2414 Calculus II
Chapter 9(2)
Power Series
Convergence of Power Series
Power Series
If x is a variable then

a x
n 0
n
n
 a0  a1 x  a2 x  a3 x ...
2
3
is called a Power Series

n
2
3
a
(
x

c
)

a

a
(
x

c
)

a
(
x

c
)

a
(
x

c
)
...
 n
0
1
2
3
n 0
is a Power Series Centered at the constant c
Radius of Convergence
For a Power Series centered at c, only one of the
following can happen:
1) The series converges only at c
R=0
2) There is a number R > 0 so that the series
converges absolutely for |x – c| < R and
R=a#
diverges for |x – c| > R
R is the Radius of Convergence
3) The series converges absolutely for all x
R=
The set of all x that make the series converge is the
Interval of Convergence
(
)
R
C
R=0
R
∞

Find the Radius of Convergence of:
n
n
!
x

n 0
Use the Ratio Test since we have factorials.
(n  1)!x n1
(n  1)(n!) x n x
Lim
 Lim
 x Lim(n  1)  
n
n
n 
n 
n 
n! x
n! x
This Diverges.
Divergence eliminates the second two choices so
you are back to converging only at c and R = 0
(1)n x 2 n1
Find the Radius of Convergence of 
n0 (2n  1)!

Since we have factorial try the ratio test.
( 1) n1 x 2 n3
( 2 n  3)!
Lim
2 n1
n 
( 1) x
( 2 n 1)!
 Lim
x2
( 2 n  3)( 2 n  2 )
n
n 
 Lim
( 1) n ( 1) x 2 n ( x 3 )( 2 n 1)!
n
2n
(
2
n

3
)(
2
n

2
)[(
2
n

1
)!
](

1
)
(
x
)( x )
n 
0
Since 0 < 1, this always converges. The
Radius of convergence is R = 
Find the Radius and Interval
of Convergence for:

 3( x  2)
n
n 0
Three tests can be used: Geometric Series, Root, Ratio
Geometric is the “easiest”
x  2  1 Converge
x  2  1 Diverge
The Radius of Convergence is R = 1
The Interval of Convergence is 1
unit from “c” or (1,3)

(2 x)

2
n
n 1
Find the Interval of Convergence of
n
Since we have powers try the ratio test.
Lim
( 2 x ) n1
( n 1) 2
n 
 Lim
n 
(2 x)
n2
n
 Lim
n2 (2 x)
n  2 n 1
2
n 
( 2 x )n ( 2 x ) n2
( 2 x ) n ( n 1) 2
 2x
By Ratio test Converges if 2x < 1. The
Radius is R = ½. The interval will be
 21 , 12 

 
Find the Interval of Convergence of
n 0
x n
2
Since we have a power try the root test.
Lim
n 
n
 
x n
2
 Lim
n 
x
2

x
2
By Root test Converges if x/2 < 1. The
Radius is R = 2. The interval will be   2 , 2 