# Thread: Proof that lim sup (sin n) = 1

1. ## 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?

2. ## 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.

Originally Posted by phys251
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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