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For $a_n=\sin(n)$, it can use the continued fraction method to construct the subsequence, so that $\lim a_{n_k}\rightarrow 0$.

But for $a_n=\sin(n^2)$, how to construct (if possible) a subsequence such that it converges to $0$ ?

MathFail
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    $\sin(314)$ is not close to zero. Can you be slightly more precise in your question? Maybe a $2\pi$ is missing somewhere? – coudy Mar 17 '23 at 09:58
  • Consider the convergents in the simple continued fraction of $\pi$. The numerators can be used for the desired subsequence. – Peter Mar 17 '23 at 10:02
  • Are you sure about your solution for $a_n =\sin(n)$? – jjagmath Mar 17 '23 at 10:49
  • oh, sorry, I have edited it @coudy jjagmath – MathFail Mar 17 '23 at 10:53
  • You need to find subsequence $b_{n_k}$ for $b_n = (n^2) mod π$. You can try to use Pigeonhole principle, there must exist distinct $p$ and $q$ such that $|p^2 - q^2 - lπ| < ε$ – rumathe Mar 17 '23 at 11:02

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