Homework 4 Solutions
Math 131A3
1. (11.2) We’ll do all five parts for each sequence in turn.
a
n
= (

1)
n
.
Two monotone subsequences are (
x
1
, x
3
, x
5
,
· · ·
) = (1
,
1
,
1
,
· · ·
), and
similarly for the even entries.
The set of subsequential limits is
{±
1
}
.
Therefore
lim sup
a
n
= 1 and lim inf
a
n
=

1. This sequence is bounded,

a
n

<
1 for all
n
, but
does not converge or diverge to
±∞
.
b
n
=
1
n
.
The sequence (
b
n
) is monotone decreasing, so any subsequence is as well;
furthermore (
b
n
) converges to 0, so any subsequence of (
b
n
) also converges to 0. There
fore the set of subsequential limits is
{
0
}
, and lim sup
b
n
= lim inf
b
n
= 0. Finally (
b
n
)
is bounded, e.g.

b
n

<
2 for all
n
.
u
n
= (

1
2
)
n
.
Any subequence (
x
n
i
) where all the
n
i
are even is monotone decreas
ing, and any subsequence (
x
n
i
) where all the
n
i
are odd is monotone increasing. The
sequence
u
n
converges to 0, so the set of subsequential limits is
{
0
}
, and lim sup
u
n
=
lim inf
u
n
= 0. Finally,
u
n
is bounded, e.g.

u
n

<
1 for all
n
.
x
n
= 5
(

1)
n
.
Possible monotone subsequences include (
x
1
, x
3
, x
5
,
· · ·
) = (
1
5
,
1
5
,
1
5
,
· · ·
)
and (
x
2
, x
4
, x
6
,
· · ·
) = (5
,
5
,
5
,
· · ·
). Note that (5
,
5
,
· · ·
,
5
,
1
5
,
1
5
,
· · ·
) is also monotone
decreasing, and (
1
5
,
1
5
,
· · ·
1
5
,
5
,
5
,
5
,
· · ·
) is monotone increasing. The set of subsequen
tial limits is
{
1
5
,
5
}
, and lim sup
x
n
= 5 while lim inf
x
n
=
1
5
. The sequence
x
n
neither
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 Fall '08
 hitrik
 Limits, Order theory, Limit of a sequence, lim sup, subsequence