(12.8) Power Series
Example.
For what
x
does the series
∑
∞
n
=0
1
3
n
x
n
= 1 +
1
3
x
+
1
9
x
2
+
1
27
x
3
+
· · ·
converge?
This is just a geometric series. Recall
∑
∞
n
=0
ar
n
converges to
a
1

r
provided

r

<
1. Here,
a
= 1 and
r
=
x/
3. So the series converges to
1
1

x/
3
=
3
3

x
provided

x/
3

<
1, or

x

<
3.
That is, the series converges exactly when

3
< x <
3.
We can think of this series as a function:
f
(
x
) = 1+
1
3
x
+
1
9
x
2
+
1
27
x
3
+
· · ·
. This function has
domain given by the interval (

3
,
3) because that’s where the series converges. Actually,
f
(
x
) =
3
3

x
, but with domain (

3
,
3).
Example.
For what
x
does the series
∑
∞
n
=1
x
n
2
n
n
=
1
2
x
+
1
8
x
2
+
1
24
x
3
+
· · ·
converge?
We can use the ratio test. Here,
a
n
=
x
n
2
n
n
. When we do the ratio test, we take absolute
values because depending on what
x
we choose, the series might not be positive. So
lim
n
→∞
a
n
+1
a
n
= lim
n
→∞
x
n
+1
2
n
+1
(
n
+ 1)
x
n
2
n
n
= lim
n
→∞
x
n
+1
2
n
+1
(
n
+ 1)
2
n
n
x
n
= lim
n
→∞
nx
2(
n
+ 1)
=

x

2
So we know the series converges if

x

2
<
1. (In fact, it converges absolutely in that case.)
So it converges absolutely if

x

<
2 or

2
< x <
2.
We also know that it diverges if

x

2
>
1, or

x

>
2. So it diverges if
x <

2 or
x >
2.