Show that $$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth\pi a+\frac{1}{2a^2}, a>0$$
I know I must use summation theorem and I calculated the residue which is:
$$Res\left(\frac{1}{z^2+a^2}, \pm ai\right)=-\frac{\pi}{2a}\coth\pi a$$
Now my question is: how do I get the last term $+\frac{1}{2a^2}$ after using the summation theorem?