<s>[INST] Evaluate $\sum_{n=1}^{\infty}\arctan(\frac{2}{n^2})$? $$\begin{align}
\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2}\right)&=\sum_{n=1}^{\infty}\arctan\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)\\
&=\sum_{n=1}^{\infty}\arctan(n+1)-\arctan(n-1)\\
&= \frac\pi4
\end{align}$$
Is this correct?
 [/INST] The series telescopes. The partial sum $S_N$ is equal to
\begin{align}
S_N &= [\arctan (2) - \arctan (0)] + [\arctan (3) - \arctan (1)] + \cdots\\
& \qquad \cdots  + [\arctan (N) - \arctan (N - 2)] + [\arctan(N + 1) - \arctan (N - 1)]\\
&= - \arctan (0) - \arctan (1) + \arctan (N) + \arctan (N + 1)\\
&= -\frac{\pi}{4} + \arctan (N) + \arctan (N + 1)
\end{align}
So for the sum $S$ we have
\begin{align}
S &= \lim_{N \to \infty} S_N\\
&= \lim_{N \to \infty} \left [- \frac{\pi}{4} +  \arctan(N) + \arctan(N + 1) \right ]\\
&= -\frac{\pi}{4} + \frac{\pi}{2} + \frac{\pi}{2}\\
&= \frac{3 \pi}{4}.
\end{align}
</s>