We can use the theorem of Mittag Leffler and some "art" to show that:

1. $\frac{{\pi }^2}{{\sin }^2(\pi z)}=\sum _{n=-\mathrm{\infty },n\ne 0}^{\mathrm{\infty }}\frac{1}{(z-n{)}^2}$
2. $\pi \cot (\pi z)=\frac{1}{z}+\sum _{n=1}^{\mathrm{\infty }}\frac{2z}{z^2-n^2}$