<s>[INST] Why is the reciprocal of the power series $1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots$ equal to $1+\frac{z^2}{6}+\frac{7z^4}{360}+\cdots$? \begin{align}
\frac1{\sin(z)}
&=\frac1z\frac{z}{\sin(z)}\\
&=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\
&=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\
&=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots
\end{align}
In particular, from lines 2 to 3, what is happening? (eg. why does reciprocating result in a change of signs?)
 [/INST] The Taylor series of $\frac{\sin z}{z}$ at the origin is straightforward to compute, the Taylor series of the reciprocal function $\frac{z}{\sin z}$ a bit less. However, we may notice that $\frac{z}{\sin z}$ is an even meromorphic function with simple poles at the elements of $\pi\mathbb{Z}\setminus\{0\}$. Additionally
$$ z\prod_{n\geq 1}\frac{1-\frac{z^2}{n^2 \pi^2}}{1-\frac{4z^2}{(2n-1)^2\pi^2}} =\tan(z)\tag{1}$$
leads to:
$$ \frac{1}{\sin z}=\frac{d}{dz}\log\tan\frac{z}{2}\\=\frac{1}{z}+\sum_{n\geq 1}\left(\frac{1}{\pi -2 n \pi -z}+\frac{1}{-2 n \pi +z}-\frac{1}{\pi -2 n \pi +z}+\frac{1}{2 n \pi +z}\right) \tag{2}$$
which simplifies into
$$ \frac{z}{\sin z}=1+2z^2\sum_{n\geq 1}\left(\frac{1}{\left[(2n-1)\pi\right]^2-z^2}-\frac{1}{\left[2n \pi\right]^2-z^2}\right)\tag{3}$$
then into:
$$\begin{eqnarray*} \frac{z}{\sin z}&=&1+2\sum_{n\geq 1}\sum_{h\geq 0}\frac{z^{2h+2}}{\pi^{2h+2}}\left(\frac{1}{(2n-1)^{2h+2}}-\frac{1}{(2n)^{2h+2}}\right)\\&=&1+\sum_{h\geq 0}\frac{z^{2h+2}}{\pi^{2h+2}}\zeta(2h+2)\left(2-\frac{1}{4^{h}}\right)\end{eqnarray*}\tag{4}$$
from which it is clear that the Taylor series of $\frac{z}{\sin z}$ at the origin only has non-negative coefficients.
That can be seen also as a consequence of $\frac{1}{\sin z}=\cot(z/2)-\cot(z)$ and the well-known identity
$$ \sum_{h\geq 1}\zeta(2h)z^{2h}=\frac{1-\pi z\cot(\pi z)}{2}.\tag{5} $$
</s>