2.9 *Upper and Lower Limits 135
Theorem 2.9.9 A bounded sequence { Xn} converges if and only if lim Xn and
n->oo
lim Xn are both real numbers and are equal. In fact,
n->oo
lim Xn = L <=> lim Xn = lim Xn = L.
n--+oo n--+oo n--+oo
Proof. Exercise 4. •
Theorem 2.9.10 Let {xn} be a bounded sequence. Then lim Xn and lim Xn
n->oo n->oo
are cluster points of {xn}; moreover, they are the minimum and maximum
cluster points of { Xn}, respectively.
Proof. Suppose {xn} is a bounded sequence. Let L = lim Xn and U =
n->oo
lim Xn· Then, Ve: > 0, Theorems 2.9.7 and 2.9.8 guarantee that the intervals
n->oo
( L - c:, L + c:) and ( U - c:, U + c:) contain Xn for infinitely many n. Thus, L and
U are cluster points of {xn} by Definition 2.6.14.
Now, let W be a cluster point of {xn}· Then 3 subsequence {xnk} of {xn}
such that Xnk ---+ W. By Theorem 2.9.9,
lim Xnk = lim Xnk.
k->oo k->oo
Then, by Theorem 2.9.6 (d), L::; lim Xnk ::; U. That is, L::; W::; U. •
k->oo
EXERCISE SET 2.9
- Find the upper and lower limits of each of the following sequences:
(a) { ~' -^2 3, 3,^1 -~ 4' 4,^1 -^4 5, 5,^1 - ~' i, ... }
(b) {sin "671"} (c) { nsin nt}
(d) { nsn+l in n^5 ~ } (e) { n + (-l~n(2n+l)}
(f) {l+(-l)n cosntn2 } (g) { ncos(nt)}
(h) { (1 +cos n;) *}
- Prove Theorem 2.9.6 (c).
- Prove Theorem 2.9.8.
- Prove Theorem 2.9.9.