This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: m such that m > 1 a ± b . But then a ± b < 1 m and therefore we get the contradiction a > b + 1 m : 7.4 (a) Let x n := p 2 n for n ³ 1, then x n is irrational since p 2 is so and lim n !1 x n = 0 ; which is rational. (b) De±ne x n = (1 + 1 n ) n for n ³ 1. Then each x n is rational and lim n !1 x n = e; which is irrational. 1 2 HOMEWORK 2: SOLUTIONS/HINTS 7.2 In the following let ± > 0. (a) Let N > 1 ± , then j a n j = ± ± ± ± n n 2 + 1 ± ± ± ± ± ± ± ± n n 2 ± ± ± = 1 n < ± for n ² N and hence lim a n = 0 : (d) Let N > 16 25 ± ± , then we get ± ± ± ± d n ³ 2 5 ± ± ± ± = ± ± ± ± 16 5(5 n + 2) ± ± ± ± < ± ± ± ± 16 25 n ± ± ± ± ± ± ± ± ± 16 25 N ± ± ± ± < ± for any n ² N and therefore the limit is just 2 5 : (e) Pick N > 1 ± , then j s n j = 1 n ± 1 N < ± for n ² N and thus lim n !1 s n = 0...
View
Full
Document
This note was uploaded on 02/12/2012 for the course MATH 1311a taught by Professor Peter during the Spring '11 term at UCLA.
 Spring '11
 Peter

Click to edit the document details