calculating the sum of a series I was wondering if there is a way to use complex-analysis to solve the sum of the following series? (just like you can use it to solve integrals of some kinds (using fourier's transform for integrals).

$$\sum^{\infty}_{n=0}\frac{1}{n^2+25}$$

and

$$\sum^{\infty}_{n=0}\frac{(-1)^n}{n^2+25}$$

And if not, how to solve them otherwise?
thanks in advance!

We can evaluate the two sums using following expansions of trigonometry functions
which can be proved using Mittag Leffler's theorem:
$$
\frac{\cos x}{\sin x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n\pi}
\quad\text{ and }\quad
\frac{1}{\sin x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{(-1)^n}{x+n\pi}
$$
Let's look at the expansion at the left first. Substitute $x$ by $\pi y i$, move the term for $n = 0$ and group the terms for $+n$ and $-n$ togather, we get
$$\frac{\cos(\pi y i)}{\sin(\pi y i)} - \frac{1}{\pi y i} = \frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{1}{y i + n} + \frac{1}{yi - n}\right)
= \frac{2 y}{\pi i}\sum_{n=1}^\infty\frac{1}{y^2+n^2}$$
This leads to
$$\sum_{n=0}^\infty \frac{1}{y^2+n^2} = \frac{1}{y^2} + \frac{\pi}{2y}\left( \frac{\cosh(\pi y)}{\sinh(\pi y)} - \frac{1}{\pi y}\right)
= \frac{1}{2y^2} + \frac{\pi}{2y}\frac{\cosh(\pi y)}{\sinh(\pi y)}
$$
The evaluation of the second sum is pretty similar, you get something like
$$
\sum_{n=0}^\infty \frac{(-1)^n}{y^2+n^2} = \frac{1}{2y^2} + \frac{\pi}{2y} \frac{1}{\sinh(\pi y)}
$$
As a result, one get
$$
\sum_{n=0}^\infty \frac{1}{25+n^2} = \frac{1}{50} + \frac{\pi}{10}\frac{\cosh(5\pi)}{\sinh(5\pi)}
\quad\text{ and }\quad
\sum_{n=0}^\infty \frac{(-1)^n}{25+n^2} = \frac{1}{50} + \frac{\pi}{10}\frac{1}{\sinh(5\pi)}$$
