A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for the value of  $$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \left(\frac{1}{n} \right)? $$
Unlike the other infinite series, this infinite series only converges conditionally.

This is just a possible way to proceed, and doesn't provide in any way a complete solution to the problem, thus I will put it as community wiki, and pray you to contribute if you have any insights.
The series converges (check!), inserting the Taylor series for $\arctan(x)$:
$$\begin{array}{ll}\sum_{n=1}^\infty(-1)^{n+1}\arctan\left(\frac{1}{n}\right)&=\sum_{n=1}^\infty(-1)^{n+1}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\frac{1}{n^{2k+1}}\\&=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n^{2k+1}}\\&=\sum_{k=0}^\infty\frac{(-1)^k(1-2^{-2k})}{2k+1}\zeta(2k+1)\end{array}$$
where in the second equality we have used the fact that everything converges, and in the third we noticed that the second series in the second line is the Dirichlet eta function.
I don't know if this leads anywhere, but since the zeta function has been studied pretty heavily maybe there's some results somewhere that we can use to finish from here.
