LESSON 23 POWER SERIES

Definition A power series in the variable of x is a series of the form
a
n0
n
xn ,
where each a n is a real number.
Definition Let c be a real number. A power series in x  c is a series of the

form
a
n0
n
( x  c ) n , where each a n is a real number.

Example Find all values of x for which the power series

n 1
xn
n 3 n is convergent.
xn
We will use the Ratio Test. Let u n  n . Then
n3
lim
n  
un
 1
un
= nlim
 
xn  1
( n  1) 3 n
 1

n3n
xn
x
1
x
n
lim
lim
=
= 3 n
1 =
3 nn 1
1
n
1
x
lim
 1.
since n  
1
3
1
n

By the Ratio Test, the series
lim
n  
un
 1
un


n 1
un  1
xn
lim
 1 . Since
n 3 n diverges if n   u n
x
3 , then the series diverges for all real number x such that
x
 1  x  3 . Note that the set of real numbers x such that x  3 is the
3
set of real numbers given by (   ,  3 )  ( 3,  ) .

By the Ratio Test, the series

n 1
un  1
xn
lim
 1.
n 3 n is absolutely convergent if n   u n
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since nlim
 
un
 1
un

x
3 , then the series is absolutely convergent for all real
x
1 
3
the open interval (  3, 3 ) .
x  3 . This is the set of real numbers given by
number x such that
The Ratio Test is inconclusive when nlim
 
un
 1
 1 . Since lim
n  
un
un
 1
un

x
3 ,
x
 1  x  3 . Thus, we must check
3

xn
the convergence or divergence of the series  n 3 n when x   3 and x  3 .
n 1
then the Ratio Test is inconclusive when


For x   3 , we obtain the series
n 1
(  3) n
n3n =


n 1
(  1) n 3 n
=
n3n

 (  1)
n
n 1
1
. You
n
can show that this series converges by the Alternating Series Test (AST).

For x  3 , we obtain the series

n 1
3n
n3n =

1
 n.
n 1
This is the divergent harmonic
series.

Thus, the power series

n 1
xn
n 3 n is convergent for all real numbers in the interval
[  3, 3 ) .

Animation of the power series

n 1
xn
[  3, 3 ) .
n 3 n converging on the interval
Answer: [  3 , 3 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

Example Find all values of x for which the power series

n 1
2 n ( x  5) n
is
n2
convergent.
2 n ( x  5) n
We will use the Ratio Test. Let u n 
. Then
n2
lim
n  
un
 1
un
= nlim
 
2n
 n 

2 x  5 lim 
n  
n

1


2
 1
( x  5) n
( n  1) 2
 1
n2
 n
2 ( x  5) n
n2
= 2 x  5 nlim
=
  ( n  1) 2
 n 


= 2 x  5 since nlim
   n  1


2

n 

=  nlim
  n  1


2
=
2



1 
 lim
 1
.
n  
1

1  
n


By the Ratio Test, the series
lim
n  
un
 1
un
Since nlim
 

n 1
2 n ( x  5) n
is absolutely convergent if
n2
 1.
un
 1
un
 2 x  5 , then the series is absolutely convergent for all real
number x such that 2 x  5  1 
x 5 
1
.
2
1
1
1
   x 5 
2
2
2
9
11
 9 11 
 x  . This is the set of real numbers given by the open interval  ,  .
2
2
2 2 
Solving this inequality, we have that x  5 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
The Ratio Test is inconclusive when nlim
 
lim
un
n  
 1
un
un
 1
un
 1 . Since
 2 x  5 , then the Ratio Test is inconclusive when 2 x  5  1 
1
1
1
9
11
 x 5   x 5  x 
or x  . Thus, we must
2
2
2
2
2
n
n

2 ( x  5)
9
x 
check the convergence or divergence of the series 
when
2
n
2
n 1
x 5 
11
.
2
and x 
9
9 10
1
  . Thus, we obtain the series
, we have that x  5  
2
2
2
2
n
1
n
n
n 1
2



n
n
 2 (  1)


