Hwk1_Solns

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/09/2009 for the course MATH 3110 taught by Professor Ramakrishna during the Fall '08 term at Cornell.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online