Math 318 HW #3 Solutions
1. Exercise 6.5.2. Find suitable coeﬃcients (
a
n
) so that the resulting power series
∑
a
n
x
n
(a) converges absolutely for all
x
∈
[

1
,
1] and diverges oﬀ of this set;
Answer.
Consider the series
∞
X
n
=1
x
n
n
2
.
Then for all
x
∈
[

1
,
1],
±
±
x
n
n
2
±
±
≤
1
n
2
; since
∑
1
n
2
converges, the Weierstrass Mtest implies
that the above series converges absolutely on [

1
,
1]. On the other hand, for any
x
∈
R
such that

x

>
1, the sequence of terms (
x
n
/n
) diverges, so we see that the above series
diverges away from [

1
,
1].
(b) converges conditionally at
x
=

1 and diverges at
x
= 1;
Answer.
Consider the series
∞
X
n
=1
x
n
n
.
We’ve already seen in class that this series converges conditionally at
x
=

1 and
diverges at
x
= 1.
(c) converges conditionally at both
x
=

1 and
x
= 1.
Answer.
Consider the series
∞
X
n
=1
(

1)
n
x
2
n
n
.
(It may look like I’m cheating by ﬁddling with the exponent on
x
, but the above is really
just shorthand for the series
∑
a
n
x
n
where
a
n
= 0 when
n
is odd and
a
n
= (

1)
n/
2
/
(
n/
2)
when
n
is even.)
Notice that this series satisﬁes the hypotheses of the Alternating Series Test for any
x
∈
[

1
,
1], so it converges on this interval. Notice that the series
∑
±
±
±
(

1)
n
x
2
n
n
±
±
±
reduces
to the harmonic series when
x
=
±
1, so the series only converges conditionally at
x
=
±
1.
(d) Is it possible to ﬁnd an example of a power series that converges conditionally at
x
=

1
and converges absolutely at
x
= 1?
Answer.
No. If there were such a series
∑
a
n
x
n
, then, when
x
=

1, we would have
∞
X
n
=0

a
n
x
n

=
∞
X
n
=0

a
n

(

1)
n

=
∞
X
n
=0

a
n

=
∞
X
n
=0

a
n
(1)
n

,
which converges by hypothesis. But this contradicts the fact that
∑
a
n
x
n
only converges
conditionally at
x
=

1.
2.
(a) Exercise 2.7.12 (Summation by Parts). Let (
x
n
) and (
y
n
) be sequences, and let
s
n
=
x
1
+
x
2
+
···
+
x
n
. Use the observation that
x
j
=
s
j

s
j

1
to verify the formula
n
X
j
=
m
+1
x
j
y
j
=
s
n
y
n
+1

s
m
y
m
+1
+
n
X
j
=
m
+1
s
j
(
y
j

y
j
+1
)
.
1
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Using the fact that
x
j
=
s
j

s
j

1
, we have
n
X
j
=
m
+1
x
j
y
j
=
n
X
j
=
m
+1
(
s
j

s
j

1
)
y
j
=
s
m
+1
y
m
+1

s
m
y
m
+1
+
...
+
s
n
y
n

s
n

1
y
n
=
s
n
y
n

s
m
y
m
+1
+
n

1
X
j
=
m
+1
s
j
(
y
j

y
j
+1
)
.
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 Spring '08
 Staff
 Power Series, lim, sj

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