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I have already finished my analysis classes with good grades. But lately I am playing with Geogebra and wonder if there exists a sequence of natural numbers $(n_k)$ such that $$|\sin{n_k}|\leq\frac{1}{n_k^2}.$$

I have done some research and found that $\{\sin{n}:n\in\mathbb{N}\}$ is dense in $[-1,1]$. However, density does not immediately imply the existence of such sequence, because such $n_k$ making $|\sin{n_k}|$ so small, could occur very lately (i.e. $n_k$ may) be very large.

I am playing with Geogebra because I am wondering if the sequence $x_n = n|\sin{n}|$ is properly divergent or not. It clearly has a properly divergent subsequence to $+\infty$ but I am wondering if there is another subsequence with limit 0.

Any help would be grateful.

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  • $\begingroup$ This is about the continued fraction of $\pi.$ It seems unlikely that there are infinitely many $n_k,$ but I doubt we can prove it yet. $\endgroup$ Commented Nov 26 at 16:30
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    $\begingroup$ It is however very likely that there is a sequence such that $n_k \sin n_k \to 0$. That would follow if the partial quotients are unbounded. In any case, there are sequences such that $n_k\sin n_k$ is bounded. $\endgroup$ Commented Nov 26 at 16:43
  • $\begingroup$ (Is there even one example of $|\sin{n}|\leq\frac{1}{n^2}$ other than $n=1$?) With first powers instead of squares in the denominator (i.e. $|\sin{n_k}|\leq\frac{1}{n_k}$), OEIS A337371 and A332095 are related. $\endgroup$ Commented Nov 27 at 0:57

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As mentioned in this answer, we do not know if the sequence $\frac{1}{n^2\sin n}$ converges to zero. However, the result you ask for is significantly weaker, though still unknown.

Suppose we have a rational $p/q$ such that $|p/q - \pi| = \epsilon$ for a very small $\epsilon>0$. Then $3q \leq p \leq 4q$ certainly. Further, we have $$|\sin(p)| = |\sin(q\pi + q\epsilon)| = |\sin(q\pi)\cos(q\epsilon)+\cos(q\pi)\sin(q\epsilon)|=|0\pm\sin(q\epsilon) |\leq q\epsilon \leq \frac{p\epsilon}{3}$$

If the irrationality measure of $\pi$ were greater than $3$, then we could achieve $\epsilon \leq p^{-3}$ for infinitely many $p$, and thus have a convergent subsequence. Conversely, we can show by a similar calculation that if we have a convergent subsequence, the irrationality measure of $\pi$ is at least $3$.

The current bounds on the irrationality measure of $\pi$, however, are that it is at least $2$ and at most $7.103\dots$, so we simply cannot answer your question.

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