(ii) Determine the interval of convergence
3.
[
−
1
/
6
,
1
/
6]
4.
(
−∞
,
∞
)
5.
[
−
1
,
1]
6.
[
−
6
,
6)
7.
converges only at
x
= 0
8.
[
−
6
,
6]
correct
Explanation:
Since
R
= 6, the given series
(i) converges when

x

<
6, and
(ii) diverges when

x

>
6.
On the other hand, at the point
x
=
−
6 and
x
= 6, the series reduces to
∞
summationdisplay
n
=1
(
−
1)
n
n
4
,
∞
summationdisplay
n
=1
1
n
4
respectively. But
∞
summationdisplay
n
=1
vextendsingle
vextendsingle
vextendsingle
(
−
1)
n
n
4
vextendsingle
vextendsingle
vextendsingle
=
∞
summationdisplay
n
=1
1
n
4
,
so by the
p
Series Test with
p
= 4 both series
converge. Consequently, the given series has
interval convergence = [
−
6
,
6]
.
010
10.0points
Find the radius of convergence,
R
, of the
power series
∞
summationdisplay
n
=1
√
n
(
x
−
5)
n
.
of the series.
garcia (pgg378) – HW14 – he – (53725)
6
3.
R
=
∞
4.
R
= 5
5.
R
= 0
Explanation:
The given series has the form
∞
summationdisplay
n
=1
a
n
(
x
−
5)
n
where
a
n
=
√
n.
Now for this series,
(i)
R
= 0 if it converges only at
x
= 5,
(ii)
R
=
∞
if it converges for all
x
,
while 0
<R<
∞
(iii) if it converges for

x
−
5

<R
, and
(iv) diverges for

x
−
5

>R
.
But
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
=
√
n
+ 1
√
n
=
radicalbigg
n
+ 1
n
,
so
lim
n
→ ∞
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
= 1
.
By the Ratio test, therefore, the given series
(a) converges for all

x
−
5

<
1, and
(b) diverges for all

x
−
5

>
1.
Consequently,
R
= 1
.
011
10.0points
Determine the interval of convergence of
the series
∞
summationdisplay
n
=1
n
3
(
x
−
5)
n
.
1.
interval convergence = (
−
6
,
−
4)
2.
converges only at
x
= 5
3.
interval convergence = (4
,
6)
correct
4.
interval convergence = [4
,
6)
5.
interval convergence = (
−∞
,
∞
)
6.
interval convergence = (
−
6
,
−
4]
Explanation:
The given series has the form
summationdisplay
n
=1
a
n
(
x
−
5)
n
with
a
n
=
n
3
. Now
lim
n
→∞
a
n
+1
a
n
=
lim
n
→∞
parenleftBig
n
+ 1
n
parenrightBig
3
= 1
.
By the Ratio Test, therefore, the given series
(i) converges when

x
−
5

<
1, and
(ii) diverges when

x
−
5

>
1.
On the other hand, at the points
x
−
5 =
−
1
and
x
−
5 = 1 the series reduces to
∞
summationdisplay
n
=1
(
−
1)
n
n
3
,
∞
summationdisplay
n
=1
n
3
respectively.
But by the Divergence Test,
both of these diverge. Consequently,
interval convergence = (4
,
6)
.
012
10.0points
Find the radius of convergence,
R
, of the
series
∞
summationdisplay
n
=1
(
−
1)
n
n
3
n
(
x
+ 1)
n
.
1.
R
=
1
3
garcia (pgg378) – HW14 – he – (53725)
7
2.
R
= 0
3.
R
= 3
correct
4.
R
=
∞
5.
R
= 1
Explanation:
The given series has the form
summationdisplay
n
=1
a
n
(
x
+ 1)
n
with
a
n
=
(
−
1)
n
n
3
n
.
Now for this series
(i)
R
= 0 if it converges only at
x
= 0,
(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
3(
n
+ 1)
=
1
3
.
By the Ratio Test, therefore, the given series
converges when

x
+1

<
3 and diverges when

x
+ 1

>
3. Consequently,
R
= 3
.
013
10.0points
Determine the radius of convergence,
R
, of
the power series
∞
summationdisplay
n
=1
(
−
2)
n
√
n
(
x
+ 5)
n
.