farr (arf2325) – HW14 – meth – (54160)
1
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001
10.0points
Find the interval of convergence of the se
ries
∞
summationdisplay
n
=1
x
n
√
n
+ 4
.
1.
interval of cgce = (
−
4
,
4]
2.
interval of cgce = [
−
4
,
4]
3.
converges only at
x
= 0
4.
interval of cgce = [
−
1
,
1)
correct
5.
interval of cgce = [
−
1
,
1]
6.
interval of cgce = (
−
1
,
1)
Explanation:
When
a
n
=
x
n
√
n
+ 4
,
then
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
x
n
+1
√
n
+ 5
√
n
+ 4
x
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=

x

parenleftbigg
√
n
+ 4
√
n
+ 5
parenrightbigg
=

x

radicalbigg
n
+ 4
n
+ 5
.
But
lim
n
→ ∞
n
+ 4
n
+ 5
= 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
+ 4
.
Applying the Integral Test with
f
(
x
) =
1
√
x
+ 4
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
+ 4
.
which is an alternating series
∞
summationdisplay
n
=1
(
−
1)
n
a
n
,
a
n
=
f
(
n
)
with
f
(
x
) =
1
√
x
+ 4
the same continuous, positive and decreasing
function as before. Since
lim
x
→ ∞
f
(
x
) =
lim
x
→ ∞
1
√
x
+ 4
= 0
,
however, the Alternating Series Test ensures
that (
‡
) converges.
Consequently, the
interval of convergence = [
−
1
,
1)
.
002
10.0points
Find the interval of convergence of the se
ries
∞
summationdisplay
n
=1
(
−
1)
n
x
n
3
n
2
+ 5
.
farr (arf2325) – HW14 – meth – (54160)
2
1.
[
−
1
,
1)
2.
converges only at
x
= 0
3.
(
−
1
,
1]
4.
[
−
3
,
5]
5.
[
−
1
,
1]
correct
6.
(
−
3
,
5]
Explanation:
When
a
n
= (
−
1)
n
x
n
3
n
2
+ 5
,
then
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
−
x
n
+1
3(
n
+ 1)
2
+ 5
3
n
2
+ 5
x
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=

x

parenleftbigg
3
n
2
+ 5
3(
n
+ 1)
2
+ 5
parenrightbigg
.
But
3(
n
+ 1)
2
+ 5 = 3
n
2
+ 6
n
+ 8
,
while
lim
n
→ ∞
3
n
2
+ 5
3
n
2
+ 6
n
+ 8
= 1
.
Thus
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
3
n
2
+ 5
.
Applying the Integral Test with
f
(
x
) =
1
3
x
2
+ 5
we see that
f
is continuous, positive, and de
creasing on [1
,
∞
); in addition,the improper
integral
I
=
integraldisplay
∞
1
f
(
x
)
dx
converges, so the infinite series (
∗
) converges
also.
On the other hand, at
x
= 1, the series
becomes
(
‡
)
∞
summationdisplay
n
=1
(
−
1)
n
3
n
2
+ 5
.