How to calculate the series $\sum\limits_{n=1}^{\infty} \arctan(\frac{2}{n^{2}})$?

Note that $\arctan(u)-\arctan(v)=\arctan\left(\frac {u-v}{1+uv}\right)$. Taking $u=n+1$ and $v=n-1$ shows that $$\arctan\left(\frac {2}{n^2}\right)=\arctan(n+1)-\arctan(n-1)$$

Thus we see that the series telescopes and $$\sum_{n=1}^{\infty}\arctan\left(\frac {2}{n^2}\right)=2\arctan(\infty)-\arctan(0)-\arctan(1)=\pi -0-\frac {\pi}4=\frac {3\pi}4$$

\begin{align*} \sum_{n=1}^\infty\arctan\left ( \frac{2}{n^2} \right ) &=-arg \prod_{n=1}^\infty\left (1-\frac{2i}{n^2} \right ) \\ &=-arg \prod_{n=1}^\infty\left (1-\frac{(\sqrt{2i})^2}{n^2} \right ) \\ &=-arg\left(\frac{\sin(\pi\sqrt{2i})}{\pi\sqrt{2i}} \right ) \\ &=-arg\left(-\frac{(1/2+i/2)\sinh\left(\pi \right )}{\pi} \right ) \\ &= \frac{3\pi}{4} \end{align*}