Evaluate $\sum_{n=1}^{\infty}\arctan(\frac{2}{n^2})$?

Question

$$\begin{align} \sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2}\right)&=\sum_{n=1}^{\infty}\arctan\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)\\ &=\sum_{n=1}^{\infty}\arctan(n+1)-\arctan(n-1)\\ &= \frac\pi4 \end{align}$$

Is this correct?

Answer 1

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}

Answer 2

$$\sum^{\infty}_{n=1}\bigg(\tan^{-1}(n+1)-\tan^{-1}(n)\bigg)$$ $$+\sum^{\infty}_{n=1}\bigg(\tan^{-1}(n)-\tan^{-1}(n-1)\bigg)$$

Convert into Telescopic Series

$$=\tan^{-1}(\infty)-\tan^{-1}(1)+\tan^{-1}(\infty)-\tan^{-1}(0)=$$

$$=\pi-\frac{\pi}{4}=\frac{3\pi}{4}.$$