Computation of the convergence radius of special power series As we know, for the computation of the convergence radius of the power series, we have two method, one is ${R=\lim _{n \rightarrow \infty }\left|{\frac {a_{n}}{a_{n+1}}}\right|}$,the other is $R = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}}$. But I find a special power series which is $$\sum_{n=1}^{\infty}\frac{\sin n}{n}x^n.$$ In this power series, $a_n$ is　$\frac{\sin n}{n}$，it is difficult to use either of the above methods to compute the convergence radius. I use the computer program, and the convergence radius seems to be 1 and the corresponding convergence domain is [-1,1], but I want to know how to compute it rigorously and analytically.
 A: One way to deal wit this  is to recall that $\{\sin n:n\in\mathbb{N}\}$ is a dense set in $[-1,1]$. This is a well known fact, a simple presentation can be found here. Also MSE has a few postings regarding this.
Let $n_k$ be a strictly increasing sequence of integers along which $\sin n_n\xrightarrow{k\rightarrow\infty}1$. Then, there is $k_0$ such that $\sin n_k\geq \frac12$ for all $k\geq k_0$. For such $k$
$$\lim_n\Big|\frac{\sin n_n}{n_k}\Big|^{1/n_k}\geq|\sin(1/2)|^{1/n_k} n^{-1/n_k}_k\xrightarrow{k\rightarrow\infty}1$$
This shows that $1\leq \limsup_n\Big|\frac{\sin n}{n}\Big|^{1/n}$. On the other hand, $\Big|\frac{\sin n}{n}\Big|\leq1$. Therefore
$$ \limsup_n\Big|\frac{\sin n}{n}\Big|^{1/n}=1$$

A related proof is obtain by Recalling that irrational rotations of the circle are dense in $\mathbb{S}^1$, that is, for any $\theta\in\mathbb{R}\setminus\mathbb{Q}$, $\{e^{i2\pi n\theta}:n\in\mathbb{N}\}$ is dense in $\mathbb{S}^1$.
Since $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ we have that
$$\frac{\sin n}{n}=\frac{e^{i2\pi n\frac{1}{2\pi}}-e^{-i2\pi n\frac{1}{2\pi}}}{2in}$$
Then along a sequence $n_k\nearrow\infty$, $e^{i2\pi n_k\frac{1}{2\pi }}\xrightarrow{k\rightarrow\infty}i$. Hence, for all $k$ large enough,
$1< |e^{i2\pi n\frac{1}{2\pi}}-e^{-i2\pi n\frac{1}{2\pi}}|\leq2$. Along the sequence $n_k$
$$
\Big|\frac{e^{i2\pi n_k\frac{1}{2\pi}}-e^{-i2\pi n_n\frac{1}{2\pi}}}{2in_n}\Big|^{1/n_k}\xrightarrow{k\rightarrow\infty}1$$
Consequently,
$$ \limsup_n\Big|\frac{\sin n}{n}\Big|^{1/n}=1$$
