Summation of the series$\sum_{n=1}^\infty\frac{1}{n^2+4}$

Question

Evaluate the sum of the following series

$$\sum_{n=1}^\infty\frac{1}{n^2+4}$$

I saw a video in youtube where it is solved using complex analysis. What other method can be used to solve this?

Answer 1

A proof from scratch. We may notice that $$ \frac{1}{n^2+4}=\int_{0}^{+\infty}\frac{\sin(nx)}{n} e^{-2x}\,dx $$ by integration by parts, so $$ \sum_{n\geq 1}\frac{1}{n^2+4} = \int_{0}^{+\infty}W(x) e^{-2x}\,dx $$ with $W(x)$ being the $2\pi$-periodic extension of the function which equals $\frac{\pi-x}{2}$ on $(0,2\pi)$.
This Fourier series allows to state $$ \sum_{n\geq 1}\frac{1}{n^2+4} = \left(1+e^{-2\pi}+e^{-4\pi}+\ldots\right)\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-2x}\,dx $$ and ultimately $$ \sum_{n\geq 1}\frac{1}{n^2+4} = \frac{1}{1-e^{-2\pi}}\left[\frac{1}{8}e^{-2x}(1-2\pi+2x)\right]_{0}^{2\pi}=\frac{2\pi\coth(2\pi)-1}{8}. $$

Answer 2

After asking the same question to my prof, he get's this:

$$\sum_{n=1}^\infty\frac{1}{n^2+4}=$$ $$\frac{1}{8}\left(2\pi coth(2\pi )-1\right)=$$ $$\frac{1}{8}\left(2\pi \left(1+\frac{2}{-1+e^{4\pi}}\right)-1\right)=$$ $$\frac{1}{8}\left(2\pi +\frac{4\pi}{e^{4\pi}-1}-1\right)=$$ $$\frac{1+2\pi +2\pi e^{4\pi}-e^{4\pi}}{8e^{4\pi}-8}$$