Showing $f(x) = \frac{x^2}{\sin(x)}$ is analytic near $0$ Problem

Show the function 
$$
f(x) = \frac{x^2}{\sin(x)}
$$
is an analytic about $x=0$.


Try
We have 
$$
f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }
$$
Letting $f(x) = \sum_{n=0}^\infty a_n x^n $, we have
$$
a_0 = 0, a_1 = 1, a_2 = 0, a_3 = 1/6, a_4 = 0, a_5 = 7/360, \cdots
$$
thus $f(x) = x - x^3/6 + \cdots$.
Now we have to decide if $\sum_{n=0}^\infty a_n x^n$ is convergent near zero.
But I cannot see the rules of $a_0, a_1, a_2, \cdots$, so how should I proceed?

It is known that 

Assume $f: = \sum_{n=0}^\infty b_n x^n$ and $g:= \sum_{n=0}^\infty c_n x^n$ be convergent near zero with radius of convergence $\rho>0$. 
If $g(0) \neq 0$, we have

$$
\frac{f}{g} = \sum_{n=0}^\infty d_n
$$
for some radius of convergence $\le \rho$. So relating this fact to my problem, 
$$
f(x) = \frac{x}{1 - x^2/3! + x^4/5! - \cdots }
$$
may be convergent with radius of convergence $\le \infty$, which is the radius of $x^2$(poylnomial) and $\sin(x)$
 A: We have $f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }=\frac{x}{1 - x^2/3! + x^4/5! - \cdots }.$
The nominator and the denominator are power series with radius of convergenc $= \infty$. The power series in the denominator is $ \ne 0$ at $x=0.$ Hence there is $ r>0$ such that
$f(x)= \sum_{n=0}^\infty d_nx^n$ for $|x|<r.$
We have $r= \pi$. Why ?
A: $f(x)=\frac x {g(x)}$ where $g(x)=\frac {\sin\, x} x$ (for $x \neq 0$, $g(0)=1$). So what you have written as 'It is known that ....' immediately gives you the result. 
A: $f$ is not defined where $\sin x=0$. Elsewhere, it is the ratio of two analytic functions, hence it is analytic in $\mathbb R\setminus\{k\pi\}$.
Note that discussing analyticity at $x=0$ would be irrelevant.
A: 
A real function $f(x)$ is $R-analytic$ at some point $x_0$ if there exists an interval $(x_0-\epsilon,x_0+\epsilon)$ s.t. $f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ valid in $(x_0-\epsilon,x_0+\epsilon)$.

It is not so hard to see that $a_n=\frac{f^{(n)}(x_0)}{n!}$ is the Taylor's coefficient.    Since $lim_{x\rightarrow 0}f(x)=1$ which assures a removable singularity at $x=0$ (though $a_0=f(0)$ is not explicitly defined in the question). On defining $f(0)=1$ you will get a nice series in  non-negative powers of $x$ valid in$(-π,π)$, which shows $f(x)$ is $R-analytic$ at $x=0$.
