I found some interesting series ,for some examples $$ \sum_{n=0}^{\infty}{\left( -1 \right) ^n\mathrm{arc}\tan \left( \tanh \left( \left( 2n+1 \right) \pi \right) \right)}=\frac{1}{2}\mathrm{arc}\tan \left( \sqrt{2\sqrt{3\sqrt{2}-4}} \right) $$ $$ \sum_{n=0}^{\infty}{\left( -1 \right) ^n\mathrm{arc}\tan \left( \sinh \left( 2n+1 \right) \pi \right)}=\frac{\mathrm{arc}\cos \left( 3-2\sqrt{2} \right)}{2} $$ $$ \sum_{n\in \mathbb{Z}}{\left( -1 \right) ^n\mathrm{arc}\tan \left( \frac{\sinh \left( n\pi \right) -i\cosh \left( n\pi \right)}{\sqrt{2}} \right)}=-\mathrm{arc}\cos \left( \sqrt{1+\frac{1}{\sqrt{2}}} \right) $$
$$ \sum_{n=0}^{\infty}{\left( -1 \right) ^n\mathrm{arc}\tan \left( \left( \sqrt{2}-1 \right) \tanh \left( \left( n+\frac{1}{2} \right) \sqrt{3}\pi \right) \right)}=\frac{1}{2}\mathrm{arc}\tan \left( \frac{\left( 2+\sqrt{3} \right) ^{\frac{3}{8}}}{\sqrt{\frac{2\left( \sqrt{2\left( \sqrt{\sqrt{3}+2}+2 \right)}+2 \right)}{\sqrt{\sqrt{\sqrt{3}+2}+2}-\sqrt[4]{\sqrt{3}+2}}}} \right) $$ How do we prove that they converge,are these new results？
The first one doesn't seem to converge.
The tanh term tends to 1, so the arctan tends to $\pi/4$.
According to Wolfram Alpha, the sum up to odd $n$ tends to about -0.00186743404 and the sum up to even $n$ tends to about 0.7835307293489.
I think the same holds for the other sums.