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.
 A: 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 $1$. 
A: I am not quite sure that this sequence converges. 
Let $ a_n = |\sin(n)|^{1/n}$. First I would like to give a reference to the following paper 
http://www.jstor.org/stable/2688681?seq=1#page_scan_tab_contents . This paper asserts that the set of limit points of $\sin(n)$ is equal to $[-1,1]$. It would be apparent from here that the set of limit points of the absolute value of this function would be equal to $[0,1]$.
Consider $b_n = \log a_n = \frac{\log(|\sin(n)|)}{n}$. If we first look at the subsequences of $b_{n_k}$ of $b_n$ for which $a_{n_k}$ has a limit point of $y\in (0,1]$, then $b_{n_k}\rightarrow 0$ which would mean that the corresponding $a_{n_k}$'s would converge to $1$. 
On the other hand, when we consider the subsequences of $b_n$ for which $a_{n_k}\rightarrow 0$, then we would need to apply L'Hopital rule to figure out the limiting value of $b_{n_k}$.
Let us abstract ourselves from the problems with absolute value as they would only correspond to minor technical details, and let us consider $b_{n_k}$ without the absolute values and take derivatives with respect to $n_k$ of both numerator and the denominator. This would get us 
$$
b_{n_k}\rightarrow -\frac{\cos(n_k)}{\sin(n_k)}.
$$
Since $\sin(n_k)$ converge to $0$, we would have $\cos(n_k)$ coverging to $1$. This would imply that $b_{n_k}\rightarrow -\infty$. This in turn allows us to conclude that $a_{n_k} = e^{b_{n_k}} \rightarrow 0$.
As a result, $a_n$ has two limit points $0$ and $1$, thus it is not convergent.
