Convergence of $(\sin x)^x$

My answer: It does not converge to anything! It must be clear that this does not converge to any non-zero $L$. So the question is: "Does it converge to zero." Let the proposition $P$, be the "converges to zero!" statement. Then $$P^c="\exists\varepsilon>0 \textrm{ s.t. } \#\{n\in\mathbb{N}\Big||\sin n|>\varepsilon^\frac{1}{n}\}=\infty"$$ But using the approximation $\varepsilon^\frac{1}{n}\approx\frac{-\log\varepsilon}{n}$ This becomes $$P^c="\exists a<\infty \textrm{ s.t. } \#\{n\in\mathbb{N}\Big||\sin n|>1-\frac{a}{n}\}=\infty"$$ Then it is easy to use the properties of the sine function to see that $$P^c=Q_-\cup Q_+$$ With $$Q_\pm="\#\Big\{n\in\mathbb{N}\Big||\{\frac{n}{2\pi}\mp\frac{1}{4}\}|<\frac{a}{\sqrt n}\Big\}=\infty"$$ From now on we focus only on the proposition $Q_+$ and try to prove it. Let $A$ be the conjecture

$A$: For any sequence $0\leq e_n\leq 1$ with infinite sum $\sum_0^\infty e_n$, any irrational number $\alpha\in(0,1)$ and any sequence $t_n\in[0, 1]$ for which the limit $\lim_{n\rightarrow\infty}t_n$ exists, the following holds

$$\#\Big\{(n, \{n\alpha\})\Big|n\in\mathbb{N}, |\{n\alpha\}-t_n|\leq e_n\Big\}=\infty$$

Then with $e_n=\min(1, \frac{a}{\sqrt n})$, $\alpha=\frac{1}{2\pi}$, $t_n=.25$ we get $$A\Rightarrow Q_+\Rightarrow P^c$$ And

I believe $A$ holds. But to solve the recent problem it is easier to use Dirichlet's Approximation Theorem. It states that that for every irrational $\alpha\in(0,1)$ there is an infinite sequence $n_k$ for which $\{n_k\alpha\}\leq\frac{1}{n_k}$. Now consider the infinite sequence $$m_k:=n_k\Big\lfloor\frac{1}{4\{n_k\alpha\}}\Big\rfloor$$ For this we have $$m_k\{n_k\alpha\}^2=n_k\{n_k\alpha\}^2\Big\lfloor\frac{1}{4\{n_k\alpha\}}\Big\rfloor\leq.25n_k\{n_k\alpha\}\leq.25$$ Or equivalently $$\{n_k\alpha\}\leq\frac{.5}{\sqrt{m_k}}$$ This guarantees $$|\{m_k\alpha\}-.25|\leq\{n_k\alpha\}\leq\frac{.5}{\sqrt{m_k}}$$ Which proves $Q_+$.