114 Chapter 2 • Sequences
Define Ek+i = L - Xnk. Then Ek+i > 0, so by (13), :J infinitely many
n EN 3 Xn E (L -Ek+1, L). Hence, :J nk+l > nk 3 Xnk+i E (L -ck+1, L). Note
that Xnk+i > L - ck+l = L - (L - Xnk) = Xnk. That is , Xnk < Xnk+,.
By mathematical induction, we have defined a subsequence { Xnk} of { Xn}
that is (strictly) monotone increasing.
Case 2, (L, +oo) contains infinitely many terms of {xn}: Exercise 19. •
Corollary 2.6.19 Every bounded sequence has a monotone subsequence.
Proof. An immediate consequence of Theorems 2.6. 16 and 2.6.18. •EXERCISE SET 2.6- Prove Lemma 2.6.3. [Hint: Use mathematical induction.]
- Prove Lemma 2.6.6.
- Complete Part (2) of the Proof of Theorem 2.6.8.
- Find each of the following limits:
(
(a) lim 1 + -^1 )2n
n->oo n(b) lim (1 + ~)n
n->oo 3n
n2
(c) n-+oo lim (1 + ~) n (d) lim (-n )n
n->oo n + 1(e) lim (1-.!.)n
n-1-00 n- In each of (a) and (b), prove or disprove that for all disjoint nonempty
sets A and B , it is possible for a sequence { Xn} of real numbers to satisfy
the given condition.
(a) {xn} is eventually in A and eventually in B.
(b) { Xn} is frequently in A and frequently in B. - Suppose that the subsequences { X2n} and { X2n+l} of even-numbered and
odd-numbered terms, respectively, of {xn} both have limit L. Prove that
Xn ~ L. (L may be finite or infinite.) - Prove Theorem 2.6.12.
- Prove Theorem 2.6.13 (a) and (b).
- Prove Theorem 2.6.13 (c) and (d).