12.8. Power Series
Find the radius of convergence and interval of convergence of the series.
10.
∞
X
n
=1
x
n
n
3
n
solution:
If
a
n
=
x
n
n
3
n
, then
lim
n
→∞
|
a
n
+1
a
n
|
= lim
n
→∞
|
x
n
+1
(
n
+ 1)3
n
+1
n
3
n
x
n
|
= lim
n
→∞
|
xn
(
n
+ 1)3
|
=
|
x
|
3
lim
n
→∞
n
n
+ 1
=
|
x
|
3
.
By the Ratio test, the series converges when
|
x
|
3
<
1. That is,
|
x
|
<
3. So,
R
= 3.
When
x
=
-
3, the series is the alternating harmonic series, which converges by the Alter-
nating Series test. When
x
= 3, it is the harmonic series, which diverges. Thus, the interval
of convergence is [
-
3
,
3).
14.
∞
X
n
=0
(
-
1)
n
x
2
n
(2
n
)!
.
solution:
a
n
= (
-
1)
n
x
2
n
(2
n
)!
, so
lim
n
→∞
|
a
n
+1
a
n
|
= lim
n
→∞
|
x
|
2
n
+2
(2
n
+ 2)!
·
(2
n
)!
|
x
|
2
n
= lim
n
→∞
|
x
|
2
(2
n
+ 1)(2
n
+ 2)
= 0
Thus, by the Ratio test, the series converges for all real
x
and we have
R
=
∞
and the
interval of convergence (
-∞
,
∞
).
12.9. Representation of Functions as Power Series
8. Find a power series representation for the function
f
(
x
) =
x
4
x
+ 1
and determine the
interval of convergence.
solution:
f
(
x
) =
x
4
x
+ 1
=
x
·
1
1
-
(
-
4
x
)
=
x
∞
X
n
=0
(
-
4
x
)
n
=
∞
X
n
=0
(
-
1)
n
4
n
x
n
+1
.
1

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2
The series converges when
| -
4
x
|
<
1 ( that is,
|
x
|
<
1
4
) since
∑
∞
n
=0
(
-
4
x
)
n
is a geometric
series. It is easy to see that the series diverges at the end points. So, the interval of conver-
gence is (
-
1
4
,
1
4
).
14. (a) Find a power series representation for


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- Fall '09
- odell
- Power Series, Mathematical Series, lim
-
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