1 1
1   2  3      3  5  7  
2 2 2
3
3 3 3


1
1
3
3
5
2 1

lim Sn 
8
8
1
1
n
1
1 2
2
3
The given series is convergent.
1 1 1
1 1 1
Again 1   2  3     3  5  
2 2 2
3 3 3
3
19
lim Sn  2  
8
8
n 
Infinite Series
1151
This series is also convergent.
Hence, the given series is absolutely convergent.
Exercise 20.12
Discuss the convergence of the following series :
1 1 1
1. 1      
2 4 8
2.
 1n – 1 n
1 2 3
4

  

2 5 10 17
n2  1
3.
1
4.
1
5.
2 3 4
  
3 32 33
x x2 x3
  
2 3 4
sin x sin 2x sin 3x



13
33
23
Ans. Absolutely convergent
Ans. Conditionally convergent
Ans. Absolutely convergent
Ans. Divergent
Ans. Absolutely convergent
20.27 POWER SERIES IN X
a0  a1x  a2x2    anxn  
is a power series in x, here a’s are independent of x.
Proof.
lim
n 
If
un  anxn, un  1  an  1xn  1
un  1
an  1 xn  1
an  1
x
 lim
 lim
n
un
an x
n
n   an
an  1
 k, then by D’Alember t’s Ratio Test
an
1
, the series is convergent.
k
1
1
Thus, the power series is convergent if   x  
k
k
1
1
Thus, the interval of the power series is  to for convergence. Outside this interval
k
k
the series is divergent.
x2 x3 x4
Example 42. Find the values of x for which the series x  2  2  2    converges.
2 3
4
If | kx |  1  | x | 
n1
xn ,
n x
u
1



1
n
n2
n  12
2
un  1
un  1
n
1
x , lim

  lim
2 xx
2
un
u
n  1
n
n
n 
1
1  n 


By D’Alembert’s Test the given series is convergent for | x |  1 and divergent if
| x |  1.
At x   1. The series becomes
Solution. un   1n – 1
Infinite Series
1152
1 1 1
  
22 32 42
This is an alternately convergent series.
At x   1. The series becomes
1
1 1
1 2  2  2 
2 3 4
This is also convergent series, p  2.
Hence, the interval of convergence is  1  x  1.
1
20.28 EXPONENTIAL SERIES
ex  1  x 
is convergent for all values of x.
xn – 1 ,
xn
un 
un  1 
Proof.
n!
n  1 !
un  1
xn
 lim
lim
n   un
n n !
xn
x2 x3



n!
2! 3!
n  1 !
x
01
 lim
n
xn – 1
n
Hence, by D’Alembert’s Test the exponential series is convergent for all values of x.
20.29 LOGARITHMIC SERIES
x2 x3 x4
xn
log 1  x  x        1n – 1  
n
2 3 4
is convergent for  1  x  1.
xn
xn  1
un   1n – 1 , un  1   1n
Proof.
n
n1
n
n

1
un  1
 1 x
n
n 
1 
 lim

  lim 
x   lim 
lim
xx

– 1 xn
n
u


n

n
1
1
1
n
n 
n
n 
n   1  1 


n 

Thus, the series is convergent for | x |  1 and divergent for | x | 1.
At x  1. The series becomes
1 1 1
1       which is convergent.
2 3 4
At x   1. The series becomes
1 1 1
 1       which is divergent.
2 3 4
20.30 BINOMIAL SERIES
n n  1 2 n n  1 n  2 3
1  xn  1  nx 
x 
x 
2!
3!
is convergent for | x |  1.
n n  1  n  r  2 r – 1
ur 
x
Proof.
r  1 !
Infinite Series
1153
n n  1  n  r  1 r
x
r!
ur  1
n  1

nr1
 lim
 1 x   x for r  n  1
x  lim 
lim
r
u
r
r
r
r
r 

If | x |  1, the series is convergent by D’Alembert’s Test.
ur  1 
Exercise 20.13
Test the convergence of the following series :
1.
1  x  2x2  3x3   nxn   
2.
xn
1 x x2 x3
   

