moseley (cmm3869) – HW06 – Gilbert – (56380)
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001
(part 1 of 2) 10.0 points
For the series
∞
summationdisplay
n
=1
(

1)
n
n
+ 5
x
n
,
(i) determine its radius of convergence,
R
.
1.
R
=
1
5
2.
R
= 5
3.
R
= 0
4.
R
= 1
correct
5.
R
= (
∞
,
∞
)
Explanation:
The given series has the form
∞
summationdisplay
n
=1
a
n
with
a
n
= (

1)
n
x
n
n
+ 5
.
Now for this series,
(i)
R
= 0 if it converges only at
x
= 0,
(ii)
R
=
∞
if it converges for all
x
,
while 0
< R <
∞
,
(iii) if it converges when

x

< R
, and
(iv) diverges when

x

> R
.
But
lim
n
→ ∞
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
=
lim
n
→ ∞

x

n
+ 5
n
+ 6
=

x

.
By the Ratio Test, therefore, the given series
converges when

x

<
1 and diverges when

x

>
1. Consequently,
R
= 1
.
002
(part 2 of 2) 10.0 points
(ii) Determine the interval of convergence
of the series.
1.
interval convergence = (

1
,
1]
correct
2.
converges only at
x
= 0
3.
interval convergence = (

1
,
1)
4.
interval convergence = (

5
,
5)
5.
interval convergence = [

5
,
5)
6.
interval convergence = (

5
,
5]
7.
interval convergence = [

1
,
1)
Explanation:
Since
R
= 1, the given series
(i) converges when

x

<
1, and
(ii) diverges when

x

>
1.
On the other hand, at the point
x
= 1 and
x
=

1, the series reduces to
∞
summationdisplay
n
=1
(

1)
n
n
+ 5
,
∞
summationdisplay
n
=1
1
n
+ 5
respectively.
But by the Alternating Series
Test, the first of these series converges.
On
the other hand, if we set
a
n
=
1
n
+ 5
,
b
n
=
1
n
,
then
lim
n
→ ∞
a
n
b
n
=
lim
n
→ ∞
n
n
+ 5
= 1
.
By the
p
series Test with
p
= 1, however,
the series
∑
n
b
n
diverges. Thus by the Limit
Comparison test, the series
∑
n
a
n
also di
verges. Consequently, the given series has
interval convergence = (

1
,
1]
.
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moseley (cmm3869) – HW06 – Gilbert – (56380)
2
keywords:
003
(part 1 of 2) 10.0 points
For the series
∞
summationdisplay
n
=1
(

1)
n
n
2
n
(
x
+ 1)
n
,
(i) determine its radius of convergence,
R
.
1.
R
= 0
2.
R
=
∞
3.
R
=
1
2
4.
R
= 1
5.
R
= 2
correct
Explanation:
The given series has the form
summationdisplay
n
=1
a
n
(
x
+ 1)
n
with
a
n
=
(

1)
n
n
2
n
.
Now for this series
(i)
R
= 0 if it converges only at
x
=

1,
(ii)
R
=
∞
if it converges for all
x
,
while if
R >
0,
(iii) it coverges when

x
+ 1

< R
, and
(iv) diverges when

x
+ 1

> R
.
But
lim
n
→ ∞
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
=
lim
n
→ ∞
n
2(
n
+ 1)
=
1
2
.
By the Ratio Test, therefore, the given series
converges when

x
+1

<
2 and diverges when

x
+ 1

>
2. Consequently,
R
= 2
.
004
(part 2 of 2) 10.0 points
(ii) Determine the interval of convergence
of the series.
1.
converges only at
x
=

1
2.
interval convergence = [

1
,
3)
3.
interval convergence = (

3
,
1]
correct
4.
interval convergence = (

3
,
1)
5.
interval convergence = (
∞
,
∞
)
6.
interval convergence = [

1
,
3]
Explanation:
Since
R
= 2, the given series
(i) converges when

x
+ 1

<
2, and
(ii) diverges when

x
+ 1

>
2.
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 Spring '07
 Sadler
 Multivariable Calculus, Mathematical Series, Moseley

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