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Math Help - Proof that lim sup (sin n) = 1

  1. #1
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    Proof that lim sup (sin n) = 1

    Very similar to the idea that lim sup (sin x) = 1, but now we are dealing with just the natural numbers, not all the reals. I have started this proof but will need help completing it.

    Part A: Show lim sup (sin x) <= 1.
    1. Observe, N \subset \Re. So \limsup (\sin n) \leq \limsup (\sin x).
    2. Recall, \limsup (\sin x) = 1.
    3. Transitively, \limsup (\sin n) <= 1.

    Part B: Show lim sup (sin x) >= 1.
    This is the part where I get stuck. Options I have tried include:
    -Letting lim sup (sin x) = 1 - epsilon, and showing that any positive epsilon creates a contradiction;
    -Creating a subsequence of n, a_n, such that sin a_1 <= sin a_2 <= ... <= 1 (note: it can be shown sin a_1 < sin a_2 < ... < 1).

    But I'm not sure what to do from there. Thoughts? Suggestions?
    Last edited by phys251; June 6th 2013 at 01:04 AM.
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  2. #2
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    Re: Proof that lim sup (sin n) = 1

    The key fact here is that \pi is irrational, i.e., the step length of the sequence 1, 2, ... is not commensurate with the period of sine.

    Quote Originally Posted by phys251 View Post
    Part B: Show lim sup (sin x) >= 1.
    This is the part where I get stuck. Options I have tried include:
    ...
    -Creating a subsequence of n, a_n, such that sin a_1 <= sin a_2 <= ... <= 1 (note: it can be shown sin a_1 < sin a_2 < ... < 1).
    You need not just to have a strictly increasing sequence \sin(a_n), but to make sure that \sin(a_n) tends to 1. You can use the auxiliary fact that \{n\pmod{2\pi}\mid n\in\mathbb{N}\} is dense in [0,2\pi]. See this post that discusses a similar fact that \{n\pi\pmod{1}\mid n\in\mathbb{N}\} is dense in [0, 1].
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