# Show that sum_(n=0)^infty 1/(n^2+a^2)=pi/(2a) coth pi a+1/(2a^2), a>0

Show that
$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{n}^{2}+{a}^{2}}=\frac{\pi }{2a}\mathrm{coth}\pi a+\frac{1}{2{a}^{2}},a>0$
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Camden Stanton
The method of residues applies to sums of the form
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)=-\sum _{k}{\text{res}}_{z={z}_{k}}\pi \mathrm{cot}\pi z\phantom{\rule{thinmathspace}{0ex}}f\left(z\right)$
where ${z}_{k}$ are poles of f that are not integers. So when f is even in n, you may express as follows:
$2\sum _{n=1}^{\mathrm{\infty }}f\left(n\right)+f\left(0\right)$
For this case, $f\left(z\right)=1/\left({z}^{2}+{a}^{2}\right)$ and the poles ${z}_{±}=±ia$ and using the fact that $\mathrm{sin}ia=i\mathrm{sinh}a$, we get
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{n}^{2}+{a}^{2}}=\frac{\pi }{a}\text{coth}\pi a$
The rest is just a little more algebra.