Modern Analysis, Homework 4
Possible solutions
1.
[12] (5 points) Prove that every sequence of reals has a monotone (ie either decreas
ing or increasing, though not necessarily
strictly
so) subsequence.
Solution:
Let
{
x
n
}
be a sequence of reals.
Suppose that lim sup
n
→∞
x
n
= +
∞
Then
choose
n
1
∈
N
so that
x
n
1
>
1.
Having chosen
n
1
,
· · ·
, n
i
∈
N
, choose
n
i
< n
i
+1
∈
N
so that
x
n
i
+1
> x
n
i
.
Then
{
x
n
i
}
forms a monotone (increasing) subsequence.
If instead
lim sup
n
→∞
x
n
=
∞
(for example
x
n
=

n
), then a similar argument shows that
x
n
has
a monotone decreasing subsequence. Now suppose that
s
*
= lim sup
n
→∞
x
n
is finite. There
are two cases to consider:
(1) The set
S
>
=
{
n
∈
N

x
n
≥
s
*
}
is infinite
(2) The set
S
<
=
{
n
∈
N

x
n
≤
s
*
}
is infinite
(notice that these cases are
not
mutually exclusive, but
at least
one of them holds). First
suppose that case 1 holds.
Choose some
n
1
∈
S
>
(this gives us our
x
n
1
).
Having chosen
n
1
,
· · ·
, n
i
∈
S
>
, choose
n
i
+1
∈
S
>
such that
n
i
+1
> n
i
and
x
n
i
+1
≤
x
n
i
(possible since we’re
assuming that
S
>
is
infinite
). In this way, we construct a monotone decreasing subsequence
{
x
n
i
}
of
{
x
n
}
.
In a similar way, if case 2 holds, we can construct a monotone
increasing
subsequence.
2.
[15] (10 points) Rudin, Ch 3, problem 16, parts (a) and (b).
Fix a positive number
α
. Choose
x
1
>
√
α
and define
x
2
, x
3
, x
4
,
· · ·
, by the recursion formula
x
n
+1
=
1
2
x
n
+
α
x
n
.
(a) Prove that
{
x
n
}
decreases monotonically and that lim
x
n
=
√
α.
Solution:
As a first step, I claim that with the above definition, we have
x
n
>
√
α
for all
n
∈
N
. Let’s prove this by induction: the statement for
n
= 1 is true by construction. Now
suppose that
x
1
,
· · ·
, x
k
>
√
α
for some
k
∈
N
. Then
x
k
+1

√
α
=
x
k
2
+
α
2
x
k

√
α
=
1
x
k
x
2
k
2
+
α
2

x
k
√
α
=
1
2
x
k
(
x
2
k

2
√
αx
k
+
α
)
=
1
2
x
k
(
x
k

√
α
)
2
>
0
So by induction, we have
x
n
>
√
α
for all
n
∈
N
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
Now with this in place, we can show that
{
x
n
}
is decreasing:
x
n

x
n
+1
=
x
n

x
n
2
+
α
2
x
n
=
x
n
2

α
2
x
n
=
x
2
n

α
2
x
n
>
0 (since
x
n
>
√
α
)
Since
{
x
n
}
is a decreasing sequence which is bounded from below, it follows that it con
verges to some limit,
L
.
By the defining equation
x
n
+1
=
1
2
x
n
+
α
x
n
,
it follows that
L
=
1
2
(
L
+
α
L
)
. Solving for
L
gives
L
2
=
α
. Now since
x
n
>
√
α >
0 for all
n
∈
N
, it follows
that
L
too must be positive. Hence
L
=
√
α
.
(b) Put
ε
n
=
x
n

√
α
, and show that
ε
n
+1
=
ε
2
n
2
x
n
<
ε
2
n
2
√
α
so that, setting
β
= 2
√
α
,
ε
n
+1
< β
ε
1
β
2
n
for all
n
∈
N
.
Solution:
In part (a), we already computed
ε
n
+1
=
x
n
+1

√
α
=
(
x
n

√
α
)
2
2
x
n
, so
ε
n
+1
=
ε
2
n
2
x
n
.
Since we know that
x
n
>
√
α
for all
n
∈
N
, it follows that
ε
n
+1
<
ε
2
n
2
√
α
= (2
√
α
)
ε
n
2
√
α
2
.
Setting
β
= 2
√
α
and applying the previous formula recursively in
n
gives
ε
n
+1
< β
ε
1
β
2
n
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Peters
 lim, en, Metric space, Compact space, Xn, ΣN

Click to edit the document details