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

  • $\begingroup$ What are your thoughts? Do you see where the problem may be? $\endgroup$ – 5xum Feb 6 '14 at 8:43
  • $\begingroup$ This is known as Flint Hill series. See this arxiv.org/abs/1104.5100 $\endgroup$ – Sungjin Kim Feb 6 '14 at 8:58
  • $\begingroup$ From currently known results, what we can prove is the convergence of $1/(n^7\sin n)$, to the limit $0$. $\endgroup$ – Sungjin Kim Feb 6 '14 at 9:07
  • $\begingroup$ 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. $\endgroup$ – Dongyu Wu Feb 6 '14 at 9:15
  • 1
    $\begingroup$ @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? $\endgroup$ – J.R. Feb 6 '14 at 9:46

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


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.

  • $\begingroup$ 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. $\endgroup$ – achille hui Feb 6 '14 at 12:35
  • 4
    $\begingroup$ 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 $\left| \xi - \frac{m}{n} \right| < \frac{1}{\sqrt{5}n^2}$. $\endgroup$ – achille hui Feb 6 '14 at 14:07
  • 5
    $\begingroup$ We don't need Hurwitz's theorem, continued fraction should be enough: $|\xi-\frac mn|\leq \frac {1}{n^2}$ $\endgroup$ – Sungjin Kim Feb 6 '14 at 22:34
  • 1
    $\begingroup$ 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? $\endgroup$ – mengdie1982 Apr 12 at 10:31
  • $\begingroup$ The other one is easy. $\endgroup$ – J.R. Apr 13 at 17:58

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.

  • $\begingroup$ 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. $\endgroup$ – Sean Eberhard Feb 6 '14 at 14:08
  • $\begingroup$ 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. $\endgroup$ – Dongyu Wu Feb 7 '14 at 11:48
  • $\begingroup$ The statement about random $x$ was only intended to inform any conjecture about $\pi$, which often behaves as though it were random. $\endgroup$ – Sean Eberhard Feb 7 '14 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.