MAT 137Y, 2009–2010 Winter Session, Solutions to Supplementary Problem Set #3
1. (SHE 12.5)
4. Given
a
k
= (

1
)
k
1
k
ln
k
, we know that
∑

a
k

=
∑
1
k
ln
k
diverges by the integral test (see 11.2 #21), but
∑
a
k
converges by the alternating series test, so the series converges conditionally and not absolutely.
6. As lim
k
→
∞
k
ln
k
H
=
∞
, the alternating series test does not apply and the series diverges.
12. The values of sin
k
π
4
alternate between 0
,
±
1
/
√
2, and
±
1. The limit as
k
→
∞
clearly does not exist, so
again the series diverges by the basic divergence test.
16. Since lim
k
→
∞
1
p
k
(
k
+
1
)
=
0 and the series alternate in sign, the series converges by the alternating series test.
Using the limit comparison test with
b
k
=
1
k
, the series
∑
(

1
)
k
√
k
(
k
+
1
)
=
∑
1
√
k
(
k
+
1
)
diverges, so the series in
question conditionally converges (and not absolutely).
28. Since every absolutely convergent series converges, it is sufficient to show that the series absolutely con
verges. Note that

a
k

=
sin
(
k
π
/
2
)
k
√
k
<
1
k
3
/
2
, so by the comparison test, the series converges absolutely.
32. With any alternating series, the error estimate of the actual series and the partial sum
s
n
is approximately
a
n
+
1
(Equation 11.4.5). Therefore the error estimate between
∑
(

1
)
k
+
1 1
k
and
s
20
is
a
21
=
1
21
.
42. If the sequence of terms
{
a
n
}
is nonincreasing instead of decreasing, the alternating series still converges.
To see this, we can make slight changes of the proof of Theorem 11.4.4.
The even partial sums
s
2
m
are now nonnegative.
Since
s
2
m
+
2
≤
s
2
m
, the sequence of even terms converges, so
s
2
m
→
L
.
Since
s
2
m
+
1
=
s
2
m

a
2
m
+
1
and
a
2
m
+
1
→
0, we have
s
2
m
+
1
→
L
. Hence
s
n
→
L
.
2. (SHE 12.8)
6. Applying the ratio test, lim
k
→
∞
2
k
+
1
x
k
+
1
k
2
k
2
2
k
x
k
=
lim
k
→
∞
2
x
k
k
+
1
2
=
2

x

.
Hence the series converges when 2

x

<
1, or

x

<
1
2
and diverges when

x

>
1
2
. (Therefore the radius of
convergence is
1
2
.) To find the interval of convergence, we check the points

x

=
1
2
. If
x
=
1
2
, then we have
∑
2
k
k
2
·
1
2
k
=
∑
1
k
2
, which is a convergent
p
series. if
x
=

1
2
, then we have
∑
(

1
)
k
k
2
which converges by the
alternating series test. Therefore the interval of convergence is
[

1
2
,
1
2
]
.
8. Applying the ratio test, lim
k
→
∞
(

1
)
k
+
1
x
k
+
1
√
k
+
1
·
√
k
(

1
)
k
x
k
=
lim
k
→
∞
r
k
k
+
1

x

=

x

.
Hence the series converges when

x

<
1 and diverges when

x

>
1. If
x
=
1, then we have
∑
(

1
)
k
√
k
which
converges by the alternating series test. If
x
=

1, then we have
∑
(

1
)
k
·
(

1
)
k
√
k
=
∑
1
√
k
, which diverges by
the
p
series test. Therefore the interval of convergence is
(

1
,
1
]
.