Solving algebraically for s we obtain s 3 Since s n 6 for all n 3 cannot be the

# Solving algebraically for s we obtain s 3 since s n 6

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Solving algebraically for s, we obtain s = ± 3. Since s n 6 for all n , - 3 cannot be the limit. We conclude that lim s n = s = 3. 18.11 Finish the proof of Theorem 18.3 for a bounded decreasing sequence. Proof. Supposse that ( s n ) is a bounded decreasing sequence. Let S denote the nonempty bounded set { s n : n N } . By the completeness axiom S has a greatest lower bound, and we let s = inf S . We claim that lim s n = s . Given any ε > 0 , s + ε is not a lower bound for S . Thus there exists an integer N such that s N < s + ε . Furthermore, since ( s n ) is decreasing and s is an upper bound for S , we have s s n s N < s + ε for all n > N . Hence ( s n ) converges to s. 4
18.13 Prove Lemma 18.11 Lemma 18.11 Every Cauchy sequence is bounded. Proof 1. By Theorem 18.12 we know that every Cauchy sequence is con- vergent and by Theorem 16.13 every convergent sequence is bounded, then every Cauchy sequence is bounded. Proof 2. Applying Definition 18.9 with ε = 1 we obtain N N so that m, n > N implies | s n - s m | < 1 In particular | s n - s N +1 | < 1 for all n > N , so | s n | < | s N +1 | + 1 for n > N . Let M = max {| s 1 | , | s 2 | , | s N | , . . . , | s N +1 | + 1 } then we have | s n | ≤ M for all n N , so ( s n ) is bounded. 5

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• Fall '08
• Akhmedov,A
• Mathematical analysis, Sn, Guillermo Cu´llar

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