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 not an answer, but a useful way to transform the expression and link it to another, more simple sum:
$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\frac{\pi}{4}-\sum_{n=1}^{\infty} \left( \arctan \frac{1}{2n}-\arctan \frac{1}{2n+1}\right)$$
$$\arctan \frac{1}{2n}-\arctan \frac{1}{2n+1}=\arctan \frac{1}{4n^2+2n+1}$$

$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\frac{\pi}{4}-\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2+2n+1}$$


From this question (ananswered by the way) we have an identity:
$$\sum_{n=0}^{N} \arctan \frac{1}{n^2+n+1}=\arctan(N+1)$$
Which means
$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^2+n+1}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$
Considering:
$$(2n-1)^2+2n-1+1=4n^2-2n+1$$
We get another identity (separating even and odd terms):
$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^2+n+1}=\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2+2n+1}+\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2-2n+1}$$
And now we have:

$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2-2n+1}$$

The convergence of these two sums is slightly better than the original

If we take geometric mean of the last two expressions, it gives excellent convergence.
