8.6. POWER SERIES
521
Solution. If we evaluate f at x = 1 we obtain the series
∞
k=0
1
2 k
which is a geometric series with ratio
1
2 . So we can sum this series and find that
f (1) =
1
1 −
1
2
= 2.
Similarly,
f (3/2) =
∞
k=0
3
4
k
=
1
1 −
3
4
= 4.
In general, f (x) is a geometric series with ratio
x
2 and
f (x) =
∞
k=0
x
2
k
=
1
1 −
x
2
=
2
2 − x
provided that −1 <
x
2 < 1 (so that the ratio is less than 1 in absolute value). Thus, the
power series that defines f converges for −2 < x < 2.
As with Taylor series, we define the interval of convergence of a power series (8.26)
to be the set of values of x for which the series converges. In the same way as we did
with Taylor series, we typically use the Ratio Test to find the values of x for which the
power series converges absolutely, and then check the endpoints separately if the radius of
convergence is finite.
Example 8.7. Let f (x) =
∞
k=1
x k
k 2 . Determine the interval of convergence of this power
series.
Solution. First we will draw graphs of some of the partial sums of this power series to get
an idea of the interval of convergence. Let
S n (x) =
n
k=1
x k
k 2
for each n ≥ 1. Figure 8.7 shows plots of S 10 (x) (in red), S 25 (x) (in blue), and S 50 (x) (in
green). The behavior of S 50 particularly highlights that it appears to be converging to