<s>[INST] 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 $\sin(x)$
 [/INST] 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 ?
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