Convergence of the sequence $\frac{1}{n\sin(n)}$

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

Solution 1:

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.

Solution 2:

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.