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# hw8solutions - n ≤ n 1 = ⇒ 1 n 1 ≤ 1 n = ⇒ 1 1 n 1...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: February 8, 2010 Assignment 8 1. Calculate lim n →∞ ( n 2 + n - n ). Here, we note that p n 2 + n - n = n n 2 + n + n = 1 p 1 + (1 /n ) + 1 . From here, it is intuitively clear that the limit is 1 / 2, but we have not yet justified pulling the limit inside of the square root. We have proved in class that you can pull the limit inside of addition and division. Thus, it suffices to show that p 1 + (1 /n ) 1. To this end, we know that p 1 + (1 /n ) 1 for all n . Moreover, by the Archimedian Principle, we know that 1 is the greatest lower bound. Indeed, if a > 1, then we can choose n N sufficiently large so that 1 /n < a 2 - 1, or 1 + (1 /n ) < a 2 or p 1 + (1 /n ) < a . Moreover, we know that p 1 + (1 /n ) is decreasing. Indeed,
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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 b-1 > 0 since b < 1. Then b n = 1 (1 + a ) n ≤ 1 (1 / 2) n ( n-1) a 2 by the binomial theorem. And, nb n ≤ 2 ( n-1) 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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