Using Fourier Series to find the sum of a numerical series

Question

I have to use a Fourier series to compute the sum of the series $$\frac{1}{2} + \sum_{n=1}^\infty (-1)^n \frac{1}{n^2 + a^2}$$ My guesses are the Fourier series $$e^{ax} = \frac{e^{a\pi} - e^{-a\pi}}{\pi}[\frac{1}{2} + \sum_{n=1}^\infty (-1)^n \frac{1}{n^2 + a^2} (a\cos nx - n \sin nx)]$$ and $$\cosh x = \frac{2}{\pi}\sinh a\pi [\frac{1}{2} + \sum_{n=1}^\infty (-1)^n \frac{a}{n^2 + a^2} (\cos nx)]$$ I try to simplify each Fourier series expansion to get a $1$ in the numerator, but I always seem to get something else - such as an unwanted $a$ or $n$. Any useful hints?

Answer 1

Let $S$ be the series given by

$$S=\frac12 + \sum_{n=1}^\infty \frac{(-1)^n}{n^2 + a^2}\tag 1$$

Then, note that the Fourier series of $\cosh(ax)$ on the interval $[-\pi,\pi]$ is given by

$$\cosh(ax)=\frac{\sinh(a\pi)}{\pi a}\left(1+2a^2\sum_{n=1}^\infty \frac{(-1)^n}{n^2+a^2}\cos(nx)\right) \tag 2$$

Setting $x=0$ in $(2)$ and solving for the series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2 + a^2}$ reveals

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2 + a^2}=\frac{\pi}{2a\sinh(\pi a)}-\frac1{2a^2} \tag 3$$

Comparing $(1)$ and $(3)$, we obtain

$$\bbox[5px,border:2px solid #C0A000]{S=\frac{a^2-1}{2a^2}+\frac{\pi}{2a\sinh(\pi a)}}$$

And we are done!