2005 MA 1A HOMEWORK 8 SOLUTIONS
(1) Suppose
a
n
> a
n
+1
≥
0 for all
n
. Define
f
(
x
) =
∞
X
n
=0
(

1)
n
a
n
x
n
.
(a) Prove that this series is absolutely convergent for

x

<
1.
Proof.
∞
X
n
=0

(

1)
n
a
n
x
n

=
∞
X
n
=0
a
n

x
n

=
∞
X
n
=0
a
n

x

n
.
Since
a
n
≤
a
0
for all
n
, we get that
a
n

x

n
≤
a
0

x

n
for all
n
.
Therefore we may apply
the comparison test, comparing
∑
∞
n
=0
a
n

x

n
to
∑
∞
n
=0
a
0

x

n
.
But the latter series is just a
geometric series, with ratio

x

. Since

x

<
1, this series is convergent. Thus by the comparison
test, we get that
∑
∞
n
=0
a
n

x

n
is convergent, so
∞
X
n
=0
(

1)
n
a
n
x
n
is absolutely convergent.
(b) Suppose from now on that
a
n
→
0 as
n
→ ∞
. This means that
∑
∞
n
=0
(

1)
n
a
n
is convergent.
Define
L
=
∑
∞
n
=0
(

1)
n
a
n
. Let
f
k
(
x
) be the partial sum used to define
f
(
x
) where
n
runs from
0 to
k
. Prove that, for 0
< x <
1,
f
2
m
(
x
)
≥
f
(
x
)
≥
f
2
m
+1
.
Proof.
Since the series is absolutely convergent, we get
f
(
x
)
=
∞
X
n
=0
(

1)
n
a
n
x
n
=
2
m
X
n
=0
(

1)
n
a
n
x
n
+
∞
X
n
=2
m
+1
(

1)
n
a
n
x
n
=
2
m
X
n
=0
(

1)
n
a
n
x
n
+
∞
X
j
=
m
+1
(

a
2
j

1
x
2
j

1
+
a
2
j
x
2
j
)
=
2
m
X
n
=0
(

1)
n
a
n
x
n

∞
X
j
=
m
+1
(
a
2
j

1
x
2
j

1

a
2
j
x
2
j
)
≤
2
m
X
n
=0
(

1)
n
a
n
x
n
=
f
2
m
(
x
),
because
a
2
j

1
x
2
j

1
≥
a
2
j
x
2
j
for all
j
.
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 Fall '08
 Borodin,A
 2J, π, 1 J, 0 2m

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