MAT 137Y 200809 Winter Session, Solutions to Problem Set 12
1
(SHE 12.2)
4. Using partial fractions, we have
∞
∑
k
=
0
1
(
k
+
1
)(
k
+
3
)
=
1
2
∞
∑
k
=
0
1
k
+
1

1
k
+
3
.
This gives us a telescoping sum; if we look at the partial sums, we have
s
n
=
1
2
1

1
3
+
1
2

1
4
+
1
3

1
5
+
···
+
1
n

1
n
+
2
+
1
n
+
1

1
n
+
3
=
1
2
1
+
1
2

1
n
+
2

1
n
+
3
Therefore,
∞
∑
k
=
0
1
(
k
+
1
)(
k
+
3
)
=
lim
n
→
∞
s
n
=
lim
n
→
∞
1
2
1
+
1
2

1
n
+
2

1
n
+
3
=
3
4
.
6.
∞
∑
k
=
0
(

1
)
k
5
k
=
∞
∑
k
=
0

1
5
k
=
1
1
+
1
5
=
5
6
.
10.
∞
∑
k
=
2
3
k

1
4
3
k
+
1
=
∞
∑
k
=
0
3
k
+
1
4
3
k
+
7
=
3
4
7
∞
∑
k
=
0
3
4
3
k
=
3
4
7
1
1

3
4
3
=
3
15616
.
20. Since lim
n
→
∞
1
4

5
4
n
is not zero, it follows by the basic divergence test that the series diverges.
28. The total length removed in the Cantor set is
1
3
+
2
9
+
4
27
+
···
=
1
3
∑
∞
k
=
0
(
2
3
)
k
=
1
3
·
1
1

2
3
=
1. Some
points that are in the Cantor Set include 0
,
1
,
1
3
,
2
3
,
1
9
,
2
9
,
7
9
,
8
9
. In fact, for all positive integers
n
, the
point 1
/
3
n
is a point on the Cantor set. This leads to the curious paradox that despite the total
length removed in the set is 1, there are infinitely many points on the Cantor set!
32.
(a) Since
∑
a
k
converges, it follows that lim
k
→
∞
a
k
=
0. Therefore, lim
k
→
∞
1
a
k
6
=
0, since the limit goes
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 Spring '08
 UPPAL
 Fractions, Mathematical Series, lim, k→∞

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