LESSON 19 CONVERGENT AND DIVERGENT SERIES

Definition Let { a n } be a sequence with a domain of { n   : n  N } , where N
is a positive integer. Then the sum of all the terms of in the sequence, denoted by

a
nN
n
, is called an infinite series, or simply a series.
NOTE: If the domain of the sequence is   , then the series of this sequence is

a
n 1
n
.
Examples The following are series.
  

n sin 

3
n
n 1




1.

4.
n
n 1
1
5/3
7.
n5
3.

5.

n
8.
n3
2
n
n 1
1
3
8

1
(  1)

n 1
n2
n

3

n
n  1 10

2.

5n
n 2  16
6.
4
n 1
1
 2n

a
Definition Given the series
nN
n
, define the sequence { S n } , where
Sn  aN  aN 1  aN  2       aN  n 1 
N  n 1
a
nN
n
. This sequence is called the
sequence of partial sums. Note that the N given above is a positive integer.

Thus, given the series
a
nN
n
, we have that
Copyrighted by James D. Anderson, The University of Toledo
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S1  a N
S 2  a N  a N  1  S1  a N  1
S3  aN  aN 1  aN  2  S2  aN  2
S4  aN  aN 1  aN  2  aN  3  S3  aN  3
S5  aN  aN  1  aN  2  aN  3  aN  4  S4  aN  4
.
.
.
Sn  aN  aN 1  aN  2       aN  n  2  aN  n 1  Sn 1  aN  n 1
.
.
.

Thus, given the series
a
n 1
n
, we have that
S1  a1
S 2  a1  a 2  S1  a 2
S 3  a1  a 2  a 3  S 2  a 3
S 4  a1  a 2  a 3  a 4  S 3  a 4
S 5  a1  a 2  a 3  a 4  a 5  S 4  a 5
.
.
.
S n  a1  a 2  a 3       a n  1  a n  S n  1  a n
Copyrighted by James D. Anderson, The University of Toledo
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.
.
.

Definition Given the series
a
nN
n
, whose sequence of partial sums is { S n } , then
S n  S , then we say that the series is convergent (or converges, or
if nlim
 
S n does not exist, then we say that the series is divergent
converges to S.) If nlim
 
(or diverges.) Note that the N given above is a positive integer.

TERMINOLOGY: S is called the sum of the series
a
nN

n
and write
a
nN
n
 S
For the examples given above, we will show in this lesson that the series




  
3
1
 and  4 are divergent, and the series  n and  2
n sin 

 2n
n 1
n3 n
n  1 10
n 1
 3n 

n
are convergent. We will show in later lessons that the series

n
divergent, and the series
n 1
3
1
,
8

n
n 1
1
5/3
n5

, and
 (  1)
n2
n
2
5n
is
 16
1
are
n 1
convergent.

Definition The series
 ar
n 1
n 1
 a  ar  ar2  ar3       arn 1     ,
where a and r are constants and a  0 , is called a geometric series.

Theorem The geometric series
 ar
n 1
n 1
converges and has a sum of S 
if r  1 . The geometric series diverges if r  1 .
Proof Will be provided later.
Copyrighted by James D. Anderson, The University of Toledo
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a
1 r

Example Determine whether the series
3
 10
n 1
n
converges or diverges. If it
converges, then find its sum.

The series
3
 10
n 1
n
, which was one of our examples given above, is a geometric

3
series since  n =
n  1 10

3  1 
 

n  1 10  10 
n 1
.
1
 1 , then by the theorem above, this geometric series converges and
10
3
a
10  3  3  1
has a sum of S 
=
1
10  1 9 3 .
1 r
1
10
Since r 
Answer: Converges;
1
3

Example Determine whether the series
n
n3
2
1
converges or diverges. If it
 2n
converges, then find its sum.
This series was one of our examples given above.
We will rewrite the fraction
1
using partial fraction decomposition.
n 2  2n
1
B
A
 1  A( n  2 )  B n
=
+
n(n  2)
n 2
n
To solve for A, choose n  0 : 1   2 A  A  
1
2
Copyrighted by James D. Anderson, The University of Toledo
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To solve for B, choose n  2 : 1  2 B  B 
Thus,
1
1
=
=
n(n  2)
n  2n
2

Thus,

n3
1
=
2
n  2n


n3
1
2
1
1
1 1
1
2
2

.

