Homework 4 Solutions
1. For each of the following sequences (
s
n
), find the set
S
of subsequential limits,
and use this to find lim inf
s
n
and lim sup
s
n
.
(a) (11.1)
s
n
= 3 + 2(

1)
n
.
Solution:
(
s
n
) = (5
,
1
,
5
,
1
, . . .
), so
S
=
{
1
,
5
}
, and therefore lim sup
s
n
=
5 and lim inf
s
n
= 1.
(b)
s
n
=
n
2
(

1 + (

1)
n
).
Solution:
(
s
n
) = (0
,

2
,
0
,

18
,
0
,

50
, . . .
), so
S
=
{∞
,
0
}
, and there
fore lim sup
s
n
= 0 and lim inf
s
n
=
∞
.
(c) (
s
n
) = (0
,
1
,
2
,
0
,
1
,
3
,
0
,
1
,
4
, . . .
).
Solution:
S
=
{
0
,
1
,
∞}
, so lim sup
s
n
=
∞
and lim inf
s
n
= 0.
2. Prove that every unbounded sequence contains a monotone subsequence that
diverges to either
∞
or
∞
.
Solution:
If (
s
n
) is unbounded above, for each choice of
K
, there exists a
sequence element
s
n
k
so that
s
n
k
> K
. The subsequence (
s
n
k
) clearly diverges
to
∞
. Proof is similar if (
s
n
) is unbounded below.
3. Suppose
x >
1. Prove that lim
x
1
/n
= 1.
Solution:
First of all,
x
1
/n
is bounded below by 1 since
x >
1.
Note the
quantity
x
1
n
(
n
+1)
>
1, so
x
1
n
+1
< x
1
n
(
n
+1)
x
1
n
+1
=
x
1
n
,
which means the sequence is decreasing.
Since it is decreasing and bounded
below by 1, a limit exists for the sequence, call it
s
. Note any subsequence must
converge to
s
.
However, the subsequence
x
1
/
2
n
converges to
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 Winter '11
 Wiley
 Differential Equations, Equations, Limits

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