<s>[INST] Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I  showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z  - \frac{z^3}{3!} + \frac{z^5}{5!} +  \cdots   } =  \frac 1 {z\big(1  - (\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots )\big)} \\[10pt]
& = \frac{1}{z} \bigg( 1 + \bigg(\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots \bigg) +  \bigg(\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots \bigg)^2 + \cdots   \bigg) \\[10pt]
& = \frac{1}{z} +\frac{z}{3!}  + \bigg( \frac{1}{3!}- \frac{1}{5!}\bigg)z^3 + \cdots
\end{align}
This equality holds if and only if:
$$\bigg|\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots \bigg| < 1
\iff \bigg|\frac{\sin z}{z}  - 1\bigg| < 1$$
How can I show that the last inequality implies $|z| <  \pi$ ?
 [/INST] Since $\text{Res}\left(\frac{1}{\sin z},z=k\pi\right) = (-1)^m $ for any $k\in\mathbb{Z}$, we have
$$ \frac{1}{\sin z} = \frac{1}{z}+\sum_{m\geq 1}(-1)^m\left(\frac{1}{z-m\pi}+\frac{1}{z+m\pi}\right) \tag{1}$$
over any compact subset of $\mathbb{C}\setminus\pi\mathbb{Z}$. In particular:
$$ \frac{1}{\sin z}-\frac{1}{z} = 2z\sum_{m\geq 1}\frac{(-1)^{m+1}\frac{1}{m^2 \pi^2}}{1-\frac{z^2}{m^2\pi^2}}=2z\sum_{m\geq 1}z^{2m-2}\frac{\eta(2m)}{\pi^{2m}} \tag{2} $$
the radius of convergence of the Laurent series at the origin is $\pi$ (due to $\eta(2m)\approx 1$ for large values of $m$), as expected since $\pi$ is the distance from the origin of the closest singularities (simple poles at $z=\pm \pi$).
</s>