2 ( x  5)
n 1
 2
2n
(

1
)



=
=
=
. Since
n2
n2
n2
n2
n 1
n 1
n 1
For x 


n 1


n 1
1
(  1) 2
n
n

=

n 1
1
, which is a convergent p-series with p  2 , then the series
n2

1
(  1) 2 is absolutely convergent. Thus, the series

n
n 1
convergent.
n

 (  1)
n 1
n
1
is
n2
11
11 10 1

 . Thus, we obtain the series
, we have that x  5 
2
2
2
2
n
1
1
n
2n  
n
n
 2  n



2 ( x  5)
1
2
2




=
=
=
2
2
2 , which is a convergent p2
n
n
n
n 1
n 1
n 1 n
n 1
series with p  2 .
For x 

Thus, the power series

n 1
2 n ( x  5) n
is convergent for all real numbers in the
n2
 9 11 
closed interval  ,  .
2 2 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
 9 11 
Answer:  , 
2 2 
COMMENT: The answers for this two examples have been intervals. The
following theorem will give an explanation for this.

Theorem If a power series
a
n0
n
x n converges for a non-zero real number c,
then it is absolutely convergent for all real numbers x such that x  c . If a

power series
a
n0
n
x n diverges for a non-zero real number d, then it diverges for
all real numbers x such that x  d .
Proof Will be provided later.
COMMENT: This theorem tells us that the real numbers for which a power series
will converge are contained in an interval. The interval can be open, closed, or
half-open, half-closed. This interval is called the interval of convergence of the
power series. When you are asked to find the interval of convergence of a power
series, you will use the Ratio Test and you will have to remember to check the
endpoints of the interval for convergence.
COMMENT: The inequality x  r is the interval (  r , r ) and is the onedimensional case of a circle whose center is the number 0 and whose radius is r.
The inequality x  c  r is the interval ( c  r , c  r ) and is the onedimensional case of a circle whose center is the number c and whose radius is r.
The number r is called the radius of convergence of the power series. When
you are asked to find the radius of convergence of a power series, you will use the
Ratio Test and you will NOT have to check the endpoints of the interval for
convergence.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Examples Find the interval and the radius of convergence for the following power
series.

1.

n 1
nxn
n 4  16
nxn
We will use the Ratio Test. Let u n  4
. Then
n  16
lim
n  
un
 1
un
( n  1) x n  1 n 4  16

= nlim
=
  ( n  1) 4  16
nxn
n  1
n 4  16 
n 1
n 4  16


x lim

  nlim
= x  nlim
 
  ( n  1) 4  16 
n  
n
n
( n  1) 4  16



We need to calculate the two limits.
lim
n  
1
n 1

1 101

= nlim
 
n
n

16
1
1 4
4
n  16
n
n  16
lim
 n = lim
lim
4
4
=
4
n


n   ( n  1 )  16
1
n   ( n  1 )  16
16
1

(
n

1
)

4
 n

n
n4
4
4
16
n4
1
4
= nlim
 
1
16 =

1    4
n
n

Thus, nlim
 
un
 1
un
10
1
1  0  4  0
n  1
n 4  16 

 = x
  lim
= x  nlim
4
   n   n   ( n  1)  16 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

By the Ratio Test, the series
lim
n  
un
 1

n 1
nxn
is absolutely convergent if
n 4  16
 1.
un
Since nlim
 
un
 1
un
 x , then the series is absolutely convergent for all real
number x such that x  1 . This is the set of real numbers given by the
open interval (  1, 1) .
Now, we will need to check for convergence at the two endpoints of x   1
and x  1 .

