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How would one go about finding the value of this summation? ${\sum_{n=1}^{\infty}}\frac{(-1)^{n+1}}{n^2+1}$

Riemann'sPointyNose
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    Is there any reason to imagine that it has a pleasant closed formula? – lulu May 30 '20 at 20:22
  • Since this is a Leibniz series then the series converges and the sum is bounded by $|S|\leq \frac{1}{2}$ (the first element). But other than that im not sure you can get a number. – Eminem May 30 '20 at 20:24
  • @lulu I saw on Wikipedia that ${\sum_{n=1}^{\infty}\frac{1}{n^2+a^2}}$ for ${a \in \mathbb{R}}$ had a nice closed form, so I just wondered if the alternating one had any known closed form – Riemann'sPointyNose May 30 '20 at 20:29
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    I believe it should be equal to$$\sum_{n=1}^\infty\frac1{n^2+1}-\frac12\sum_{n=1}^\infty\frac1{n^2+(1/2)^2}=\frac12\left(1-\frac{\pi}{\sinh{(\pi)}}\right)$$ – Peter Foreman May 30 '20 at 20:56

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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 $s=ia$ $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n^2+a^2} = -\frac{1}{2a^2}\left(\frac{\pi i a}{\sin(\pi i a)}-1\right) $$

$$\boxed{\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n^2+1} = \frac{1}{2}\left(1-\frac{\pi}{\sinh(\pi)}\right)}$$

Stefan Lafon
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