<s>[INST] Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2  \right)
$$ 
I tried it lot and got a result but then stuck! (2)
(1) http://i.stack.imgur.com/26hA4.jpg
  (2) http://i.stack.imgur.com/g2vBb.jpg
 [/INST] One may observe that summing
$$
u_{r+1}-u_{r-1}
$$ may be simplified  as a telescoping sum:

$$
\sum_{r=1}^N\left(u_{r+1}-u_{r-1}\right)=\sum_{r=1}^N\left(u_{r+1}-u_{r}\right)+\sum_{r=1}^N\left(u_{r}-u_{r-1}\right)=u_{N+1}+u_N-u_1-u_0.
$$

From the identity
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2  \right)
$$ by telescoping you then obtain
$$
\begin{align}
\sum_{r=1}^N\arctan \left(\frac{4}{r^2+3} \right)&=\arctan \left(\frac{N+1}2 \right)+\arctan \left(\frac{N}2 \right)
\\\\&-\arctan \left(\frac12  \right)-\arctan \left(\frac{1-1}2  \right),
\end{align}
$$ giving, as $N \to \infty$ ,

$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)=2\arctan \left(\infty \right)-\arctan \left(\frac12 \right)=\frac{\pi}2+\arctan 2.
$$

</s>