# Does $a_n = n\sin n$ have a convergent subsequence?

I know that $\sin n$ has a convergent subsequence since it is bounded, but does the unbounded sequence $n\sin n$ have a convergent subsequence?

Given a subsequence of $\sin n$ that tends to zero, it is still possible that when we multiply the convergent subsequence $\sin(n_k)$ by $n_k$, the limit may not be $0$.

Since $\pi$ is irrational, we have infinitely many rationals $p/q$ (Dirichlet's approximation theorem) such that $$\left| \pi - \frac{p}{q} \right | < \frac{1}{q^2}$$ This implies $$|\sin p | = |\sin(q\pi - p) | < |q\pi - p| < \frac{1}{q}$$ Hence $|p\sin p| < \frac{p}{q}$, which is bounded.
Thus we have obtained a bounded subsequence of $\{n\sin n\}$, from which we can further select a convergent subsequence.
• (+1) btw I don't think you actually require $\pi$ irrational – Tim kinsella Feb 25 '18 at 6:37
• @Timkinsella I am sorry, my bad. $|\alpha - p/q|<1/q^2$ indeed also has infinitely many pairs satisfying it when $\alpha$ is rational. But this is not a crucial point. – pisco Feb 25 '18 at 7:14