Prove $\sum_{n=1}^{\infty}\arctan{\left(\frac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)\cdots\right)}$ 
show that:
  $$\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdot\dfrac{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}+e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}-e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}\right)}-\dfrac{\pi}{8}$$

This relust is nice.(maybe have some wrong)
becasue I know this famous problem
$$\sum_{n=1}^{\infty}\arctan{\dfrac{2}{n^2}}=\dfrac{3\pi}{4}$$
and follow AMM( E3375) problem
$$\sum_{n=1}^{\infty}\arctan{\dfrac{1}{n^2}}=\arctan{\left(\dfrac{\tan{\frac{\pi}{\sqrt{2}}}-th{\dfrac{\pi}{\sqrt{2}}}}{\tan{\dfrac{\pi}{\sqrt{2}}}+th{\dfrac{\pi}{\sqrt{2}}}}\right)}$$
Follow is AMM solution:
My try:  my problem I want use this methods,But at last failure it.
 Thank you  for you help.
This problem is similar this:we konw this

$$\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$
  and then little hard problem:
  $$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{1}{2}(\pi\coth{\pi}-1)$$

 A: 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.
A: 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.
