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?

"Recall" this trigonometric identity:
$$
\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}
$$
Now think about applying that:
\begin{align}
& \arctan \frac{an+b}{cn+d} + \arctan \frac{en+f}{gn+h} \\[10pt]
= {} & \arctan \frac{\dfrac{an+b}{cn+d} + \dfrac{en+f}{gn+h}}{1 - \dfrac{an+b}{cn+d} \cdot \dfrac{en+f}{gn+h}} \\[10pt] = {} & \arctan \frac{(an+b)(gn+h) + (en+f)(cn+d)}{(cn+d)(gn+h) - (an+b)(en+f)} \\[10pt]
= {} & \arctan \frac{(ag+eh)n^2 + (bg+ah+ed+fe)n + (bh + fd)}{(cg-ae)n^2 + (ch+dg-af-be)n + (dh-bf)}
\end{align}
You'd want to choose $a,b,c,d,e,f,g,h$ in such a way that $ag+eh=0,$ $cg-ae=1,$ etc.
Then you've got two terms, $\dfrac{an+b}{cn+d} + \arctan \dfrac{en+f}{gn+h},$ where originally you had one.
