Prove $\sum_{n=1}^{\infty}\arctan{\left(\frac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)\cdots\right)}$

Solution 1:

For any $\alpha > 0$, let $u + iv = \pi\sqrt{\alpha^2 + i}$, we have $$\begin{align} & \tan\left\{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{n^2+\alpha^2}\right)\right\} =\tan\left\{\sum_{n=1}^\infty\arg\left(1+\frac{i}{n^2+\alpha^2}\right)\right\}\\ = &\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{i}{n^2+\alpha^2}\right)\right\} =\tan\left\{\arg\prod_{n=1}^\infty \left( \frac{1+\frac{\alpha^2+i}{n^2}}{1+\frac{\alpha^2}{n^2}} \right)\right\}\\ = &\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{\alpha^2+i}{n^2}\right)\right\} =\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{(u+iv)^2}{n^2\pi^2}\right)\right\}\\ = & \tan\left\{\arg\left(\frac{\sinh(u+iv)}{u+iv}\right)\right\} =\tan\left\{\arg\left(\frac{\tanh u + i\tan v}{u+iv}\right)\right\}\\ = & \tan\left\{\arg(\tanh u + i\tan v) - \arg(u+iv)\right\}\\ = & \tan\left\{\tan^{-1}\left(\frac{\tan v}{\tanh u}\right)-\frac12\arg(\alpha^2 + i)\right\}\\ = & \tan\left\{\tan^{-1}\left(\frac{\tan v}{\tanh u}\right)-\frac12\tan^{-1}\left(\frac{1}{\alpha^2}\right)\right\} \end{align}$$ For $\alpha = 1$, we have $u + iv = \pi\sqrt{\frac{\sqrt{2}+1}{2}} + i\pi\sqrt{\frac{\sqrt{2}-1}{2}}$, so

$$\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{n^2+1}\right) = \tan^{-1}\left(\frac{\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)}{\tanh\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)}\right)- \frac{\pi}{8} + N \pi$$

for some integer $N$ to be determined. Numerically, the RHS excluding the unknown term $N\pi$ is about $1.0373$. On the LHS, we know it is a number around $1$. So the unknown constant $N$ is $0$ and we are done.

Solution 2:

Here is a closed form

$$ -\frac{1}{4}-\frac{1}{4}\,{\frac {\pi \,\cot \left( \pi \,\sqrt {-1+i} \right) }{\sqrt {-1 +i}}}-\frac{1}{4} \,{\frac {\pi \,\cot \left( \pi \,\sqrt {-1-i} \right) }{ \sqrt {-1-i}}}\sim 0.9676963204 .$$

Maybe one can simplify further.