+
= 
2  n  2 n 
n  2
n

1 1
1
1

  =
2n  2 n
2


n3

1
  n  2

We will find the sequence { S n } of partial sums for the series
where a n 
1

n 


n3

1
  n  2

1
,
n 
1
1
 . Thus,
n 2 n
S1  a 3  1 
1
3
S 2  a 3  a 4  S1  a 4  1 
1 1 1
 
3 2 4
S3  a3  a4  a5  S2  a5  1 
1 1 1 1 1
1 1 1
    1  
3 2 4 3 5
2 4 5
S4  a3  a4  a5  a6  S3  a6  1 
1 1 1 1 1
1 1 1
    1  
2 4 5 4 6
2 5 6
S5  a3  a4  a5  a6  a7  S4  a7  1 
1 1 1 1 1
1 1 1
    1  
2 5 6 5 7
2 6 7
.
.
.
Sn  a3  a4  a5       an  2  1 
1
1
1
3
1
1


 

2 n 1 n  2 2 n 1 n  2
Copyrighted by James D. Anderson, The University of Toledo
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.
.
.
3
1
1 
3
3
 =

00 = .
S n = lim  
Then nlim
n   2
 
n 1 n  2
2
2

 1
1 3

  . Thus,


Thus,

n  2
n3 n  2



n3
1
1
=
n 2  2n
2


n3

1
  n  2

1
 =
n 
13
3
  =
22
4
Answer: Converges;
3
4

COMMENT: The series
n
n3

Theorem If a series
a
nN
n
2
1
is called a telescoping series.
 2n
an  0 .
is convergent, then nlim
 
Proof Will be provided later.
The contrapositive statement of this theorem gives us a test for divergence.
a n  0 , then the series
Test for Divergence: If nlim
 

a
nN
n
is divergent.
COMMENT: The Test for Divergence is the second most misused statement by
Calculus students. Students want to apply the converse of the previous theorem,
a n  0 , then the series
which is the statement if nlim
 

a
nN
However, this is NOT true.
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n
is convergent.

1
 n.
1
 0.
n
n 1
However, we will show in a later lesson that this series is DIVERGENT. The

1
series  is called the (divergent) harmonic series.
n 1 n
An easy example to keep in mind is the series
We have that nlim
 
Examples Use the Divergence Test to show that the following series diverge.

1.
  
 n sin  3n 
n 1


This series was one of our examples given above. We want to show that
  
  0 .
lim n sin 
n  
3
n


  
 =   0
lim n sin 
n  
3
n


  

sin 
3
n
  


n sin
Since n sin   =
, then we can write nlim
1
 
3
n


n
  

sin 
3
n


0
lim
,
which
has
an
indeterminate
form
of
.
n  
1
0
n
 
sin 
 3x
lim
We will apply L’Hopital’s Rule to x  
1
x



. Thus,
Copyrighted by James D. Anderson, The University of Toledo
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  

 =
3
n


 
sin 
 3x
lim
x  
1
x
  
 
  D x 
cos 
 3x 
 3x
lim
= x
1
Dx  
x






   
1
  D x  
cos 
x
 3x  3
lim
= x
=
1
Dx  
x
  


1




 =
lim cos 
cos 
lim
cos 0  ( 1) 

=
3 x
3
3
3
3
 3 xx
 3x 
 
sin 
 3n
lim
Thus, n  
1
n



   

   0 .
lim
n
sin
. Thus, n  
3
n

 3
Answer: Divergent (by the Divergence Test)

2.
4
n 1
This series was one of our examples given above.
lim 4  4  0
n  
Answer: Divergent (by the Divergence Test)

Theorem If
a
nN

n
and
b
nN
n
are convergent series with sums A and B,
respectively, then

1.
 (a
nN
n
 b n ) is a convergent series and has of sum of A  B .
Copyrighted by James D. Anderson, The University of Toledo
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
2.
if c is a constant, then
 ca
nN
n
is a convergent series and has of sum of
c A.

 (a
3.
nN
n
 b n ) is a convergent series and has of sum of A  B .
Proof Will be proved later.

