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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 $0$.

Iced Palmer
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    Compare: https://math.stackexchange.com/questions/164390/does-leftn2-sin-n-right-have-a-convergent-subsequence/164402 (a more general question, despite the title) – Micah Feb 25 '18 at 06:22

1 Answers1

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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.

pisco
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    It seems like you are using some version of Dirichlet's approximation theorem, which (at least to me) is a somewhat nontrivial result. – shalop Feb 25 '18 at 06:35
  • (+1) btw I don't think you actually require $\pi$ irrational – Tim kinsella Feb 25 '18 at 06:37
  • @Timkinsella Could you elaborate what do you mean? – pisco Feb 25 '18 at 06:40
  • It seems true that for all $\alpha \in \mathbb{R}$ there are infinitely many $p,q\in \mathbb{Z}$ such that $|\alpha - p/q|<1/q^2$. Anyway, its not so important. – Tim kinsella Feb 25 '18 at 06:43
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    @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 07:14
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    @pisco yeah, I guess its just the totally trivial point that your proof also works for functions with rational period. – Tim kinsella Feb 25 '18 at 07:16
  • It impossible to apply the above solution to the sequence $n^2\sin n$ because of the index 2 in the Dirichlet theorem. What to do in that case? – pabodu Feb 26 '18 at 15:58
  • @pabodu Then this becomes an unsolved problem due to limited knowledge on irrational exponent of $\pi$, see the answer of https://math.stackexchange.com/questions/164390 for details. – pisco Feb 27 '18 at 04:02