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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 $0$.

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Yes.

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.

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    $\begingroup$ It seems like you are using some version of Dirichlet's approximation theorem, which (at least to me) is a somewhat nontrivial result. $\endgroup$ – Shalop Feb 25 '18 at 6:35
  • $\begingroup$ (+1) btw I don't think you actually require $\pi$ irrational $\endgroup$ – Tim kinsella Feb 25 '18 at 6:37
  • $\begingroup$ @Timkinsella Could you elaborate what do you mean? $\endgroup$ – pisco Feb 25 '18 at 6:40
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    $\begingroup$ @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. $\endgroup$ – pisco Feb 25 '18 at 7:14
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    $\begingroup$ @pisco yeah, I guess its just the totally trivial point that your proof also works for functions with rational period. $\endgroup$ – Tim kinsella Feb 25 '18 at 7:16

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