Hwk3_Solns

# Hwk3_Solns - Math 3110 Spring 09 Homework 3 Selected...

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Math 3110 Spring ’09 Homework 3: Selected Solutions [3.1 - 1 (a,c,e)]: Claim : (a) lim n →∞ sin n - cos n n = 0, (c) lim n →∞ n n 2 + 3 n + 1 = 0, (e) lim n →∞ p n 2 + 2 - n = 0 Proof Fix any small ² > 0. (a) | sin n - cos n | n 2 n < ² as soon as n > 2 ² . (c) n n 2 + 3 n + 1 < n n 2 = 1 n < ² when n > 1 ² . (e) n 2 + 2 - n = n 2 + 2 - n · ˆ n 2 + 2 + n n 2 + 2 + n ! = n 2 + 2 - n 2 n 2 + 2 + n = 2 n 2 + 2 + n < 2 n < ² when n > 2 ² . [3.2 - 3 (a)]: Claim : The sequence a n = 2 n X k = n +1 1 k has a limit. Proof It suffices to show that a n is bounded and increasing monotonically. We immediately see that 0 < a n < (2 n - n - 1) 1 n + 1 < n n + 1 < 1 for all n . And for all n , a n +1 - a n = 1 2 n + 1 + 1 2 n + 2 - 1 n + 1 = 1 2 n + 1 - 1 2 n + 2 > 0 . So a n is a bounded monotone sequence and hence has a limit. [3.4 - 2]: Claim : For any r > 0, sequence a n = r n n ! 0 as n → ∞ Proof When r 1 this follows immediately from Theorem 3.4 in your text. For r > 1 notice that a n +1 = r n + 1 a n = a n + k = k Y j =1 r n + j a n < r n + 1 k a n 1

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So the sequence starts decreasing when n > r - 1 and is halved at each step once n > 2 r - 1. Since a n is a fixed constant and, by Theorem 3.4 in your text, lim k →∞ r n
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