TITLE: Convergent subsequence of $\sin(n)$
QUESTION [7 upvotes]: According to theorem every bounded sequence in $R$ has a convergent subsequence.Can anybody make a convergent subsequence of $\sin{n}$.

REPLY [3 votes]: Here's a semi-explicit one.  The convergents of the continued fraction of $\pi$ are a sequence of rational approximations $p_n/q_n$ with $p_n$, $q_n$ positive integers tending to $\infty$, 
$$ \left| \pi - \dfrac{p_n}{q_n}\right| < \dfrac{1}{q_n^2}$$
and thus $$|\sin(p_n)| = |\sin(p_n - \pi q_n)| \le |p_n - \pi q_n| < 1/q_n \to 0 \ \text{as}\ n \to \infty$$