it implies
s
n
≤
t
m
for all
n, m
∈
N
. In particular, taking
m
= 3, we get
s
n
≤
b
3
=
256
81
<
4
.
Therefore lim
s
n
=
e
exists.
Appendix. Some problems discussed in the class
4.4.1. True or false:
(a) A sequence
(
s
n
)
converges to
s
iff every subsequence of
(
S
n
)
converges to
s
.
(b) Every bounded sequence is convergent.
(c) Let
(
s
n
)
be a bounded sequence. if
(
s
n
)
oscillates, then the set
S
of subse
quential limits of
s
n
contains at least two points.
(d) Let
(
s
n
)
be a bounded sequence and let
m
= lim sup
s
n
. Then for every
>
0
, there exists
N
∈
N
such that
n
≥
N
implies that
s
n
> m

.
(e) If
(
s
n
)
is unbounded above, then
(
s
n
)
contains a subsequence that has
+
∞
as a limit.
Solution:
(a) True. (=
⇒
) By Theorem 4.4.4. (
⇐
=) The (
s
n
) itself is also a subsequence.
(b) False. For example,
s
n
= (

1)
n
.
(c) True.
lim sup s
n
and lim
inf s
n
.
(d) False. For example,
s
n
= (

1)
n
,
m
= lim
sup s
n
= 1, take
=
1
2
.
(e) True. We can choose (
s
n
k
) such that
n
1
< n
2
< n
3
< ...
and
s
n
k
≥
k
for all
k
∈
N
.
4.4.5. Find the limits:
85
(a)
s
n
=
1 +
1
2
n
2
n
.
(b)
s
n
=
1 +
1
n
2
n
.
(c)
s
n
=
1 +
1
n
n

1
.
(d)
s
n
=
n
n
+1
n
.
(e)
s
n
=
1 +
1
2
n
n
.
(f)
s
n
=
n
+2
n
+1
n
+3
.
(g)
s
n
=
1

1
n
n
.
(h)
s
n
=
1

2
n
n
.
Solution:
(a)
s
n
=
1 +
1
2
n
2
n
→
e.
(b)
s
n
=
1 +
1
n
2
n
=
1 +
1
n
n
2
→
e
2
.
(c)
s
n
=
1 +
1
n
n

1
=
1 +
1
n
n
1 +
1
n
→
e
1
=
e.
(d)
s
n
=
n
n
+ 1
n
=
1
1 +
1
n
n
→
e.
(e)
s
n
=
1 +
1
2
n
n
=
s
1 +
1
2
n
2
n
→
√
e.
86
(f)
s
n
=
n
+ 2
n
+ 1
n
+3
=
1 +
1
n
+ 1
n
+1
·
n
+ 2
n
+ 1
2
→
e
·
1 =
e.
(g)
1

1
n
n
=
n

1
n
n
=
1
(
n
n

1
)
n
=
1
(1 +
1
n

1
)
n
=
1
(1 +
1
n

1
)
n

1
·
1
1 +
1
n

1
→
1
e
·
1 =
1
e
.
(h)
1

2
n
n
=
1

2
n
n
2
·
2
=
1

2
n
n
2
2
→
1
e
2
=
1
e
2
.
We used the result (g) in the last step.
4.4.8.
If
(
s
n
)
is a subsequence of
(
t
n
)
and
(
t
n
)
is a subsequence of
(
s
n
)
, can we
conclude that
(
s
n
) = (
t
n
)?
Prove or give a counterexample.
Proof:
Not true. For example, (
s
n
) = (0
,
1
,
0
,
1
,
0
,
1
, ...
) and (
t
n
) = (1
,
0
,
1
,
0
,
1
,
0
, ...
).
4.4.13. Le
(
s
n
)
and
(
t
n
)
be bounded sequences. Prove
lim
sup
(
s
n
+
t
n
)
≤
lim
sup
(
s
n
) + lim
sup
(
t
n
)
,
and find an example to show that the equality may not hold.
Proof:
By Corollary 4.4.12, there is a subsequence (
s
n
k
+
t
n
k
) of (
s
n
+
t
n
) such that
lim
k
(
s
n
k
+
t
n
k
) = lim
n
sup
(
s
n
+
t
n
)
.
Since (
s
n
k
) is bounded, there is a convergent subsequence (
s
n
km
). Since (
t
n
km
) is bounded,
there is a convergent subsequence, still denoted by the same notation. Then we use Theorem
4.2.1 to imply
lim
m
s
n
km
+ lim
m
t
n
km
= lim
m
(
s
n
km
+
t
n
km
)
.
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 Fall '08
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 Limit of a sequence, subsequence, Snk, lim snkm