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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/   
    
Associated Topics:
High School Sequences, Series

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