Math 3110
Spring ’09
Homework 5:
Selected Solutions
[7.21]:
Claim
:
∞
X
0
1
(2
n
+ 1)
2
=
π
2
8
Proof
We know that
∞
X
0
1
n
2
=
π
2
6
, and
∞
X
0
1
(2
n
+ 1)
2
=
∞
X
0
1
n
2

∞
X
0
1
(2
n
)
2
. So
∞
X
0
1
(2
n
+ 1)
2
=
π
2
6

1
4
∞
X
0
1
n
2
=
π
2
6

1
4
π
2
6
=
π
2
8
[7.21]:
Claim
: If
∞
X
0
a
n
is absolutely convergent and
{
b
n
}
is bounded, then
∞
X
0
a
n
b
n
is convergent.
Proof
Since
{
b
n
}
is bounded, there is a
B
such that

b
n
 ≤
B
for all
n
. So
∞
X
0

a
n
b
n
 ≤
B
∞
X
0

a
n

,
which is ﬁnite since
∞
X
0
a
n
is absolutely convergent. Thus
∞
X
0
a
n
b
n
also converges.
[7.21]:
Claim
: Let
{
a
n
}
be a sequence and
{
a
n
i
}
any subsequence. (a) If
∞
X
0
a
n
is absolutely
convergent, then
∞
X
0
a
n
i
is absolutely convergent. (b) It is not the case that
∞
X
0
a
n
convergent
implies that
∞
X
0
a
n
i
is convergent.
Proof
(a) Since
{
a
n
i
}
is a subsequence and
∞
X
0
a
n
is absolutely convergent, for some
L <
∞
we
clearly have
k
X
i
=0

a
n
i
 ≤
n
k
X
n
=0

a
n
 ≤
∞
X
0

a
n
 ≤
L
for all
k
. So
∞
X
0
a
n
i
is absolutely convergent as
claimed. (b) On the other hand, the series
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 Fall '08
 RAMAKRISHNA
 Math

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