This preview shows pages 1–2. Sign up to view the full content.
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 RAMAKRISHNA
 Math

Click to edit the document details