Limit of $\left|\sin(n)\right|^{1/n}$

I'm having trouble showing rigorously what is the limit of $x_n=|\sin(n)|^{1/n}$ in a rigorous manner. What I have shown is that, $x_n$ cannot converge to $0$ and is bounded by $1$, and that should suffice to show that $x_n$ effectively converges to $1$.

However, I can't figure out how to formalize this proof, and show it in a rigorous manner. My guess would be to try and show that the limit of $|a_n|^{1/n}$ can be $1$ if $|a_n|$ is bounded by $1$ and does not converge to $0$. I don't know if this more general statement holds, and if it would simplify or complexify the problem.

Hei,

the idea is to bound $\sin(n)$ for $n\in \mathbb{N}$ from below in such a way that you see that $\sin(n)$ is so far away from $0$ that $\left|\sin(n)\right|^\frac{1}{n}$ goes to $1$. Therefore we have to show that natural numbers have a certain distance to multiples of $\pi$.

For this, you can use the fact that $\pi$ is not a Liouville number (see http://mathworld.wolfram.com/LiouvilleNumber.html).

So, there is an $n_o\in\mathbb{N}$ such that $\left|\pi-\frac{p}{q}\right|\geq \frac{1}{q^{n_o}}$ for all $p,q\in \mathbb{N}$, or, equivalently $\left|q\pi-p\right|\geq \frac{1}{q^{n_o-1}}$.

Now choose $p=n$, and $q$ in a way that $q\pi$ is close to $n$, i.e. $q\in[\frac{n-\frac{\pi}{2}}{\pi}, \frac{n+\frac{\pi}{2}}{\pi}]$.

As now $q\leq \frac{n+\frac{\pi}{2}}{\pi}$ and $\left|q\pi-p\right|\leq\frac{\pi}{2}$, and as for $x\in[0,\frac{\pi}{2}]$ there is the estimate $\sin(x)\geq \frac{x}{2}$, we get the following series of inequalities: $$ |\sin(n)|=|\sin(q\pi-n)|=\sin|q\pi-n|\geq \frac{1}{2}|q\pi-n|\geq\frac{1}{2q^{n_o-1}}\geq \frac{1}{2}\cdot\left(\frac{\pi}{(n+\frac{\pi}{2})}\right)^{n_o-1}. $$ Taking the $n$-th root, we obtain $$ |\sin(n)|^{\frac{1}{n}}\geq \frac{1}{2^{\frac{1}{n}}}\cdot\left(\frac{\pi^{\frac{1}{n}}}{(n+\frac{\pi}{2})^{\frac{1}{n}}}\right)^{n_o-1}. $$ As the limit of $n^{\frac{1}{n}}$ for $n\rightarrow\infty$ is $1$ and $n_o$ is fixed, the right hand side goes to $1$ for $n\rightarrow\infty$. As the left hand side is bounded from above by $1$ aswell, it has to converge to $1$.