I know there is such a subsequence for $b_n = \sin(n)$. What about $a_n = n\sin(n)$?
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1can you please clarify on what it is you are trying to ask? – Ittay Weiss May 20 '13 at 23:33
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@Abel, $n$ has to be a natural number. – dfeuer May 20 '13 at 23:36
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1This was asked recently: http://math.stackexchange.com/questions/390452/is-zero-a-cluster-point-of-n-sin-n?lq=1 See also the last comment in the following answer: http://math.stackexchange.com/a/221168/405 – Samuel May 20 '13 at 23:42
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The basic question is whether $\sin i$ is close enough to zero often enough. – dfeuer May 20 '13 at 23:44
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2Can anyone confirm this is an open problem? – Pedro May 20 '13 at 23:51
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At its core, the fundamental question is the following: given $\epsilon > 0$ small, what is the expected growth rate of $n_{\epsilon}$, where $n_{\epsilon}$ is the first integer such that $n_{\epsilon} \equiv \pm \epsilon$ mod $\pi$. – Christopher A. Wong May 20 '13 at 23:51
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Christopher A. Wong, there won't usually be any $n_\epsilon$ like that. What do you actually mean? – dfeuer May 20 '13 at 23:58
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2Computing with $n$ from 1 to 2 million there exists only $15$ values of $n$ such that $-1<n\cdot \sin(n)<1$. Of course this is not a proof, but sugests that maybe this subsequence does not exists. – Integral May 21 '13 at 00:38
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Going up to $5$ million and we still have only $15$ values. – Integral May 21 '13 at 00:46
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@dfeuer, Sorry, I meant that $n_{\epsilon} \le \epsilon$ mod $\pi$; you can show using the pigeonhole principle that for $\epsilon > 0$, there exists an integer $n$ such that $n$ is at most $\epsilon$ away from some multiple of $\pi$. – Christopher A. Wong May 21 '13 at 00:50
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1@Integral, how are you doing these computations? They need to be rather absurdly precise to give the right results. – dfeuer May 21 '13 at 01:02
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1why rather absurdly? It's necessary to be only 1 digit precise. – Integral May 21 '13 at 01:25
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2The computation of $sin(n)$ will need to be precise since $n$ is so large and will "move" many decimals to the left. – Pedro May 21 '13 at 19:09
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Possible duplicate of Does $a_n = n\sin n$ have a convergent subsequence? – nathan.j.mcdougall Jan 10 '19 at 21:50