(
B
) If an infinite series
n
a
n
converges,
then
lim
n
→ ∞
a
n
= 0
.
But for the given series in (
B
),
lim
n
→ ∞
a
n
=
lim
n
→ ∞
√
n

8
√
n
+ 4
= 1
.
Consequently, the series in (
B
) does not con
verge.
(
C
) After division,
8
n
+ 4
7
n

6
=
8 +
4
n
7

6
n
,
so the inequality
8
n
+ 4
7
n

6
n
≥
8
7
n
holds for all
n
. But the series
∞
n
= 1
8
7
n
is a geometric series whose common ratio
r
=
8
7
.
Now
r >
1, so this geometric series does
not converge. Hence by the comparison test
the series in (
C
) does not converge.
Consequently, of the given infinite series,
only
A
converges.
015
10.0 points
Which of the following series
(
A
)
∞
n
= 1
(5 sin
n
)
2
6
n
2
+ 2
(
B
)
∞
n
= 2
5
n
8
n
2
ln
n
+ 2
(
C
)
∞
n
= 1
5 sin
n
π
6
n
+ 8
diverge(s)?
1.
B
only
correct
2.
A
only
3.
B
and
C
4.
A, B,
and
C
5.
A
and
C
Explanation:
(
A
) Since (sin
θ
)
2
≤
1, the inequality
0
≤
(5 sin
n
)
2
6
n
2
+ 2
<
25
1
n
2
cadena (jc59484) – HW11 – lawn – (55930)
11
holds for all
n
. But by the Integral test, the
infinite series
∞
n
= 1
1
n
2
converges. Consequently, the comparison test
ensures that the series
∞
n
= 1
(5 sin
n
)
2
6
n
2
+ 2
converges.
(
B
) After division,
5
n
8
n
2
ln
n
+ 2
=
5
8
n
ln
n
+
2
n
.
But
n
ln
n
5
8
n
ln
n
+
2
n
→
5
8
>
0
,
so
n
ln
n
5
n
8
n
2
ln
n
+ 2
→
5
8
>
0
.
On the other hand, by the Integral test the
infinite series
∞
n
= 2
1
n
ln
n
does not converge, hence by the limit compari
son test, the given series in (
B
) does converge.
(
C
) As the Integral test shows, the series
∞
n
= 1
5
6
n
+ 8
diverges.
However, sin
n
π
=
0 for all
n
.
Hence every term in the given series in (
C
)
reduces to 0,
i.e.
, the series becomes
0 + 0 + 0 +
. . .
+ 0 +
. . . ,
which obviously converges (and has sum 0).
Hence the given series in (
C
) actually con
verges.
Consequently, of the given infinite series,
only
B
diverges.
016
10.0 points
Find all values of
p
for which the infinite
series
∞
n
= 1
7
n
5
8
n
8
+ 1
p
converges.
1.
p <

3
4
2.
p >
3
3.
p >
3
4
4.
p >
1
3
correct
5.
p <

1
3
Explanation:
After division,
7
n
5
8
n
8
+ 1
=
7
8
n
3
+
1
n
5
,
so
n
3
7
n
5
8
n
8
+ 1
→
7
8
>
0
as
n
→ ∞
. Thus by the Limit Comparison
test, the infinite series
∞
n
= 1
7
n
5
8
n
8
+ 1
p
will converge if and only if the infinite series
∞
n
= 1
1
n
3
p
converges. But by the Integral test we know
that the series
∞
n
= 1
1
n
p
cadena (jc59484) – HW11 – lawn – (55930)
12
converges if and only if
p >
1. Consequently,
the given series will converge if and only if
3
p >
1,
i.e.
, when
p >
1
3
.
017
10.0 points
Determine all values of
p
for which the series
∞
n
= 4
3
n
p
ln
n
converges.
1.
p >
2
2.
p <
0
3.
p >
0
4.
p
≥
1
5.
p >
1
correct
6.
p <

2
Explanation:
By the Divergence Test, the series
∞
n
= 4
3
n
p
ln
n
diverges when
p <
0 because then
lim
n
→ ∞
1
n
p
ln
n
=
∞
.
Now suppose
p
≥
0. If 0
≤
p
≤
1, then
3
n
ln
n
≤
3
n
p
ln
n
,
so by the Comparison Test the given series
diverges because, by the Integral Test with
f
(
x
) =
3
x
ln
x
,
4
≤
x <
∞
,
the series
∞
n
= 4
3
n
ln
n
diverges.
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 Fall '11
 Gramlich
 Accounting, Calculus, Mathematical Series, lim