K for any given sequence s n s n has a convergent

(k) For any given sequence ( s
n
), ( s
n
) has a convergent subsequence ( s
n
k
).

(l) For any bounded sequence ( s
n
), lim sup s
n
= sup { s
n
} .
1 1 1
α is a subsequential limit of a bounded sequence ( s
n
), then α ≤
lim sup s
n
.
True: Assume α ∈ S where S is the set of subsequential limits of
( s
n
). Then, α ≤ sup S . By definition, lim sup s
n
= sup S . Therefore,
α ≤ lim sup s
n
.
(n) If every subsequence of a sequence ( s
n
) is convergent, then ( s
n
) itself
must be convergent.

(o) If ( s
n
) is a divergent sequence, then some subsequence of ( s
n
) must
diverge.
True: Assume ( s
n
) is a divergent sequence. Assume for the sake of
contradiction that all subsequences of ( s
n
) converge. Then ( s
n
) itself con-
verges (since ( s
n
) is identical to the subsequence ( s
n
k
) where n
k
= k ).
This is a contradiction, since we assumed ( s
n
) is a divergent sequence.
Therefore, some subsequence of ( s
n
) must diverge.
(p) If ( s
n
) is unbounded above, then ( s
n
) has a nondecreasing subsequence
( s
n
k
) which diverges to + ∞ .

2. In each case, prove that the given sequence has an infinite limit.
(a) s
n
=
4 n
2
+ 7
n - 3
, where n ∈ N . Prove s
n
→ + ∞ .