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Thread: 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, $\displaystyle N \subset \Re.$ So $\displaystyle \limsup (\sin n) \leq \limsup (\sin x).$
    2. Recall, $\displaystyle \limsup (\sin x) = 1.$
    3. Transitively, $\displaystyle \limsup (\sin n) <= 1.$

    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; Jun 6th 2013 at 12:04 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 $\displaystyle \pi$ 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 $\displaystyle \sin(a_n)$, but to make sure that $\displaystyle \sin(a_n)$ tends to 1. You can use the auxiliary fact that $\displaystyle \{n\pmod{2\pi}\mid n\in\mathbb{N}\}$ is dense in $\displaystyle [0,2\pi]$. See this post that discusses a similar fact that $\displaystyle \{n\pi\pmod{1}\mid n\in\mathbb{N}\}$ is dense in [0, 1].
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