wang (kw28364) – HW14 – staron – (53735)
1
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001
10.0points
Find the interval of convergence of the se
ries
∞
summationdisplay
n
=1
x
n
√
n
+ 3
.
1.
converges only at
x
= 0
2.
interval of cgce = [
−
3
,
3]
3.
interval of cgce = [
−
1
,
1]
4.
interval of cgce = (
−
1
,
1)
5.
interval of cgce = [
−
1
,
1)
correct
6.
interval of cgce = (
−
3
,
3]
Explanation:
When
a
n
=
x
n
√
n
+ 3
,
then
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
x
n
+1
√
n
+ 4
√
n
+ 3
x
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=

x

parenleftbigg
√
n
+ 3
√
n
+ 4
parenrightbigg
=

x

radicalbigg
n
+ 3
n
+ 4
.
But
lim
n
→ ∞
n
+ 3
n
+ 4
= 1
,
so
lim
n
→ ∞
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=

x

.
By the Ratio Test, therefore, the given series
(i) converges when

x

<
1,
(ii) diverges when

x

>
1.
We have still to check what happens at the
endpoints
x
=
±
1.
At
x
=
1 the series
becomes
(
∗
)
∞
summationdisplay
n
=1
1
√
n
+ 3
.
Applying the Integral Test with
f
(
x
) =
1
√
x
+ 3
we see that
f
is continuous, positive, and de
creasing on [1
,
∞
), but the improper integral
I
=
integraldisplay
∞
1
f
(
x
)
dx
diverges, so the infinite series (
∗
) diverges
also.
On the other hand, at
x
=
−
1, the series
becomes
(
‡
)
∞
summationdisplay
n
=1
(
−
1)
n
√
n
+ 3
.
which is an alternating series
∞
summationdisplay
n
=1
(
−
1)
n
a
n
,
a
n
=
f
(
n
)
with
f
(
x
) =
1
√
x
+ 3
the same continuous, positive and decreasing
function as before. Since
lim
x
→ ∞
f
(
x
) =
lim
x
→ ∞
1
√
x
+ 3
= 0
,
however, the Alternating Series Test ensures
that (
‡
) converges.
Consequently, the
interval of convergence = [
−
1
,
1)
.
002
10.0points
Determine the radius of convergence,
R
, of
the series
∞
summationdisplay
n
=1
(
−
6)
n
n
+ 3
x
n
.
003
10.0points
wang (kw28364) – HW14 – staron – (53735)
3
But as
n
→ ∞
, neither of
√
2
n,
(
−
1)
n
√
2
n
−→
0
holds, so both of
∞
summationdisplay
n
=1
√
2
n,
∞
summationdisplay
n
=1
(
−
1)
n
√
2
n
diverge. Consequently, the
interval of convergence = (
−
1
,
1)
.