Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
1
In Section 9.7, we saw that Taylor polynomials can be used to approximate the values of a
function. In this section and in the next two sections, we will see that several important
types of functions can be represented exactly by an infinite series, called a power series.
Power Series
Def.: A power series centered at x = c (where c is a constant) is a series of the form
Def.: A power series (where c = 0) is a series of the form

Exercise 1:
xn
is a power series centered at

n 0 n!
What are the an’s?

Exercise 2:
1
 n ( x  2)
n
.
is a power series centered at
.
n 0
What are the an’s?
A Power Series is a Function of x

A power series in x, f ( x)   a n ( x  c) n , is a function of x.
n 0
Def.: The domain of f is the set of
for which the
.
Note 1: The center, c, is always in the domain of f. In other words, every power series
converges at its center c (i.e., for x = c) because …

f (c )   a n (c  c ) n =
n 0
Note 2: In the context of power series, we conveniently consider 00 to be equal to 1.
However, to be very precise and to avoid the 00 difficulty, we should define a power series
as
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed

2
Exercise 3: Consider the power series f ( x)   (3 x) n . This power series is centered at 0.
n 0
Show that x = 0 and x = 0.1 are in the domain of f but that x = 0.5 and x = 1 are not.

f (0)   (3  0) n =
n 0

f (0.1)   (3  0.1) n =
n 0

f (0.5)   (3 
)n =
n 0

f (1)   (3  ) n =
n 0
Radius of Convergence; Interval of Convergence
Theorem: For a power series centered at c, exactly one of the following is true.
1. The series converges at
.
2. There exists a real number, R > 0, such that the series
3. The series
.
Def.: The number, R, is the radius of convergence of the power series.
If R = 0, then the series converges at
If R =  , then the series converges for
Def.: The interval of convergence is the
.
.
of the power series—that is, the
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
3
Important Note: To determine the radius of convergence of a power series, use the
That is,
u
n
converges absolutely if
n

Exercise 4: Determine the radius of convergence for
 (3 x )
n
.
n 0
u n 1
=
un
Therefore, lim
n 
u n 1
= lim 3|x| =
n
un

Thus,
 (3 x )
n
converges if 3|x| <
; that is, if
.
n 0

Exercise 5: Determine the radius of convergence for
u n 1
=
un
Therefore, lim
n 
u n 1
= lim
n
un
=
(3x) n
.

n!
n 0
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
4

(2n)! x n
Exercise 6: Determine the radius of convergence for 
.
n!
n 0
u n 1
=
un
Therefore, lim
n 
u n 1
= lim (4n  2) x =
n
un
This power series converges only at its center, c =
. A single point is considered to be an
interval of length zero. Therefore, the radius of convergence is
.
Important Note: For a power series whose radius of convergence is a finite number R > 0, the
theorem in the middle of page 2 says nothing about the convergence of the power series at the
endpoints of the interval of convergence. Except for geometric power series (see Section 9.9),
whose intervals of convergence are always open intervals, you will always have to test the
endpoints to determine if the power series converges for none of, one of, or both of those two
numbers. As a result, the interval of convergence of a power series can take any one of the
following six forms.
Radius R = 0
Interval of Convergence:
c
Radius R = 
Interval of Convergence:
c
Radius = R, 0 < R < 
c
c
c
c
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed

Exercise 7: Determine the interval of convergence for
( x  2) n 1
 (n  1) 3
n 0
5
n 1
.
u n 1
=
un
Therefore, lim
n 
u n 1
n 1
= lim
x2 =
n  3( n  2)
un
The series converges if
< 1 ; that is, if
Now, test the endpoints.
(1  2) n 1
=

n 1
n  0 ( n  1) 3

x = -1:
(5  2) n 1
=

n 1
n  0 ( n  1) 3

x = 5:
( x  2) n 1
converges for

n 1
n  0 ( n  1) 3

Thus,
 the interval of convergence
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
6

Theorem: If the function given by f ( x)   a n ( x  c) n =
n 0
has a radius of convergence R > 0, then, on the interval c - R < x < c + R , then f is
continuous, differentiable, and integrable. Also,
1. f (x) =
2.
 f ( x) dx =
Important Notes: The center and the radius of convergence of the series obtained by
differentiating or integrating a power series are the same as the center and the radius of
convergence of the original power series.
However, the interval of convergence of the series obtained by differentiating or integrating
a power series and the interval of convergence of the original power series may differ at
The theorem above suggests that, in many ways, a function defined by a power series
behaves like a polynomial. It is continuous in its interval of convergence, and both its
derivative and its antiderivative can be determined by differentiating and integrating the
given power series term by term.
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
(1) n1 ( x  1) n
.
n
n 1

Exercise 8a: Determine the interval of convergence for f ( x)  
u n 1
=
un
Therefore, lim
n 
u n 1
=
un
The series converges if
< 1 ; that is, if
Now, test the endpoints.
(1) n 1 ( x  1) n
converges for

n
n 1

Thus,
 the interval of convergence
7
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
8
Exercise 8b: Determine the interval of convergence for f (x) , where f is defined in Exercise 8a.
Its interval of convergence is 0 < x < 2 and possibly also one or both of the endpoints.
f (x) =
Test the endpoints:
Therefore, the interval of convergence for f (x) is
.
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 8 Completed
9
Exercise 8c: Determine the interval of convergence for  f ( x) dx , where f is defined in Exercise 8a.
Its interval of convergence is 0 < x < 2 and possibly also one or both of the endpoints.
 f ( x) dx =
Test the endpoints:
Therefore, the interval of convergence for  f ( x) dx is
.