TITLE: How would one find the value of this summation?
QUESTION [0 upvotes]: How would one go about finding the value of this summation? ${\sum_{n=1}^{\infty}}\frac{(-1)^{n+1}}{n^2+1}$

REPLY [1 votes]: You can verify that if $s\notin \mathbb Z$,
$$\cos{st} = \frac{\sin{\pi s}}{\pi s} \left [1+2 s^2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \cos{n t}}{n^2-s^2} \right ]$$
So if $t=0$, $$1 = \frac{\sin{\pi s}}{\pi s} \left [1+2 s^2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n^2-s^2} \right ]$$
which gives, with $s=ia$
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n^2+a^2} = -\frac{1}{2a^2}\left(\frac{\pi i a}{\sin(\pi i a)}-1\right) $$
$$\boxed{\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n^2+1} = \frac{1}{2}\left(1-\frac{\pi}{\sinh(\pi)}\right)}$$
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