TITLE: Prove $\sum_{n=1}^{\infty}\arctan{\left(\frac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)\cdots\right)}$
QUESTION [15 upvotes]: 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:
  $$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{1}{2}(\pi\coth{\pi}-1)$$

REPLY [3 votes]: 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.