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Does the sequence $$\frac{1}{n\sin(n)}$$ converge to $0$ or not? If not, what's the upper limit?

user642796
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Dongyu Wu
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  • What are your thoughts? Do you see where the problem may be? – 5xum Feb 06 '14 at 08:43
  • This is known as Flint Hill series. See this http://arxiv.org/abs/1104.5100/ – Sungjin Kim Feb 06 '14 at 08:58
  • From currently known results, what we can prove is the convergence of $1/(n^7\sin n)$, to the limit $0$. – Sungjin Kim Feb 06 '14 at 09:07
  • Well, I've seen the link and thanks for your help. But the link is about series. I think it might be a little different. The discussion here may be a little easier. – Dongyu Wu Feb 06 '14 at 09:15
  • The paper also deals with sequence. It gives possible values of $u, v$ such that $1/n^u \sin^v n$ converges to $0$. The last part explains it. However, the paper does not cover your sequence $1/n \sin n$. I just wanted to show you related paper. – Sungjin Kim Feb 06 '14 at 09:19
  • Sorry for my negligence. This is the best result I have ever got. Thanks again for your help. – Dongyu Wu Feb 06 '14 at 09:26
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    @i707107 Oh, I could have saved myself the work of writing it down, had I seen your comments.. Why did nobody post it as an answer? – J.R. Feb 06 '14 at 09:46
  • @TooOldForMath Your work is not in vain. Actually a very good explanation of what the paper did. Thanks for that. I wouldn't put it as an answer with only link indicated. – Sungjin Kim Feb 06 '14 at 09:56

2 Answers2

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The question is how close $\sin(n)$ can get to $0$.

The sequence will converge to $0$, if we are not able to find a subsequence $(n_k)_k$ such that $\sin(n_k)$ gets to $0$ so quickly, that it outperforms $n$ in going to $\infty$.

The zeros of $\sin$ are all integer multiples of $\pi$. The question is therefore: how well can we approximate $\pi$ by rationals?

By Dirichlet's approximation theorem (which is very simple to prove using the pidgeonhole principle), there exists a sequence $n_k\rightarrow\infty$ and $q_k$ such that

$$\left|\frac{n_k}{q_k}-\pi\right|<\frac{1}{q_k^2}$$

Since $|\sin(x)|= |\sin(x+k\pi)|$ and $|\sin(x)|\le |x|$ for $x\in\mathbb{R}$, $k\in\mathbb{Z}$, we have

$$\frac{1}{|n_k\sin(n_k)|}=\frac{1}{|n_k\sin(n_k-q_k\pi)|}\ge \frac{1}{n_k|n_k-q_k\pi|}\ge \frac{q_k}{n_k}\rightarrow \frac{1}{\pi}$$

That means, there is a subsequence that stays away from $0$. But clearly there also is a subsequence $n_k\rightarrow\infty$ such that $|\sin(n_k)|>1/2$ (e.g. approximate odd multiples of $\pi/2$). Then $1/|n_k\sin(n_k)|\le 2/n\rightarrow 0$. Therefore the subsequence converges to $0$.

This yields the

Conclusion: The sequence does not converge.

J.R.
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  • If $\mu = 2$, above argument shows that the sequence has a sub-sequence bounded away from $0$. If one find another sub-sequence which converges to $0$, then the limit doesn't exist. – achille hui Feb 06 '14 at 12:35
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    We don't need to use irrationality measure, we can use Hurwitz's theorem instead. For every irrational number $\xi$, there are infinitely many relative prime integers $m, n$ such that $\left| \xi - \frac{m}{n} \right| < \frac{1}{\sqrt{5}n^2}$. – achille hui Feb 06 '14 at 14:07
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    We don't need Hurwitz's theorem, continued fraction should be enough: $|\xi-\frac mn|\leq \frac {1}{n^2}$ – Sungjin Kim Feb 06 '14 at 22:34
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    your argument depends on the critical fact: there exist two subsequences $n_k$ and $q^k$ not only satisfying your inequality but also tending to $+\infty$. How does Dirichlet's approximation theorem guarantee the latter one？ – WuKong Apr 12 '19 at 10:31
  • The other one is easy. – J.R. Apr 13 '19 at 17:58
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The sequence does not converge.

Look at the fractional parts of the multiples $q\pi$ as $q$ ranges over $\{1,\dots,N\}$. Some two $q_1\pi$, $q_2\pi$ must differ by at most $1/N$, so we have an integer $q=|q_1-q_2|<N$ such that $q\pi$ differs from a positive integer $p_N\leq N\pi$ by at most $1/N$. Hence $|\sin p_N| = |\sin(p_N-q\pi)|\leq 1/N$, so $|p_N\sin p_N|\leq \pi$. Moreover since $|\sin p_N|\leq 1/N$ and $\pi$ is irrational we must have $p_N\to\infty$. Thus $\limsup 1/|n\sin n|\geq1/\pi$.

On the other hand it's easy to see $\liminf 1/|n\sin n|=0$, so the sequence does not converge.

EDIT. With an analysis similar to the above one can verify that $\limsup 1/|n\sin n|<\infty$ if and only if there is a constant $c>0$ such that $|\pi-p/q|>c/q^2$ for all $p/q\in\mathbf{Q}$. Such numbers are called "badly approximable". In terms of continued fraction expansions, a number is badly approximable iff the terms of its continued fraction expansion are bounded, so $\limsup 1/|n\sin n|=\infty$ iff the terms in the continued fraction expansion of $\pi$ are unbounded. I would guess this is unknown, but I am unsure.

To put this into a greater context, the number $e$ is not badly approximable, and almost all $x\in\mathbf{R}$ are not badly approximable. I doubt anybody would conjecture that $\pi$ is badly approximable.

There is a great wealth of relevant information on this Wikipedia page: http://en.wikipedia.org/wiki/Diophantine_approximation.

Sean Eberhard
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  • It seems likely that the upper limit is infinite (even if the irrationality measure of $\pi$ is $2$, which it probably is). One can show that if $x$ is random then the upper limit of $1/|n\sin(x\pi n)|$ is almost surely infinity. – Sean Eberhard Feb 06 '14 at 14:08
  • I'm afraid I can only show that $\frac{1}{n^{1+\epsilon}\sin (xn\pi)}\to 0$ $a.s.$ $\forall \epsilon>0 $, and even if your statement is true, it doesn't seem to help. – Dongyu Wu Feb 07 '14 at 11:48
  • The statement about random $x$ was only intended to inform any conjecture about $\pi$, which often behaves as though it were random. – Sean Eberhard Feb 07 '14 at 13:03