Unformatted text preview: n ≤ n + 1 = ⇒ 1 n + 1 ≤ 1 n = ⇒ 1 + 1 n + 1 ≤ 1 + 1 n = ⇒ p 1 + (1 / ( n + 1)) ≤ p 1 + (1 /n ) . Since this sequence is monotonically decreasing and bounded below, it must converge to its greatest lower bound, which is 1. For this latter step, you could also cite Theorem 3.2.10 of the text. 2. Let b ∈ R satisfy 0 < b < 1. Show that lim nb n = 0. We write b = 1 1+ a . Then a = 1 b1 > 0 since b < 1. Then b n = 1 (1 + a ) n ≤ 1 (1 / 2) n ( n1) a 2 by the binomial theorem. And, nb n ≤ 2 ( n1) a 2 . The right side clearly tends to 0 as n → ∞ , and thus, by the squeeze lemma, nb n → 0. 1...
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 Spring '10
 MetCalfe
 Calculus, Order theory, greatest lower bound, Archimedian Principle

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