Compute $\limsup_{n\to \infty} |\sin n|^n$

Question

$$ \limsup_{n\to \infty} |\sin n|^n $$

Consider $n$ is an integer. Since $n$ is an integer, I know that $\sin n$ is never equal to $1$. But still it is unclear if the lim sup is equal to $1$ or $0$.

Answer 1

In this post : About the limit $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n \cdot \cos (n)} $, you will see a proof that $|\cos(n)| \leqslant \frac{3}{n}$ for infinitely many $n$ (based on the Diophantine approximation) and in this case, $|\sin(n)| = \sqrt{1 - \cos^2(n)} \geqslant \sqrt{1 - \frac{9}{n^2}}$ hence, still for infinitely many $n$, $$ |\sin(n)|^n \geqslant \left(1 - \frac{9}{n^2}\right)^{\frac{n}{2}} = \exp\left(\frac{n}{2}\ln\left(1 - \frac{9}{n^2}\right)\right) = \exp\left(-\frac{9}{2n} + \mathrm{o}\!\left(\frac{1}{n}\right)\right) \rightarrow 1. $$ We deduce that $\limsup |\sin(n)|^n = 1$.

Answer 2

I can give a heuristic explanation of why the result ought to be true, but I don't think this argument can be extended.

I aim to show that for a given $\epsilon > 0, \epsilon << 1$, we should expect $$\sup_{1 \leq n \leq 1/\epsilon} |\sin(n)|^{1/\epsilon} \geq 1-\epsilon.$$ Firstly, note that if $y \geq 1-\epsilon^2$, Bernoulli's inequality implies that $$y^{1/\epsilon} \geq 1-\epsilon^2 \cdot \frac{1}{\epsilon} = 1-\epsilon.$$ Now, keep in mind that $y = \sin(n)$. Apply Taylor's theorem to $\sin(x)$ at $x = \pi/2$ and conclude $$\sin\left(\frac{\pi}{2} + t\right) \geq 1- \frac{t^2}{2}$$ for all $t$. We want to think of $t$ as the distance between $\pi/2$ and $n \mod \pi$, i.e., $$t = \left|\frac{\pi}{2} - (n \mod \pi)\right|.$$

We can see that if $|t| \leq \epsilon$, then $$\sin(\pi/2 + t) \geq 1 - \frac{\epsilon^2}{2} \implies |\sin(\pi/2 + t)|^{1/\epsilon} \geq 1-\epsilon.$$

Now, the question is basically how $n \mod \pi$ is distributed in $[0,\pi]$. Intuitively, if there we look through all $1 \leq n \leq 1/\epsilon$, the average spacing of these numbers in the interval $[0,\pi]$ should be $\pi/(1/\epsilon) = \pi \epsilon$. Hopefully, this means we can find an $n \mod \pi$ within $\pi \epsilon$ of $\pi/2$ for $1 \leq n \leq 1/\epsilon,$ which would roughly prove our result.

This is the case that agrees with my hastily constructed code $(\epsilon = 10^{-1}, 10^{-2}, ...,10^{-7},$ but it would probably be easier to prove something like $$\sup_{1 \leq n \leq 1/\epsilon} |\sin(n)|^{1/\epsilon} \geq 1-\epsilon^{1/2}.$$ You can follow the steps above to get a better bound with Bernoulli's inequality, ($y \geq 1-\epsilon^{3/2}$, suffices to find $|t| \leq \epsilon^{3/4}$, $t = |\pi/2 - n \mod \pi|$, average distance between numbers is $\pi \epsilon$ but only need to fit in an interval of length $2\epsilon^{3/4}$.)

At that point, you would hope to use some version of the Equidistribution Theorem that says $n \mod \pi$ is uniformly distributed in $[0,\pi]$, but this theorem doesn't actually give a rate of convergence.