(
A
) The Ratio Test is the better one to use
because
a
n
+1
a
n
= 3
n
!
(
n
+ 1)!
(
n
+ 1)
n
+1
n
n
.
Now
n
!
(
n
+ 1)!
=
1
n
+ 1
,
while
(
n
+ 1)
n
+1
n
n
= (
n
+ 1)
n
+ 1
n
n
.
Thus
a
n
+1
a
n
= 3
n
+ 1
n
n
→
3
e >
1
as
n
→ ∞
, so series (
A
) diverges.
(
B
) The Root Test is the better one to
apply because

a
n

1
/n
=
6
n
+ 2
→
0
as
n
→ ∞
, so series (
B
) converges.
(
C
) Again the Root Test is the better one
to apply because of the
n
th
powers. For then

a
n

1
/n
=
6
5
3
n
2
n
+ 4
→
9
5
>
1
as
n
→ ∞
, so series (C) diverges.
Consequently, of the given infinite series,
only
A
and
C
diverge.
019
10.0 points
Decide which of the following series diverge.
(
A
)
∞
n
= 1
n
7
+ 8
n
+ 2
2
7
n
(
B
)
∞
n
= 1
7
n
+ 8
n
2
+ 7
n
(
C
)
∞
n
= 1
√
n

7
√
n
+ 8
8
7
n
1.
B
only
2.
C
only
correct
3.
all of them
4.
A
only
5.
A
and
B
6.
A
and
C
7.
B
and
C
8.
none of them
Explanation:
We compute one of
lim
n
→ ∞
a
n
+1
a
n
,
lim
n
→ ∞
(
a
n
)
1
/n
for each of the given series.
(
A
) The ratio test is the better one to use:
a
n
+1
a
n
=
2
7
(
n
+ 1)
7
+ 8
n
7
+ 8
n
+ 2
(
n
+ 1) + 2
.
But
(
n
+ 1)
7
+ 8
n
7
+ 8
→
1
as
n
→ ∞
, while
n
+ 2
(
n
+ 1) + 2
→
1
.
Thus
lim
n
→ ∞
a
n
+1
a
n
=
2
7
,
so series (
A
) converges.
rabinovich (ahr454) – HW12 – um – (55915)
11
(
B
) The root test is the better one to use:
(
a
n
)
1
/n
=
7
n
+ 8
n
2
+ 7
→
0
as
n,
→ ∞
, so series (
B
) converges.
(
C
) Again the ratio test is the better one to
use:
a
n
+1
a
n
=
8
7
√
n
+ 1

7
√
n

7
√
n
+ 8
√
n
+ 1 + 8
.
But
√
n
+ 1

7
√
n

7
,
√
n
+ 8
√
n
+ 1 + 8
→
1
as
n
→ ∞
. Thus series (
C
) diverges.
Consequently, of the given infinite series,
only
C
diverges
.
020
10.0 points
Which of the following infinite series con
verges conditionally?
1.
∞
n
= 1

7
6
n
2.
∞
n
= 1
7(

n
)
n
(
n
+ 3)
n
3.
∞
m
= 1
8
7
m

6
4.
∞
n
= 1
cos(
n
π
)
3
n
8
n
+ 6
n
5.
∞
k
= 1
(

1)
k
6
8
k
ln(
k
) + 7
correct
Explanation:
The question requires us to find which of
the five series has the property that
n
a
n
converges but
n

a
n

does not converge.
(i) In the case
a
k
= (

1)
k
6
8
k
ln(
k
) + 7
the alternating series test applies since
6
8
k
ln(
k
) + 7
≥
6
8(
k
+ 1) ln(
k
+ 1) + 7
and
lim
k
→ ∞
6
8
k
ln(
k
) + 7
= 0
.
Thus
∞
k
= 1
(

1)
k
6
8
k
ln(
k
) + 7
converges.
On the other hand, if we set
b
k
=
1
k
ln(
k
)
,
then
lim
k
→ ∞
b
k

a
k

=
8
k
ln(
k
) + 7
6
k
ln(
k
)
=
4
3
= 0
,
while by the integral test,
∞
k
= 2
b
k
=
∞
k
= 2
1
k
ln(
k
)
diverges.
The Limit Comparison test thus
says that
∞
k
= 1

a
k

diverges. Consequently, the series
∞
k
= 1
(

1)
k
6
8
k
ln(
k
) + 7
converges conditionally but not absolutely.
(ii) In the case
a
n
=
7(

n
)
n
(
n
+ 3)
n
rabinovich (ahr454) – HW12 – um – (55915)
12
we see that
lim
n
→ ∞

a
n

= 7
e

3
= 0 ;
but then
a
n
oscillates between

7
e

3
and
7
e

3
as
n
→ ∞
, so cannot converge to 0.
Consequently, neither of
∞
n
= 1
a
n
,
∞
n
= 1

a
n

converges.
(iii) Since

a
m

=
a
m
in the case
a
m
=
8
7
m

6
,
both series converge or both do not converge.
But by the Comparison and
p
series tests with
p
= 1, the series
∞
m
= 1
8
7
m

6
does not converge, so neither of
∞
m
= 1
a
m
,
∞
m
= 1

a
m

converges.
(iv) In the case
a
n
= cos(
n
π
)
3
n
8
n
+ 6
n
= (

1)
n
3
n
8
n
+ 6
n
we apply the root test to the series
∑
n

a
n

.
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 Fall '11
 Gramlich
 Accounting, Mathematical Series, lim