Theorem If

 (a
nN
n

nN

a n is a convergent series and
b
nN
n
is a divergent series, then
 b n ) is a divergent series.
Proof Will be provided later.
Examples Determine whether the following series converge or diverge. If the
series converges, then give its sum.
1.
7
7
7
7

      (  1) n  1 n  1     
5 25
5

 1
NOTE: This series can also be written as  7   
 5
n 1
n 1
.
1
1
This is a geometric series where a  7 and r   . Since r   1 ,
5
5
then the geometric series converges and has a sum of
S 
7
35
35
a


=
1 51
6 .
1 r
1
5
Copyrighted by James D. Anderson, The University of Toledo
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Answer: Converges;

2.
5
2 

8
n 1
35
6
n 1
This is a geometric series where a  2 and r 
5
5
. Since r   1 , then
8
8
the geometric series converges and has a sum of
S 
2
16
16
a


=
5 85
3 .
1 r
1
8
Answer: Converges;

3.
3
n
4n 1
n
n 1
16
3
n 1

3
4
n 1

=

n 1
4n 1
=
3n


n 1
1 4n 1

=
3 3n  1
This is a geometric series where a 

14
 

n 1 3  3 
1
4
4
and r  . Since r   1 , then
3
3
3
the geometric series diverges.
Answer: Diverges

4.
n 1
3n  4
 2n  3
n 1
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4
3n  4
3 0
3
n
lim
lim
 0
=
=
=
n  
3
n   2n  3
2

0
2
2 
n
3

Thus, the series
3n  4
 2n  3
n 1
diverges by the Divergence Test.
Answer: Diverges

5.
n
n 1
2
1
 11n  30
We will rewrite the fraction
1
using partial fraction
n 2  11 n  30
decomposition.
1
B
A
 1  A( n  6 )  B ( n  5)
=
+
( n  5) ( n  6 )
n 6
n 5
To solve for A, choose n   5 : 1  A
To solve for B, choose n   6 : 1   B  B   1
Thus,
1
1
1
1

=
=
.
( n  5) ( n  6 )
n 5 n 6
n  11 n  30
2

Thus,

n 1
1
=
2
n  11n  30
 1
1 



 n  5 n  6 
n 1 


We will find the sequence { S n } of partial sums for the series
 1
1 



 n  5 n  6  . Thus,
n 1 


Copyrighted by James D. Anderson, The University of Toledo
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S1 
1 1

6 7
S2 
1 1 1 1 1 1
    
6 7 7 8 6 8
NOTE: S 2  S 1 
S3 
1 1

7 8
1 1 1 1 1 1
    
6 8 8 9 6 9
NOTE: S 3  S 2 
1 1

8 9
.
.
.
Sn 
1
1

6 n 6
.
.
.
1
1 
1
1
 =
0 = .
S n = lim  
Then nlim
n   6
 
n  6
6
6

 1
1  1
  . Thus,

Thus,  
n

5
n

6
n 1 
 6



1
1

  n  5  n  6 
n 1


=

n
n 1
2
1
=
 11n  30
1
6
Copyrighted by James D. Anderson, The University of Toledo
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Answer: Converges;

6.
n
n3
2
1
6
1
 n  2
We will rewrite the fraction
n2
1
using partial fraction
 n  2
decomposition.
B
1
A
 1  A ( n  2 )  B ( n  1)
=
+
n 2
( n  1) ( n  2 )
n 1
To solve for A, choose n   1 : 1   3 A  A  
To solve for B, choose n  2 : 1  3 B  B 
1
3
1
3
1
1
1
1
3
3
=
=
+
=
( n  1) ( n  2 )
n 1
n  2
 n  2

Thus,
n2
1 1
1 

.

3  n  2 n  1 

Thus,

n3
1
=
n2  n  2
1 1
1 
1   1
1 







=
 n  2 n  1
 n  2 n  1 
3
3
n3
n3




We will find the sequence { S n } of partial sums for the series


n3

1
1

  n  2  n  1  .

