Summation Contest Conceptual Problem-4

Using complex analysis we see that $\displaystyle f(z)=\frac{1}{(z^2+4)^2}$ has double poles at $z=\pm 2i$ . Since integrating around the upper half plane the pole $z=2i$ only comes into consideration so that,

$\displaystyle \text{Res(2i) } = \lim_{z\to 2i} \frac{d}{dz} \frac{\pi \cot (\pi z)}{(z+2i)^2} = -\frac{1}{32}(\pi\coth(2\pi)+2\pi^2 (\text{csch} (2\pi))^2)$

We deduce directly from summation lemma that : $\displaystyle \sum_{n=-\infty}^{\infty} \frac{1}{(n^2+2^2)^2} = \frac{1}{32}(\pi\coth(2\pi)+2\pi^2 (\text{csch}(2\pi))^2)$

Then simply considering it symmetric over the point $0$ the summation is reduced to,

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(n^2+4)^2} = \frac{1}{32}(-1+\pi\coth(2\pi)+2\pi^2 (\text{csch}(2\pi))^2)$