Limits and Sequencing
Date: 5/3/96 at 10:15:15 From: Anonymous Subject: Limits and sequencing This is an approaching limit question, from a maths limits question this one's supposed to be pretty hard... Find the limit a, and find an expression for the number N(epsilon) such that |a(n) - a| < epsilon for all n > N(epsilon) a(n) = 1/n sin^2 (n.pi/4) where a(n) is where the n is a subscript
Date: 10/17/96 at 8:5:49 From: Doctor Jerry Subject: Re: Limits and sequencing If a(n) = (1/n)*sin^2(pi*n/4), which is what I think you may intend, the question is not very hard (although it looks as if it might be). The factor sin^2(pi*n/4) always lies between 0 and 1. Consequently, 0<=a(n)<=1/n. Since 1/n goes to zero as n becomes infinite, so does a(n). Specifically, since 1/n < epsilon when n > 1/epsilon, we may take N(epsilon) the first positive integer greater than 1/epsilon. When n > N, |a(n) - 0| < epsilon. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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