4. Let (
s
n
) be bounded. Then there exists a subsequence (
s
n
k
) such that
s
n
k
→
lim
sup s
n
,
and there exists a subsequence (
s
n
l
) such that
s
n
l
→
lim
inf s
n
.
This can be done by the definitions of
sup
(
S
) and
inf
(
S
) (see Theorem 4.4.11).
82
5. If (
s
n
) is unbounded above, we define lim
sup
(
s
n
) = +
∞
. If (
s
n
) is unbounded below,
we define lim
inf
(
s
n
) =
∞
.
6. Let
s
n
=
1+(

1)
n
+1
2
and (
s
n
) = (1
,
0
,
1
,
0
,
1
,
0
, ...
).
Then 0 and 1 are subsequencial
limits, lim sup
s
n
= 1 and lim inf
s
n
= 0.
7. There are many important applications on lim sup
s
n
or lim inf
s
n
. For examples,
the formula for the radius of convergence for a power series
∑
∞
h
=0
a
n
x
n
is given by:
r
=
1
lim sup
n
p

a
n

.
8. Let
s
n
= (

1)
n n
+1
n
and (
s
n
) = (

1
2
,
3
2
,

4
3
,
5
4
,

6
5
, ...
). Then

1 and 1 are sunsequencial
limits, lim sup
s
n
= 1, and lim inf
s
n
=

1.
9. Let
s
n
=

2 +
1
n
,
n
= 3
k

2
,
1
n
,
n
= 3
k

1
,
1 +
1
n
,
n
= 3
k.
Then

2
,
0 and 1 are subsequenctial limits, lim sup
s
n
= 1, and lim inf
s
n
=

2.
10. Let
s
n
=
cos
nπ
3
, and
(
s
n
) =
1
2
,

1
2
,

1
,

1
2
,
1
2
,
1
, ...
.
Then
1
,
1
2
,

1
2
,

1
are subsequenctial limits, lim sup
s
n
= 1, and lim inf
s
n
=

1.
11. Let
s
n
=
n sin
2
nπ
2
. Then lim sup
s
n
=
∞
, and lim inf
s
n
= 0.
12. Let
s
n
=
n
2
cos nπ
= (

1)
n
n
2
. Then lim sup
s
n
=
∞
, and lim inf
s
n
=
∞
.
13. Let
s
n
=

2 +
1
n
,
n
= 3
k

2
,
1
n
,
n
= 3
k

1
,
1 +
1
n
,
n
= 3
k, n
≥
1
.
Then lim sup
s
n
= 1, sup(
s
n
) =
4
3
, lim
inf s
n
=

2 and inf(
s
n
) =

2.
83
14. Let
s
n
=
sin
nπ
6
.
Then lim sup
s
n
= 1, sup(
s
n
) = 1, lim inf
s
n
=

1, and
inf(
s
n
) =

1.
15. Let
s
n
=
sin
nπ
6
+
1
n
. Then lim sup
s
n
= 1, sup(
s
n
) =
3
2
, lim inf
s
n
=

1, and
inf(
s
n
) =

1.
Theorem 0.5
The limit
lim(1 +
1
n
)
n
exists.
Proof:
We want to show
s
n
=
1 +
1
n
n
is increasing sequence. In fact, by the inequality
of arithmetic and geometric means: Let
a
1
, ..., a
n
be positive real numbers. Then
a
1
+
a
2
+
...
+
a
n
n
≥
n
√
a
1
a
2
...a
n
,
i.e.,
a
1
+
a
2
+
...
+
a
n
n
n
≥
a
1
a
2
...a
n
.
In particular, if
a
1
=
a
2
=
...
=
a
n
=
x
and
a
n
+1
=
y
, we have
nx
+
y
n
+ 1
n
+1
≥
x
n
y.
(2)
Let
x
= 1 +
1
n
and
y
= 1. Then from (2), we find
n
(1 +
1
n
) + 1
n
+ 1
n
+1
≥
1 +
1
n
n
.
i.e.,
s
n
+1
=
1 +
1
n
+ 1
n
+1
≥
1 +
1
n
n
=
s
n
.
Also we let
t
n
=
1 +
1
n
n
+1
. We claim that (
t
n
) is decreasing. In fact, taking
x
= 1

1
n
and
y
= 1 in (2), we get
n
(1

1
n
) + 1
n
+ 1
n
+1
≥
1

1
n
n
,
84
i.e.,
n
n
+ 1
n
+1
≥
n

1
n
n
,
which implies
n
n

1
n
≥
n
+ 1
n
n
+1
,
i.e.,
1 +
1
n

1
)
n
≥
1 +
1
n
n
+1
,
i.e.,
t
n

1
≥
t
n
for all
n
. We also notice
s
n
≤
t
n
so that
s
1
≤
s
2
≤
s
3
≤
...
≤
t
3
≤
t
2
≤
t
1
.
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 Fall '08
 Staff
 Limit of a sequence, subsequence, Snk, lim snkm