Let a n 
1
1

Thus,
n 2 n 1
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
S1  1 
1
4
NOTE: S 1  a 3
S2  1 
1 1 1
 
4 2 5
NOTE: S 2  S 1  a 4 = S 1 
S3  1 
1 1 1 1 1
   
4 2 5 3 6
NOTE: S 3  S 2  a 5 = S 2 
S4  1 
1 1

4 7
1 1 1 1 1 1 1
1 1 1 1 1
     
= 1    
2 5 3 6 7 5 8
2 3 6 7 8
NOTE: S 5  S 4  a 7 = S 4 
S6  1 
1 1

3 6
1 1 1 1 1 1 1
1 1 1 1 1
     
= 1    
4 2 5 3 6 4 7
2 5 3 6 7
NOTE: S 4  S 3  a 6 = S 3 
S5  1 
1 1

2 5
1 1

5 8
1 1 1 1 1 1 1
1 1 1 1 1
     
= 1    
2 3 6 7 8 6 9
2 3 7 8 9
NOTE: S 6  S 5  a 8 = S 5 
1 1

6 9
.
.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
.
Sn  1 
1 1
1
1
1
 


=
2 3 n 1 n  2 n 3
6 3 2
1
1
1
11
1
1
1
  





=
6 6 6 n 1 n  2 n 3
6
n 1 n  2 n 3
.
.
.
 11
1
1
1 
11
 =


S n = lim  
Then nlim
.
n  
 
n  1 n  2 n  3
6
 6
 1
1  11

  . Thus,


Thus,

n  1  6
n3 n  2


n
n3
2
1
=
 n  2
1   1
1 
1  11 
11




=   =


3 n3 n  2 n  1
3 6 
18
Answer: Converges;

7.

n 1
11
18
1
n 2  9 n  18
We will rewrite the fraction
n
2
1
using partial fraction
 9 n  18
decomposition.
1
B
A
 1  A( n  6 )  B ( n  3)
=
+
( n  3) ( n  6 )
n 6
n  3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
To solve for A, choose n   3 : 1  3 A  A 
1
3
To solve for B, choose n   6 : 1   3 B  B  
Thus,
n2
1
3
1
1

1
1
3
3
=
=
+
=
( n  3) ( n  6 )
n  3
n  6
 9 n  18
1 1
1 

 . Thus,

3  n  3 n  6 


n 1
1
=
n 2  9 n  18
1 1
1 
1   1
1 







=
n  3 n  6
 n  3 n  6 
3
3
n 1
n 1 




We will find the sequence { S n } of partial sums for the series
 1
1
1
1 
a






. Let n
Thus,
n 3 n 6
n  6 
n 1  n  3

S1 
1 1

4 7
NOTE: S 1  a 1
S2 
1 1 1 1
  
4 7 5 8
NOTE: S 2  S 1  a 2 = S 1 
S3 
1 1

5 8
1 1 1 1 1 1
    
4 7 5 8 6 9
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
NOTE: S 3  S 2  a 3 = S 2 
S4 
1 1 1 1 1 1 1
1
1 1 1 1 1
1
      
    
=
4 7 5 8 6 9 7 10
4 5 8 6 9 10
NOTE: S 4  S 3  a 4 = S 3 
S5 
1
1

7 10
1 1 1 1 1
1
1
1
1 1 1 1
1
1
    
 
   

=
4 5 8 6 9 10 8 11
4 5 6 9 10 11
NOTE: S 5  S 4  a 5 = S 4 
S6 
1 1

6 9
1
1

8 11
1 1 1 1
1
1
1
1
1 1 1
1
1
1
   

 
  


=
4 5 6 9 10 11 9 12
4 5 6 10 11 12
NOTE: S 6  S 5  a 6 = S 5 
1
1

9 12
.
.
.
Sn 
1 1 1
1
1
1
  


=
4 5 6 n  4 n 5 n 6
15 12 10
1
1
1
37
1
1
1








=
60 60 60 n  4 n  5 n  6
60 n  4 n  5 n  6
.
.
.
 37
1
1
1 
37

 =



S n = nlim
Then nlim
.
  60
 
n  4 n  5 n  6
60

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
 1
1  37
 

Thus,  
. Thus,
n

3
n

6
n 1 
 60



n 1
1
=
n  9 n  18
2
1   1
1 
1  37 
37

 =   =


3 n3 n  3 n  6 
3  60 
180
Answer: Converges;
37
180
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860