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1:
(a) Find the radius and interval of convergence of the power series
∞
X
n
=1
(2

3
x
)
n
n
.
Solution: Let
a
n
=
(2

3
x
)
n
n
. Then
±
±
±
±
a
n
+1
a
n
±
±
±
±
=

2

3
x

n
+1
n
+ 1
n

2

3
x

n
=
n
+ 1
n

2

3
x
 → 
2

3
x

= 3
±
±
x

2
3
±
±
.
By the R.T, the power series converges when 3
±
±
x

2
3
±
±
<
1, that is when
±
±
x

2
3
±
±
<
1
3
, and diverges when
±
±
x

2
3
±
±
>
1
3
, so the radius of convergence is
R
=
1
3
. The endpoints of the interval of convergence are the
points where
±
±
x

2
3
±
±
=
1
3
, that is
x
=
1
3
and
x
= 1. When
x
=
1
3
, we have
a
n
=
1
n
and so
X
a
n
diverges.
When
x
= 1 we have
a
n
=
(

1)
n
n
and so
X
a
n
converges by the A.S.T. Thus the interval of convergence is
I
=
(
1
3
,
1
²
.
(b) Find the set of all values of
x
such that the series
∞
X
n
=1
(
x
2
+
x

1)
n
n
converges.
Solution: Let
a
n
=
(
x
2
+
x

1)
n
n
. Then
±
±
±
±
a
n
+1
a
n
±
±
±
±
=

x
2
+
x

1

n
+1
n
+ 1
·
n
5

x
2
+
x

1

=
n
+ 1
n

x
2
+
x

1

and so
lim
n
→∞
±
±
±
±
a
n
+1
a
n
±
±
±
±
=

x
2
+
x

1

. By the R.T, the series converges (absolutely) when

x
2
+
x

1

<
1 and diverges
when

x
2
+
x

1

>
1. From the graph of
y
=
x
2
+
x

1 we see that

x
2
+
x

1

<
1 when
x
∈
(

2
,

1)
∪
(0
,
1)
and

x
2
+
x

1

>
1 when
x
∈
(
∞
,

2)
∪
(

1
,
0)
∪
(1
,
∞
). We check the endpoints of these intervals. When
x
=

2
,
1 we have
a
n
=
1
n
so
∑
a
n
converges, and when
x
=

1
,
0
a
n
=
(

1)
n
n
so
∑
a
n
converges. Thus
the series converges when
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.
 Spring '08
 MARSHMAN
 Power Series

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