mei (dym252) – HW13 – schultz – (53990)
1
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001
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
3
2
n
parenleftbigg
n
−
4
n
parenrightbigg
n
2
.
1.
ρ
= 3
e
−
4
2.
ρ
= 9
3.
ρ
=
∞
4.
ρ
= 3
e
4
5.
ρ
= 9
e
−
4
correct
Explanation:
After division,
n
−
4
n
= 1
−
4
n
,
so
(
a
n
)
1
/n
=
parenleftBigg
3
2
n
parenleftbigg
1
−
4
n
parenrightbigg
n
2
parenrightBigg
1
/n
= 3
2
parenleftbigg
1
−
4
n
parenrightbigg
n
.
But
lim
n
→ ∞
parenleftbigg
1
−
4
n
parenrightbigg
n
=
e
−
4
.
Consequently,
ρ
= 9
e
−
4
.
002
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
ρ
for the series
∞
summationdisplay
n
=1
parenleftbigg
7 tan
−
1
n
6
parenrightbigg
n
.
1.
ρ
=
3
7
2.
ρ
=
7
12
3.
ρ
=
7
6
4.
ρ
=
7
π
12
correct
5.
ρ
=
3
π
7
Explanation:
For the given series
(
a
n
)
1
/n
=
7 tan
−
1
n
6
.
But
lim
n
→ ∞
tan
−
1
n
= tan
−
1
∞
=
π
2
.
Consequently,
ρ
=
7
π
12
.
003
10.0points
Determine whether the series
∞
summationdisplay
n
=0
(
−
6)
n
(2
n
)!
mei (dym252) – HW13 – schultz – (53990)
2
is absolutely convergent, conditionally con
vergent, or divergent.
1.
conditionally convergent
2.
divergent
3.
absolutely convergent
correct
Explanation:
We use the Ratio Test with
a
n
=
(
−
6)
n
(2
n
)!
.
For then
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
(
−
6)
n
+1
(2
n
+ 2)!
(2
n
)!
(
−
6)
n
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
(
−
6)
n
+1
(2
n
+ 2)!
(2
n
)!
(
−
6)
n
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
−
6
(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
→ ∞
6
(2
n
+ 1)(2
n
+ 2)
= 0
<
1
.
Consequently, the
series is absolutely convergent
.
004
10.0points
Which one of the following properties does
the series
∞
summationdisplay
n
=1
Explanation:
With
a
n
=
n
(
−
8)
n
4
n
−
1
we see that
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
(
n
+ 1)8
n
+1
4
n
parenleftbigg
4
n
−
1
n
8
n
parenrightbigg
=
n
+ 1
n
parenleftbigg
8
4
parenrightbigg
.
Thus
lim
n
→ ∞
vextendsingle
vextendsingle
vextendsingle
vextendsingle
a
n
+1
a
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
8
4
,
so by the Ratio Test the given series is
divergent
.
005
10.0points
Which one of the following properties does
the series
∞
summationdisplay
n
=2
(
−
1)
n
(3
n
+ 2) ln
n
have?
1.
divergent
2.
absolutely convergent
3.
conditionally convergent
correct
(3
x
+ 2) ln
x
=
3
,
mei (dym252) – HW13 – schultz – (53990)
3
so by the Limit Comparison Test the series
∞
summationdisplay
n
=2
1
(3
n
+ 2) ln
n
and
∞
summationdisplay
n
=2
1
n
ln
n
either both converge or both diverge.
But
the second series diverges by the integral test
because
integraldisplay
∞
2
dx
x
ln
x
=
lim
t
→∞
[ln(ln
x
)]
t
2
= +
∞
.
We conclude that the original series is not
absolutely convergent.
However it is an al
ternating series and so it may still converge
conditionally. To check whether or not this is
the case first we have that
lim
n
→∞
f
(
n