nguyen (avn365) – HW13 – staron – (52970)
1
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printout
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have
24
questions.
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001
10.0points
Determine whether the series
∞
summationdisplay
k
=1
(
−
1)
k
−
1
cos
2
(
k
)
4
k
is absolutely convergent, conditionally con
vergent or divergent
Determine whether the series
2
3
−
2
4
+
2
5
−
2
6
+
2
7
−
. . .
is conditionally convergent, absolutely con
vergent or divergent.
1.
absolutely convergent
correct
2.
divergent
3.
conditionally convergent
Explanation:
To check for absolute convergence we have
to decide if the series
∞
summationdisplay
k
=1
cos
2
(
k
)
4
k
is convergent. For this we can use the Com
parison Test with
a
k
=
cos
2
(
k
)
4
k
,
b
k
=
1
4
k
.
For then
0
≤
a
k
≤
b
k
,
since 0
≤
cos
2
(
k
)
≤
1. Thus the series
∞
summationdisplay
k
=1
cos
2
(
k
)
4
k
converges if the series
∞
summationdisplay
k
=1
1.
series is divergent
2.
series is absolutely convergent
3.
series is conditionally convergent
cor
rect
Explanation:
In summation notation,
2
3
−
2
4
+
2
5
−
2
6
+
2
7
−
. . .
=
∞
summationdisplay
n
=1
(
−
1)
n
−
1
f
(
n
)
,
with
f
(
x
) =
2
x
+ 2
.
Now the improper integral
integraldisplay
∞
1
f
(
x
)
dx
=
integraldisplay
∞
1
2
x
+ 2
dx
is divergent, so by the Integral Test, the given
series is not absolutely convergent.
On the
other hand,
f
(
n
) =
2
n
+ 2
>
2
n
+ 1 + 2
=
f
(
n
+ 1)
,
while
lim
n
→∞
2
n
+ 2
= 0
.
Consequently, by the Alternating Series Test,
the given series
is conditionally convergent
.
1
4
k
converges. But this last series is a geometric
series with
r
=
1
4
<
1
,
hence convergent.
Consequently, the given
series is
absolutely convergent
.
002
10.0points
nguyen (avn365) – HW13 – staron – (52970)
2
keywords: alternating series, Alternating se
ries test, conditionally convergent, absolutely
convergent, divergent
003
10.0points
To apply the root test to an infinite series
summationdisplay
n
a
n
the value of
ρ
=
lim
n
→ ∞

a
n

1
/n
has to be determined.
Compute the value of
ρ
for the series
∞
summationdisplay
n
=1
5
n
+ 2
n
parenleftbigg
7
4
parenrightbigg
n
.
1.
ρ
=
4
7
2.
ρ
=
8
7
3.
ρ
=
7
2
4.
ρ
=
35
4
5.
ρ
=
7
4
correct
Explanation:
After division,
5
n
+ 2
n
= 5
parenleftbigg
1 +
2
5
n
parenrightbigg
,
so
(
a
n
)
1
/n
=
parenleftbigg
5
parenleftbigg
1 +
2
5
n
parenrightbiggparenrightbigg
1
/n
7
4
.
But
lim
n
→ ∞
5
1
/n
parenleftbigg
1 +
2
5
n
parenrightbigg
1
/n
= 1
as
n
→ ∞
. Consequently,
ρ
=
7
4
.
004
10.0points
Determine whether the series
∞
summationdisplay
n
=0
(
−
3)
n
(2
n
)!
is absolutely convergent, conditionally con
vergent, or divergent.
1.
absolutely convergent
correct
2.
divergent
3.
conditionally convergent
Explanation:
We use the Ratio Test with
a
n
=
(
−
3)
n
(2
n
)!
.
For then
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
(
−
3)
n
+1
(2
n
+ 2)!
(2
n
)!
(
−
3)
n
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
(
−
3)
n
+1
(2
n
+ 2)!
(2
n
)!
(
−
3)
n
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
−
3
(2
n
+ 1)(2
n
+ 2)
vextendsingle
vextendsingle
vextendsingle
.
Thus
lim
n
→ ∞
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
=
lim
n
→ ∞
3
(2
n
+ 1)(2
n
+ 2)
= 0
<
1
.
Consequently, the
series is absolutely convergent
.
005
10.0points
Which one of the following properties does
the series
∞
summationdisplay
n
=1
(
−
1)
n
n
+ 2
have?
nguyen (avn365) – HW13 – staron – (52970)
3
1.
divergent
2.
conditionally convergent
correct
3.
absolutely convergent
Explanation:
This series is not absolutely convergent as
shown:
∞
summationdisplay
n
=1
(
−
1)
n
n
+ 2
=

∞
summationdisplay
n
=1
(
−
1)
n
n
+ 2

=
∞
summationdisplay
n
=1
(1)
n
n
+ 2
∞
summationdisplay
n
=1
1
n
+ 2
where p
≤
1, this series is divergent.