Hwk3_Solns

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

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Unformatted text preview: 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 > 0. (a) | sin n- cos n | n 2 n < as soon as n > 2 . (c) n n 2 + 3 n + 1 < n n 2 = 1 n < when n > 1 . (e) n 2 + 2- n = n 2 + 2- n n 2 + 2 + n n 2 + 2 + n ! = n 2 + 2- n 2 n 2 + 2 + n = 2 n 2 + 2 + n < 2 n < when n > 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 < 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 > . 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 So the sequence starts decreasing when...
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This note was uploaded on 11/09/2009 for the course MATH 3110 taught by Professor Ramakrishna during the Fall '08 term at Cornell.

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

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