2 3 4 5
n2
3.
1
5.
3
7.
1
1
1



1! 2! 3!
Ans. Convergent if  1  x  1
Ans. Convergent for  1  x  1
Ans. Convergent
32 33 34
     Ans. Convergent
2 3 4
1
1
1



1  x 2 1  x2 3 1  x3
4. 1 
4
42 43



1! 2! 3!
6. 
2 
2
2

23  2
3

22

4
Ans. Convergent
Ans. Convergent
Ans. Convergent if x  1
20.31 UNIFORM CONVERGENCE
If for a given   0, a number N can be found independent of x, such that for every x in
the interval a, b, the series is said to be uniformly convergent in the interval a, b.
Example 43. Discuss the uniform convergence of the series
1  x  x2   
1  xn
Sn  1  x  x2    xn 
Solution.
1x
S x  lim
n
1  xn
1

for | x |  1
1x 1x
1  xn   xn  | x |n
 1


| S x  Sn x |  


1  x 1  x  1  x 1  x
| x |n   1  x
if
Let
| x |N   1  x  N 
log  1  x
log | x |

log
2
1
1
In the interval  ,  , N has maximum value. N 
1
 2 2
log
2
1 1
Hence, the given series is uniformly convergent in the interval  ,  .
 2 2
Note. The series is convergent in  1, 1 but not uniformly convergent.
20.32. ABEL’S TEST
If vn x be either monotonic decreasing in n for each fixed x in a, b or monotonic
increasing in n for each fixed x in a, b, then  an x vn x is uniformly convergent in
a, b if
Infinite Series
1154
(i)  an x is uniformly convergent in a, b.
(ii) There exists k such that | vn x |  k for all n when a  x  b.
Example 44. Prove that
Solution. (i)
Hence,
xn
is uniformly convergent in  1, 1.
n3
1
 n3 is uniformly convergent.
(ii) | xn |  k for all n when  1 k  1
xn
 n3 is uniformly convergent by Abel’s Test.
Exercise 20.14
Prove that the following series are uniformly convergent in  1, 1.
xn
2. 
1.
 n2
4.
Prove that the series
xn
n n  1
3. 
x2n
x  n2
2n
e– 4x
e– 6x
e– 2x
 2
 2
 
2
2 1 4 1 6 1
is uniformly convergent with regard to x in x  0.
1
20.33 BRIEF PROCEDURE FOR TESTING A SERIES FOR CONVERGENCE
There are four types of series :
(1) Alternating series
(2) Positive term series
(3) Power series
(4) Geometric series
(1) Alternating Series. Apply Leibnitz’s Test.
(2) Positive Term Series.
(a) Cauchy’s Fundamental Test. Find un, if lim un  0, the series is divergent.
n
(b) Comparison Test. Apply Comparison Test lim
n
un
1
where  vn is p .
vn
n
(c) D’Alembert’s Ratio Test. If Comparison Test is not applicable, apply D’Alembert’s
un  1
.
Ratio Test, lim
un
n 
 un

 1.
(d) Raabe’s Test. If Ratio Test fails, apply Raabe’s Test lim n 
u

1
n
n 


(e) Cauchy’s Root Test. If Raabe’s Test fails apply Cauchy’s Root Test lim un1  n.
n 
un 

.
( f ) Logarithmic Test. If Cauchy’s Root Test fails, apply lim n log
un  1 
n  
(g) De Morgan’s and Bertrand’s Test. Apply De Morgan’s and Bertrand’s Test.
(3) Power Series. Apply D’Alembert’s Ratio Test. If this test fails, apply all the tests of
(2).
(4) Geometric Series
Note. Gauss’s Test and Integral Test are also useful, if they can be applied easily.
Infinite Series
1155
20.34 LIST OF THE TESTS FOR CONVERGENCE
1. Cauchy’s Fundamental Test
lim un  0,  un is divergent.
n
2. Comparison Test
un
 k,  un and  vn are of the same nature.
lim
n   vn
1
 np ,
p  1, convergent; p  1, divergent.
3. D’Alembert’s Test
u 
lim n 1  k, if k  1,  un is convergent; if k  1,  un is divergent.
un
n
4. Raabe’s Test
 un

