How would you evaluate $\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$ How would you evaluate the limit inferior of the sequence $n\,|\mathopen{}\sin n|$? That is,
$$\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$$

Edit. Let $\mu$ be the irrationality measure of $\pi$. Since $\mu$ is not known, I will split the question:


*

*Assuming $\mu > 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$?

*Assuming $\mu = 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$?



I kind of like its graph... 

 A: The short answer is: If $\mu > 2$, then $\liminf\limits_{n\to \infty}\: n\lvert\sin n\rvert = 0$, and if $\mu = 2$, then it may be $0$ or it may be strictly positive.
The value depends on the simple continued fraction expansion of $\pi$. Let $[a_0,a_1,a_2,\dotsc]$ be the simple continued fraction expansion of $\pi$. We denote the numerators of the convergents by $p_k$ and the denominators by $q_k$, and let $\alpha_k$ be the $k^{\text{th}}$ complete quotient, i.e. $\alpha_k = [a_k, a_{k+1}, a_{k+2}, \dotsc]$. Then by the general theory of simple continued fractions we have
$$\biggl\lvert \pi - \frac{p_k}{q_k}\biggr\rvert = \biggl\lvert \frac{p_k\alpha_{k+1} + p_{k-1}}{q_k\alpha_{k+1} + q_{k-1}} - \frac{p_k}{q_k}\biggr\rvert = \frac{\lvert p_{k-1}q_k - p_kq_{k-1}\rvert}{q_k^2\bigl(\alpha_{k+1} + \frac{q_{k-1}}{q_k}\bigr)} = \frac{1}{q_k^2\bigl(\alpha_{k+1} + \frac{q_{k-1}}{q_k}\bigr)}.$$
Since $a_{k+1} < \alpha_{k+1} < a_{k+1}+1$ and $0 < \frac{q_{k-1}}{q_k} < 1$, we have the inequalities
$$\frac{1}{(a_{k+1}+2)q_k^2} < \biggl\lvert \pi - \frac{p_k}{q_k}\biggr\rvert < \frac{1}{a_{k+1}q_k^2}.$$
Given $\varepsilon \in (0,1)$, there is a $\delta > 0$ such that $(1-\varepsilon)\varphi < \sin \varphi < \varphi$ for $0 < \varphi < \delta$, and for large enough $k$ we have $\frac{1}{q_k} < \delta$, whence
$$\frac{(1-\varepsilon)p_k}{(a_{k+1}+2)q_k} < p_k\lvert\sin p_k\rvert = p_k\lvert \sin (p_k-q_k\pi)\rvert < \frac{p_k}{a_{k+1}q_k}.$$
If the sequence $(a_k)$ of partial quotients is unbounded, this shows
$$0 \leqslant\liminf_{n\to\infty} \: n\lvert\sin n\rvert \leqslant \liminf_{k\to \infty}\: p_k\lvert\sin p_k\rvert = 0,$$
and if the sequence of partial quotients is bounded, then
$$\frac{\pi}{s+2} \leqslant \liminf_{k\to \infty} \: p_k\lvert\sin p_k\rvert \leqslant \frac{\pi}{s}$$
for $s = \limsup\limits_{k\to\infty} a_k$.
Now let $n \geqslant 50$, and $m$ the closest integer to $\frac{n}{\pi}$. If $\bigl\lvert\pi - \frac{n}{m}\bigr\rvert \geqslant \frac{1}{2m^2}$, then
$$n\lvert \sin n\rvert = n\lvert \sin (n - m\pi)\rvert \geqslant n\frac{2}{\pi} \lvert n - m\pi\rvert \geqslant \frac{n}{\pi m} \geqslant \frac{3}{\pi}.$$
If $\bigl\lvert\pi - \frac{n}{m}\bigr\rvert < \frac{1}{2m^2}$, then $\frac{n}{m}$ is a convergent of $\pi$, i.e. there is a $k$ and an $r$ such that $n = r\cdot p_k$ and $m = r\cdot q_k$. From the above we obtain first
$$\frac{1}{(a_{k+1}+2)q_k^2} < \biggl\lvert \pi - \frac{p_k}{q_k}\biggr\rvert = \biggl\lvert\pi - \frac{n}{m}\biggr\rvert < \frac{1}{2r^2q_k^2},$$
so $r < \sqrt{\frac{a_{k+1}+2}{2}}$, and then
$$\frac{r}{(a_{k+1}+2)q_k} < \lvert m\pi - n\rvert < \frac{r}{a_{k+1}q_k},$$
which shows $\lvert m\pi - n\rvert < \frac{\pi}{2}$, and consequently
$$n\lvert\sin n\rvert = n\lvert \sin (m-m\pi)\rvert \geqslant n\frac{2}{\pi}\lvert n - m\pi\rvert > \frac{2r^2p_k}{\pi(a_{k+1}+2)q_k}.$$
Since $r \geqslant 2$ implies $a_{k+1} > 6$, we see that $n\lvert\sin n\rvert > p_k\lvert\sin p_k\rvert$ when $n = r\cdot p_k$, $r \geqslant 2$.
As $\pi$ is not a quadratic irrational, its continued fraction expansion cannot become eventually periodic, so if $s = \limsup_{k\to\infty} a_k < +\infty$, then by the above computations we have
$$\liminf_{n\to\infty} \: n\lvert\sin n\rvert \geqslant \frac{\pi}{s+2}.$$
We summarise

$$\liminf_{n\to\infty} \: n \lvert \sin n\rvert = 0$$ if and only if the sequence $(a_n)$ of partial quotients in the simple continued fraction expansion of $\pi$ is unbounded.

If the irrationality measure $\mu$ of $\pi$ is larger than $2$, the sequence of partial quotients is unbounded, then for every $\varepsilon \in (0,\mu - 2)$ we even have $a_{k_m+1} > q_{k_m}^{\varepsilon}$ for a strictly increasing sequence $(k_m)_{m\in\mathbb{N}}$.
If $\mu = 2$, the sequence of partial quotients can still be unbounded, but as far as I'm aware, it is unknown whether it is bounded or not.
