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?