92.305 Homework 4
Solutions
October 9, 2009
Exercise 2.4.1.
Complete the proof of Theorem 2.4.6 by showing that if the series
∑
∞
n
=0
2
n
b
2
n
diverges, then so does
∑
∞
n
=1
b
n
. Example 2.4.5 may be a useful reference.
Proof:
We will show that if
∑
∞
n
=0
2
n
b
2
n
diverges then
∑
∞
n
=1
b
n
does too by exploiting a
relationship between the partial sums
s
m
=
b
1
+
b
2
+
· · ·
+
b
m
and
t
k
=
b
1
+ 2
b
2
+
· · ·
+ 2
k
b
k
.
Because
∑
∞
n
=0
2
n
b
2
n
diverges, its monotone sequence of partial sums (
t
k
) must be unbounded.
To show that (
s
m
) is unbounded it is enough to show that for all
k
∈
N
, there is term
s
m
satisfying
s
m
≥
t
k
/
2. This argument is similar to the one for the forward direction, only to
get the inequality to go the other way we group the terms in
s
m
so that the last (and hence
smallest) term in each group is of the form
b
2
k
.
Given an arbitrary
k
, we focus our attention on
s
2
k
and observe that
s
2
k
=
b
1
+
b
2
+ (
b
3
+
b
4
) + (
b
5
+
b
6
+
b
7
+
b
8
) +
· · ·
+ (
b
2
k

1
+1
+
· · ·
+
b
2
k
)
≥
b
1
+
b
2
+ (
b
4
+
b
4
) + (
b
8
+
b
8
+
b
8
+
b
8
) +
· · ·
+ (
b
2
k
+
· · ·
+
b
2
k
)
=
b
1
+
b
2
+ 2
b
4
+ 4
b
8
+
· · ·
+ 2
k

1
b
2
k
=
1
2
(2
b
1
+ 2
b
2
+ 4
b
4
+ 8
b
8
+
· · ·
+ 2
k
b
2
k
)
=
b
1
/
2 +
t
k
/
2
.
Because (
t
k
) is unbounded, the sequence (
s
m
) must also be unbounded and cannot converge.
Therefore,
∑
∞
n
=1
b
n
diverges.
squaresolid
Exercise 2.4.2.
(a) Prove that the sequence defined by
x
1
= 3 and
x
n
+1
=
1
4

x
n
converges.
Proof:
We will show that this sequence is decreasing and bounded (which will also settle
the issue of welldefinedness: since
x
1
is 3 and subsequent terms are even smaller,
x
n
never
equals 4, so the denominator 4
−
x
n
never vanishes).
First let’s use induction to show that this sequence is decreasing. Observe that
x
1
= 3
>
1 =
x
2
. We need to prove that if
x
n
> x
n
+1
then
x
n
+1
> x
n
+2
. Note that
x
n
> x
n
+1
implies
that
−
x
n
<
−
x
n
+1
. Adding 4 to both sides of the inequality gives 4
−
x
n
<
4
−
x
n
+1
. It
follows that
1
4
−
x
n
>
1
4
−
x
n
+1
,
1
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which is precisely what we need to conclude
x
n
+1
> x
n
+2
.
Thus by induction, (
x
n
) is
decreasing.
The argument above shows that (
x
n
) is bounded above by 3, so now we’ll use this fact in
proving by induction that (
x
n
) is bounded below, and more specifically, that
x
n
>
0 for all
n
.
Clearly
x
1
>
0. Now assume
x
n
>
0. Because (
x
n
) is decreasing, we know that
x
n
≤
x
1
= 3,
which implies that
x
n
+1
=
1
4

x
n
is positive. By induction, (
x
n
) is bounded below by 0 for all
positive integers
N
.
Therefore this sequence converges by the Monotone Convergence Theorem.
squaresolid
(b) Now that we know lim
x
n
exists, explain why lim
x
n
+1
must also exist and equal the same
value.
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 Spring '10
 MetCalfe
 Mathematical analysis, Limit of a sequence, Xn

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