proving existence of a sequence such that the limit exists? Can anyone prove the existence of a sequence $(n_{k})_{k\in \mathbb{N}}$ of distinct positive integers such that the limit: $\lim_{k\rightarrow \infty }\sin(n_{k})$ exists in $\mathbb{R}$
I can definitely construct a sequence $(n_{k})_{k\in \mathbb{N}}$ such that $\frac{1}{2}\leq \sin(n_{k})\leq 1$, but this doesn't imply that this sequence is convergent. Any suggestions?
 A: Every bounded sequence has convergent subsequence, see Bolzano-Weierstrass theorem.
If you apply this to the sequence $(\sin n)_{n=0}^\infty$, you get the desired result.
(Or you can mimic the proof of Bolzano-Weierstrass theorem, if you prefer.)
A: I turn my comment into an answer:
It follows directly from http://en.wikipedia.org/wiki/Irrational_rotation
A: As @MartinSleziak noted, existence of such a sequence is easy.  Slightly less obvious is how to construct the sequence more-or-less explicitly.  Suppose $p$ and $q$ be positive integers such that $\left| \dfrac{p}{q} - \pi\right| < \dfrac{\epsilon}{q}$.  Then $|p - q \pi| < \epsilon$ so $|\sin(p)| < \epsilon$.  So we could use a sequence of good rational approximations of $\pi$.  For example, we could take $p_k/q_k$ to be the convergents of
the continued fraction of $\pi$, which have $|p_k/q_k - \pi| < 1/q_k^2$, and thus
$|\sin(p_k)| < 1/q_k$.
A: A couple of things to think about:
1) Let $\phi: \mathbb{R} \to \mathbb{R} / 2\pi\mathbb{Z}$ be the projection map. Then $\phi(\mathbb{N})$ is dense in $\mathbb{R} / 2\pi\mathbb{Z}$.
2) Let $n_k$ be an increasing sequence of (distinct) positive integers defined by picking $n_k$ such that $n_k$ $($ mod $2\pi ) < \frac{1}{k}$.
