The Sequence Sin(n)Date: 02/20/2002 at 06:59:44 From: Sean McGrath Subject: The sequence sin(n) Dear Dr. Math, I am trying to prove that the sequence sin(n), for n, a natural number, does not converge. I know that if I find 2 subsequences in the intervals [-1,1/2] and [1/2,1], then my proof will be complete by a theorem in my text. And I know that there are subsequences that exist in those intervals, but how do I find them? Thanks, Sean McGrath Date: 02/20/2002 at 08:57:01 From: Doctor Floor Subject: Re: The sequence sin(n) Hi, Sean, Thanks for your question. We know that sin x >= 0.5 for x in [pi/6 + 2u*pi, 5pi/6 + 2u*pi], where u is an integer. In this case we restrict ourselves to natural numbers u. Since the width of each interval is 2pi/3, which is more than 2, for each natural number u the interval contains at least two natural numbers. From these we can create a subsequence. In the same way we can find a subsequence of natural numbers for which the sines < -1/2. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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