 (  1)
For x   1 , we obtain the series

series

n 1

n
(  1) 4
=
n  16

n
n 1
n 1
n
n
. You can show that the
n  16
4
n
is convergent using the Limit
n 4  16

Comparison Test (LCT) using the convergent p-series

alternating series
 (  1)
n
n 1
1
n
n 1
3
. Thus, the
n
is absolutely convergent. Thus, it is
n  16
4
convergent.

n
. You can show that this positive
 16
n 1
term series is convergent using the Limit Comparison Test (LCT) using the

1
convergent p-series  3 .
n 1 n
For x  1 , we obtain the series

Thus, the power series

n 1
n
4
nxn
n 4  16 is convergent for all real numbers in the
closed interval [  1, 1 ] .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860


Animation of the power series
Answer:
n 1
nxn
converging on the interval [  1, 1] .
n 4  16
Interval of Convergence: [  1, 1 ]
Radius of Convergence: 1
( x  4) n
(  1)

n 6n
n 1

2.
n
( x  4) n
We will use the Ratio Test. Let u n  (  1)
. Then
n 6n
n
lim
n  
un
 1
un
( x  4) n  1
n 6n

= nlim
  ( n  1) 6 n  1
( x  4) n
x  4
1
lim
n  
1 =
6
1
n
x 4
6
since nlim
 
=
x 4
n
lim
=
n   n  1
6
1
1
1 = 10 1
1
n
( x  4) n
By the Ratio Test, the series  (  1 )
is absolutely convergent if
n 6n
n 1

n
lim
n  
un
 1
un
Since nlim
 
 1.
un
 1
un

x  4
6 , then the series is absolutely convergent for all
real number x such that
x 4
 1 . Rewriting this inequality, we have that
6
x  4
 1  x  4  6   6  x  4  6   10  x  2 . This is
6
the set of real numbers given by the open interval (  10 , 2 ) .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Now, we will need to check for convergence at the two endpoints of
x   10 and x  2 .
( x  4) n
For x   10 , we obtain the series  (  1 )
=
n 6n
n 1

n
(  1) n 6 n
(  1)

=
n 6n
n 1

n

 (  1)
n 1
2n
1
=
n

1
n
(  6) n
 (  1) n 6 n =
n 1

n
. This is the divergent harmonic
n 1
series.
( x  4) n
For x  2 , we obtain the series  (  1 )
=
n 6n
n 1

n

6n
(  1)

n 6n =
n 1
n

1
. You can show that this alternating series is convergent using
n
n 1
the Alternating Series Test (AST).
 (  1)
n
( x  4) n
Thus, the power series  (  1 )
is convergent for all real numbers
n 6n
n 1
in the interval (  10 , 2 ] .

n
Answer:
Interval of Convergence: (  10 , 2 ]
Radius of Convergence: 6
x 2n  1
 (  1 ) ( 2 n  1)!
n0

3.
n
x 2n  1
We will use the Ratio Test. Let u n  (  1)
. Then
( 2 n  1)!
n
lim
n  
un
 1
un
x 2n  3
( 2 n  1 )!
( 2 n  1 )!
x 2 lim

= nlim
=
=
2
n

1
n   ( 2 n  3 )!
  ( 2 n  3 )!
x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
x 2 lim
n  
( 2 n  1 )!
1
2
= x nlim
=
  ( 2 n  3) ( 2 n  2 )
( 2 n  3 ) ( 2 n  2 ) ( 2 n  1 )!

1 
1 
1
  lim
  0 since lim
 0 and
x 2  lim
n   2n  3
n   2n  3
n   2n  2



lim
n  
1
0
2n  2
x 2n  1
By the Ratio Test, the series  (  1 ) ( 2 n  1 )! is absolutely convergent if
n0

n
lim
n  
un
 1
un
Since nlim
 
 1.
un
 1
un
 0 , then the series is absolutely convergent for all real
numbers. This is the open interval (   ,  ) .
x 2n  1
Thus, the power series  (  1 ) ( 2 n  1 )! is convergent for all real numbers.
n0

n
x 2n  1
Animation of the power series  (  1 ) ( 2 n  1 )! converging on the open
n0
interval (   ,  ) .

