77.
To use the alternating series test, consider
a
n
=
f
(
n
)
, where
f
(
x
) = arctan(
a/x
)
. We need to show that
f
(
x
)
is
decreasing. Since
f
0
(
x
) =
1
1 + (
a/x
)
2

a
x
2
,
we have
f
0
(
x
)
<
0
for
a >
0
, so
f
(
x
)
is decreasing for all
x
. Thus
a
n
+1
< a
n
for all
n
, and as
lim
n
→∞
arctan(
a/n
) = 0
for all
a
, by the alternating series test,
∞
X
n
=1
(

1)
n
arctan(
a/n
)
converges.
78.
The
n
th
partial sum of the series is given by
S
n
= 1

1
2
+
1
3
 · · ·
+
(

1)
n

1
n
,
so the absolute value of the first term omitted is
1
/
(
n
+ 1)
. By Theorem 9.9, we know that the value,
S
, of the sum differs
from
S
n
by less than
1
/
(
n
+1)
. Thus, we want to choose
n
large enough so that
1
/
(
n
+1)
≤
0
.
01
. Solving this inequality
for
n
yields
n
≥
99
, so we take
99
or more terms in our partial sum.
79.
The
n
th
partial sum of the series is given by
S
n
= 1

2
3
+
4
9
 · · ·
+ (

1)
(
n

1)
2
3
(
n

1)
,
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so the absolute value of the first term omitted is
(2
/
3)
n
. By Theorem 9.9, we know that the value,
S
, of the sum differs
from
S
n
by less than
(2
/
3)
n
. Thus, we want to choose
n
large enough so that
(2
/
3)
n
≤
0
.
01
. Solving this inequality
for
n
yields
n
≥
11
.
358
, so taking 12 or more terms in our partial sum is guaranteed to be within
0
.
01
of the sum of the
series.
Note: Since this is a geometric series, we know the exact sum to be
1
/
(1+2
/
3) = 0
.
6
.
The partial sum
S
12
is
0
.
595
,
which is indeed within
0
.
01
of the sum of the series. Note, however, that
S
11
= 0
.
6069
,
which is also within
0
.
01
of the
exact sum of the series. Theorem 9.9 gives us a value of
n
for which
S
n
is guaranteed to be within a small tolerance of
the sum of an alternating series, but not necessarily the
smallest
such value.
80.
The
n
th
partial sum of the series is given by
S
n
=
1
2

1
24
+
1
720
 · · ·
+
(

1)
n

1
(2
n
)!
,
so the absolute value of the first term omitted is
1
/
(2
n
+ 2)!
. By Theorem 9.9, we know that the value,
S
, of the sum
differs from
S
n
by less than
1
/
(2
n
+2)!
. Thus, we want to choose
n
large enough so that
1
/
(2
n
+2)!
≤
0
.
01
. Substituting
n
= 2
into the expression
1
/
(2
n
+ 2)!
yields
1
/
720
which is less than 0.01. We therefore take 2 or more terms in our
partial sum.
81.
Since
0
≤
c
n
≤
2

n
for all
n
, and since
∑
2

n
is a convergent geometric series,
∑
c
n
converges by the Comparison
Test. Similarly, since
2
n
≤
a
n
, and since
∑
2
n
is a divergent geometric series,
∑
a
n
diverges by the Comparison Test.
We do not have enough information to determine whether or not
∑
b
n
and
∑
d
n
converge.
82.
(a) The sum
∑
a
n
·
b
n
=
∑
1
/n
5
, which converges, as a
p
series with
p
= 5
, or by the integral test:
Z
∞
1
1
x
5
dx
= lim
b
→∞
x

4
(

4)
b
1
= lim
b
→∞
b

4
(

4)
+
1
4
=
1
4
.
Since this improper integral converges,
∑
a
n
·
b
n
also converges.
(b) This is an alternating series that satisfies the conditions of the alternating series test: the terms are decreasing and
have limit
0
,
so
∑
(

1)
n
/
√
n
converges.
(c) We have
a
n
b
n
= 1
/n
, so
∑
a
n
b
n
is the harmonic series, which diverges.