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Solutions to Midterm 2
1. Consider the power series
∞
X
n
=1
(4
x
)
n
n
.
a) (10 points) Find the radius of convergence.
b) (10 points) Find the interval of convergence.
Answer:
R
= 1
/
4, interval [

1
4
,
1
4
)
.
Solution:
a) Let us apply the Ratio Test. We have
a
n
=
(4
x
)
n
n
,
a
n
+1
=
(4
x
)
n
+1
n
+ 1
,
so
lim
n
→∞
±
±
±
±
a
n
+1
a
n
±
±
±
±
= lim
n
→∞
±
±
±
±
±
(4
x
)
n
+1
(4
x
)
n
±
±
±
±
±
·
n
n
+ 1
=

4
x

.
Therefore the series converges if

4
x

<
1
⇔ 
x

<
1
4
, and diverges, if

x

>
1
4
,
so the radius of convergence is equal to
1
4
.
b) Let us check the endpoints. For
x
=
1
4
we get a series
∞
X
n
=1
(4
/
4)
n
n
=
∞
X
n
=1
1
n
.
This is a divergent
p
series with
p
= 1.
For
x
=

1
4
we get a series
∞
X
n
=1
(

4
/
4)
n
n
=
∞
X
n
=1
(

1)
n
n
.
It converges by the Alternating Series Test, as 1
/n
is a decreasing sequence
converging to 0.
Grading: 1 point, if the Ratio Test is correct, but the radius is incorrect;
2 points for correct series in each endpoint and 3 points for the convergence
test in each endpoint
2. Consider the power series
∞
X
n
=1
(

1)
n
(
x/
2)
n
n
2
.
1
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View Full Documenta) (10 points) Find the radius of convergence.
b) (10 points) Find the interval of convergence.
Answer:
R
= 2, interval [

2
,
2]
.
Solution:
a) Let us apply the Ratio Test. We have
a
n
= (

1)
n
(
x/
2)
n
n
2
,
a
n
+1
= (

1)
n
+1
(
x/
2)
n
+1
(
n
+ 1)
2
,
so
lim
n
→∞
±
±
±
±
a
n
+1
a
n
±
±
±
±
= lim
n
→∞
±
±
±
±
±
(
x/
2)
n
+1
(
x/
2)
n
±
±
±
±
±
·
n
2
(
n
+ 1)
2
=

x/
2

.
Therefore the series converges if

x/
2

<
1
⇔ 
x

<
2, and diverges, if

x

>
2,
so the radius of convergence is equal to 2.
b) Let us check the endpoints. For
x
= 2 we get a series
∞
X
n
=1
(

2
/
2)
n
n
2
=
∞
X
n
=1
(

1)
n
n
.
It converges by the Alternating Series Test, as
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 Fall '07
 GuanYuShi
 Calculus, Power Series

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