n
Answer:
Interval of Convergence: (   ,  )
Radius of Convergence: 
x 2n  1
NOTE: In a later lesson, we will show that sin x   (  1 ) ( 2 n  1 )! for all
n0

n
real numbers x.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

4.
 (  1)
n  1
n 1
( x  1) n
n
We will use the Ratio Test. Let u n  (  1)
un
lim
n  
( x  1) n  1
n

= nlim
 
n 1
( x  1) n
 1
un
By the Ratio Test, the series
un
n  
= x  1 nlim
 
n
=
n 1
1
1
lim
1
x

1
=
since
=
 
n


1
1
1

0
1
1
n
n

lim
( x  1) n
. Then
n
1
x  1 lim
n
n  1
 1

(  1) n
 1
n 1
( x  1) n
is absolutely convergent if
n
 1.
un
Since nlim
 
un
 1
un
 x  1 , then the series is absolutely convergent for all
real number x such that x  1  1 . Rewriting this inequality, we have that
x  1  1   1  x  1  1  0  x  2 . This is the set of real numbers
given by the open interval ( 0 , 2 ) .
Now, we will need to check for convergence at the two endpoints of x  0
and x  2 .

For x  0 , we obtain the series

=
 (  1)
2n  1
n 1
1
= 
n


n 1
 (  1)
n 1
n  1
( x  1) n
=
n

 (  1)
n 1
n  1
(  1) n
n
1
. This multiple of the divergent harmonic series
n
is divergent.
2n  1
2 1
1
2n
1
NOTE: (  1)
= (  1) (  1) = [ (  1) ] (  1) = ( 1 ) (  1 ) =  1 .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

For x  2 , we obtain the series

 (  1)
n  1
n 1
( x  1) n
=
n

 (  1)
n 1
n  1
(1)n
n =
1
. You can show that this alternating series is convergent using
n
n 1
the Alternating Series Test (AST).
 (  1)
n  1

Thus, the power series
 (  1)
n  1
n 1
( x  1) n
is convergent for all real
n
numbers in the interval ( 0 , 2 ] .

Animation of the power series
 (  1)
n  1
n 1
( x  1) n
converging on the
n
interval ( 0 , 2 ] .
Answer:
Interval of Convergence: ( 0 , 2 ]
Radius of Convergence: 1
NOTE: In a later lesson, we will show that ln x 

 (  1)
n 1
n  1
( x  1) n
for
n
all real numbers in the interval ( 0 , 2 ] .

5.
x 2n
(  1)

( 2 n )!
n0
n
x 2n
We will use the Ratio Test. Let u n  (  1)
. Then
( 2 n )!
n
lim
n  
un
 1
un
x 2n  2
( 2 n )!
( 2 n )!
 2 n = x 2 lim
= nlim
=
n   ( 2 n  2 )!
  ( 2 n  2 )!
x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
x 2 lim
n  
( 2 n )!
1
2
= x nlim
=
  ( 2 n  2 ) ( 2 n  1)
( 2 n  2 ) ( 2 n  1 ) ( 2 n )!

1 
1
  lim
x 2  lim
 n   2n  2   n   2n 
lim
n  

1
  0 since lim
 0 and
n   2n  2
1
1
0
2n  1

x 2n
By the Ratio Test, the series  (  1) ( 2 n )! is absolutely convergent if
n0
n
lim
n  
un
 1
un
Since nlim
 
 1.
un
 1
un
 0 , then the series is absolutely convergent for all real
numbers. This is the open interval (   ,  ) .

x 2n
Thus, the power series  (  1) ( 2 n )! is convergent for all real numbers.
n0
n

x 2n
Animation of the power series  (  1) ( 2 n )! converging on the open
n0
interval (   ,  ) .
n
Answer:
Interval of Convergence: (   ,  )
Radius of Convergence: 

x 2n
NOTE: In a later lesson, we will show that cos x   (  1) ( 2 n )! for all
n0
n
real numbers x.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860