To evaluate a certain limit, I need to calculate the first term in the regular part of the Laurent expansion of the function $$\frac{\pi}{\sin \frac{\pi(s+1)}{2}}$$ around -1 (should be the same thing as the regular part of $\frac{\pi}{\sin \frac{\pi s}{2}}$ with $s\to (s+1)$ in the argument, I guess). The principal part of the series is obvious. However, the only two ways I know to calculate the regular part is to calculate recursively the multiplicative inverse to the series $$\frac{1}{\pi} \sin \frac{\pi s}{2}=\frac{1}{\pi}\sum_{k=0}^{\infty}(-1)^{k} \left(\frac{\pi s}{2}\right)^{2k+1}\frac{1}{(2k+1)!}$$ using the fact that $$\left(\frac{1}{\pi}\sum_{k=0}^{\infty}(-1)^{k} \left(\frac{\pi s}{2}\right)^{2k+1}\frac{1}{(2k+1)!} \right)\left( \frac{2}{s} + c_{0} + c_{1}s + c_{2} s^2 + \cdots \right)=\left( \frac{0}{s} + 1 + 0 \cdot s + 0 \cdot s^2 + \cdots \right)$$
or to calculate the $k$-th coefficient by definition, using the contour integral.
I wonder if there is a faster way to calculate the regular part of the Laurent series. People are usually concerned only with calculating the principal part, so I have not found much material on the topic.
By Mittag-Leffler's theorem, we have the expansion
\begin{aligned} {\pi\over\sin(\pi z)} &=\frac1z+\sum_{n\ge1}(-1)^n\left({1\over z-n}+{1\over z+n}\right) \\ &=\frac1z+\sum_{n\ge1}(-1)^n{2z\over z^2-n^2}. \end{aligned}
When $|z|<1$, the properties of geometric series indicates that
$$ {1\over z^2-n^2}=-{1\over n^2}\cdot{1\over1-z^2/n^2}=-\sum_{k\ge0}{z^{2k}\over n^{2k+2}}. $$
Consequently, we have
\begin{aligned} \sum_{n\ge1}(-1)^n{2z\over z^2-n^2} &=2\sum_{n\ge1}(-1)^{n-1}\sum_{k\ge0}{z^{2k+1}\over n^{2k+2}} \\ &=2\sum_{k\ge0}z^{2k+1}\sum_{n\ge1}{(-1)^{n-1}\over n^{2k+2}}. \end{aligned}
Using the identity
$$ \sum_{n\ge1}{(-1)^{n-1}\over n^s}=(1-2^{1-s})\zeta(s), $$
we conclude that
$$ {\pi\over\sin(\pi z)}=\frac1z+\sum_{k\ge0}(2-2^{-2k})\zeta(2k+2)z^{2k+1}. $$