Math 152 Section 10.5
J. Lewis

A power series about a is  cn ( x  a ) n where a and all cn s are constants.
n0

Example : f ( x )   x n , | x | 1, is a power series about 0 for f ( x ) 
n0
1
1 x
on the interval (  1,1).

Theorem : A power series  cn ( x  a ) n converges for :
n0
i) x  a only,
ii) on an interval centered at a ,
( a  R , a  R ) and possibly
at one or both endpoint s ,
or
iii) on (  ,  ).
R is called the radius of convergenc e. In case i, R  0 and in case iii, R  .
The interval on which t he series converges is the interval of convergenc e.
Example : Find the the radius and interval of convergenc e for the power series,

( x  1) n
n 1
n2n

.
Step 1 Apply the ratio test : |
a n 1
n | x 1|
|
an
n 1 2
n | x 1| | x 1|

1
n n  1
2
2
lim
for | x  1 | 2 . This describes the open interval of radius R  2 and center a  1,
(1  2,1  2 )  (  1, 3).
| x 1|
 1, that is at the endpoints, so test thes e separately .
2
 1
x 1
 1, the series is  which diverges.
2
n 1 n
Step 2 The ratio test fails for
When x  3,
When x   1,
 1
x 1
  1, the series is  (  1) n which converges by the alternatin g
2
n 1 n
series test.
Thus the interval of convergenc e is [  1, 3).
Math 152 Section 10.5
More Examples: Find the radius and the interval of convergence for each power seris.
1.
2.
3.
4.
5.

2 n ( x  2) n
n 1
3n n 2

(1  2 x ) n
n0
3n



x2n
n0
4n


( 2 x  6) n
n0
6n

(6  3 x ) n
n0
23n n


J. Lewis