 1  k, if k  1,  un is convergent; if k  1,  un is divergent.
lim n 
u
n
 n1

5. Cauchy’s Root Test
lim un1  n  k, if k  1,  un is convergent; if k  1,  un is divergent.
n
6. Logarithmic Test
lim n log
n
un
un  1
 k, if k  1,  un is convergent; if k  1,  un is divergent.
7. De Morgan’s and Bertrand’s Test

  un


 1  1 log n  k.
lim n 
n

n
1
n    



If k  1, convergent; k  1, divergent.
8. Gauss’s Test
un
 
  2
n n
un  1
(i) (a) If   1,  un is convergent
(b) If   1,  un is divergent.
   1 ,  un is convergent
(ii) If   1 and 
   1 ,  un is divergent
9. Cauchy’s Condensation Test
  n and  an  an are either both convergent or both divergent.
10. Leibnitz’s Test
Alternately positive and negative is convergent if
(ii) lim un  0
(i) | un  1 |  | un |
n
Exercise 20.15
Test the following series for convergence :
1.
1
1
1
1



n n  1
12 23 34
Ans. Convergent
Infinite Series
1156
2.
1
1
1
1




123 234 345
nn  1n  2
3.
 

 
 

4.
 tan n
6.
1
1
1
1




1 2
2 3 
3 4 
4
1

2
2

3
1
4

5
Ans. Divergent
Ans. Divergent
7.  101  n  1
9.
3

4
5.
Ans. Divergent
8.  [

n3 ]
n3  1  
Ans. Divergent
2n2  1
3
 4
3n3  2n  5


1
12.
 sin2 n
13.
 12 a  b  22 a  b  32 a  b   n2 a  b  
14.
 x  d2  x  2d2  x  3d2  
1
n
Ans. Divergent
1
n
11.
1
Hint. sin2
2
1
1

Ans. Convergent
Ans. Divergent

1
3
n
1
 sin– 1 n
Ans. Divergent
1 1
1
 
 
n  n 3 ! n3


2
Ans. Convergent
Ans. Divergent
1
Ans. Convergent
n
1 x  1 
2n – 1  x  2 
Ans. Convergent if x < 1 and x > 5;
Divergent if 1 < x > 5
2
3
2x
1 2x  1  2x 
 


x  4 2 x  4  3  x  4 
x  22
x  24
x  26
1
17.




135 246 357 468
|2
|3
|


4
1
18.
 2 3 4
10 10 10 10
|2
 |3

19. 1  2  2  
2 3
16.
20.
ax a2x2 a3x3
anxn


 2

5
10
2
n 1
21.
 nxn
22.
  1n – 1 2n  1 !
23.
  n2  1  xn
1
Ans. Convergent if 
4
 x  4; divergent if x  4
3
Ans. Convergent if  3  x  1 and x   1.
Ans. Divergent
Ans. Divergent
Ans. Convergent if  1  ax  1; divergent if | ax |  1
Ans. Convergent if 1  x; divergent if x  1
x2n – 1
 n2  1 

n3  a
24.  n
2 a
25.
1
 n tan n
Ans. Convergent
10.
15.
Ans. Convergent
x2
1
x4



2 3
2 4 
3
Ans. Convergent
Ans. Convergent if x  1; divergent if x  1
Ans. Convergent
Ans. Convergent if x2  1; divergent if x2  1
Infinite Series
1157
np
n!
3
5
7
27. 1 



2! 3! 4!
26.

Ans. Convergent
123n n
x
3  5    2n  1
Ans. Convergent if x  2; divergent if x  2
22 4
2 2  42
22  42  62
x 
x6 
x8  
34
3456
345678
30. x2 
2
2
 1  4
1
31.    

3
   3  6
32.
Ans. Convergent
22 32 42



2! 3! 4!
28. 1 
29.
Ans. Convergent
12
2
4

33. 1 
Ans. Convergent if x2  1; divergent if x2  1

1  4  7 
 1  4  7    3n  2 
Ans. Convergent

     3  6  9    3n   
6
9
3






2
12  52 12  52  92


42  82 42  82  122
 1 ! 2
 2 ! 2 2  3 ! 2 3
x
x 
x 
2!
4!
6!
Ans. Convergent
Ans. Convergent if x  4; divergent if x  4