Evaluation of $\sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$ Find the following sum
$$S= \sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$$
I am not able to make it telescopic series. Could someone help me with this?

The $k$th partial sum is
\begin{align*}
\sum_{n = 1}^k (\tan^{-1} \sqrt{n} - \tan^{-1} \sqrt{n + 1}) &= (\tan^{-1} 1 - \tan^{-1} \sqrt{2}) + (\tan^{-1} \sqrt{2} - \tan^{-1} \sqrt{3}) \\
&\quad + (\tan^{-1} \sqrt{3} - \tan^{-1} \sqrt{4}) + \dots + (\tan^{-1} \sqrt{k} - \tan^{-1} \sqrt{k + 1}) \\
&= \tan^{-1} 1 - \tan^{-1} \sqrt{k + 1} \\
&= \frac{\pi}{4} - \tan^{-1} \sqrt{k + 1}.
\end{align*}As $k$ goes to infinity, $\tan^{-1} \sqrt{k + 1}$ approaches $\frac{\pi}{2},$ so the limit of the sum as $n$ goes to infinity is $\frac{\pi}{4} - \frac{\pi}{2} = \boxed{-\frac{\pi}{4}}.$
