(k) For any given sequence (
s
n
),
(
s
n
)
has a convergent subsequence
(
s
n
k
).
Sometimes true:
The sequence
s
n
=
⎧
⎨
⎩
1
,
n
odd
1
/n,
n
even
has a con-
vergent subsequence
s
2
n
→
0.
The sequence
s
n
=
n
has no convergent subsequences.
Similarly, the
sequence
s
n
=
n
2
has no convergent subsequences.
4

(l) For any bounded sequence (
s
n
),
lim sup
s
n
= sup
{
s
n
}
.
Sometimes true:
See the sequences in (j).
(m) It
α
is a subsequential limit of a bounded sequence
(
s
n
),
then
α
≤
lim sup
s
n
Always true:
By definition: lim sup
s
n
is the least upper bound of the
set of all subsequential limits.
(n) If every subsequence of a sequence
(
s
n
)
is convergent, then
(
s
n
)
itself
must be convergent.
Always true:
If every subsequence of
(
s
n
)
is convergent, then
(
s
n
)
must be convergent since
(
s
n
) is a subsequence of itself.
5

(o) If
(
s
n
)
is a divergent sequence, then some subsequence of
(
s
n
)
must
diverge.
Always true:
If every subsequence of
(
s
n
) is convergent, then (
s
n
) is
convergent as shown immediately above.
(p) If
(
s
n
)
is unbounded above, then
(
s
n
)
has an increasing subsequence
(
s
n
k
) which diverges to +
∞
.
Always true:
This is Theorem 3, Section 19.
6

2. Let
(
s
n
) be a positive sequence such that
lim
n
→∞
s
n
+1
s
n
=
L >
1
.
Prove that
s
n
→
+
∞
.
Hint:
lim
n
→∞
x
n
= +
∞
for any number
x
such that
x >
1.
Choose a number
c
such that
1
< c < L
. Let
ϵ
=
L
−
c
.
Since
lim
n
→∞
s
n
+1
s
n
=
L
There is a positive integer
N
such that
s
n
+1
s
n
−
L <
ϵ
for all
n > N
which implies
−
ϵ
<
s
n
+1
s
n
−
L <
ϵ
and
L
−
ϵ
<
s
n
+1
s
n
< L
+
ϵ
for all
n > N.

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- Fall '08
- Staff
- Mathematical analysis, Limit of a sequence, Limit superior and limit inferior, Sn, subsequence