Calculus- Early Transcendentals, 2021a

9.8. Power Series 363
Definition 9.46: Power Series
A power series is a series of the form
where each a n is a real number.
∞
∑ a n x n ,
n=0
As we did in the section on sequences, we can think of the a n as being a function a(n) defined on the
non-negative integers. Note, however, that the a n do not depend on x.
Example 9.47
Determine whether the power series
∞
∑
n=1
x n
n converges.
Solution. We can investigate convergence using the ratio test:
|x| n+1 n
lim
n→∞ n + 1 |x| n = lim |x| n
n→∞ n + 1 = |x|.
Thus when |x| < 1 the series converges and when |x| > 1 it diverges, leaving only two values in doubt.
When x = 1 the series is the harmonic series and diverges; when x = −1 it is the alternating harmonic
series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think
of
∞
∑
n=1
x n
n
as a function from the interval [−1,1) to the real numbers.
♣
A bit of thought reveals that the ratio test applied to a power series will always have the same nice
form. In general, we will compute
|a n+1 ||x| n+1
lim
n→∞ |a n ||x| n = lim |x| |a n+1| |a n+1 |
= |x| lim = L|x|,
n→∞ |a n | n→∞ |a n |
assuming that lim|a n+1 |/|a n | exists. Then the series converges if L|x| < 1, that is, if |x| < 1/L, and
diverges if |x| > 1/L. Only the two values x = ±1/L require further investigation. Thus the series will
always define a function on the interval (−1/L,1/L), that perhaps will extend to one or both endpoints as
well. Two special cases deserve mention: if L = 0 the limit is 0 no matter what value x takes, so the series
converges for all x and the function is defined for all real numbers. If L = ∞, then no matter what value x
takes the limit is infinite and the series converges only when x = 0. The value 1/L is called the radius of
convergence of the series, and the interval on which the series converges is the interval of convergence.
We can make these ideas a bit more general. Consider the series
∞
(x + 2)
∑
n
n=0
3 n
This looks a lot like a power series, but with (x + 2) n instead of x n . Let’s try to determine the values of x
for which it converges. This is just a geometric series, so it converges when
|x + 2|/3 < 1