Hwk1_Solns

Hwk1_Solns - Math 3110 Spring 09 Homework 1 Selected...

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Math 3110 Spring ’09 Homework 1: Selected Solutions [1.2 - 1 (e)]: a n = sin ( 1 n ) is strictly decreasing . Note that d dx (sin x ) = cos x > 0 for all x (0 , 1]. So sin x is strictly increasing as x grows from 0 to 1; thus sin x is strictly decreasing as x decreases from 1 to 0. So for any strictly decreasing sequence { x n } ∈ (0 , 1], b n = sin x n is also strictly decreasing. Certainly x n = 1 n (0 , 1] is strictly decreasing, thus a n = sin ( 1 n ) is strictly decreasing as claimed. [1.3 -1 (a)]: a n = n 2 - 1 n is increasing, bounded above by 1 and lim n →∞ a n = 1. Observe that n 2 - 1 n = q n 2 - 1 n 2 = q 1 - 1 n 2 . It is clear that ± 1 - 1 ( n +1) 2 ² > ( 1 - 1 n 2 ) since 0 < 1 ( n - 1) 2 < 1 n 2 1. So a n +1 > a n , and { a n } is an increasing sequence bounded above by 1. What follows is an unnecessarily formal demonstration that lim n →∞ a n = 1: Suppose there were a very small ± > 0 such that a n (1 - ± ) for all n ∈ { 1 , 2 ,... } . Certainly for any such
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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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Hwk1_Solns - Math 3110 Spring 09 Homework 1 Selected...

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