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

We already have plenty of slick answers through creative telescoping, so I will go for the overkill. 
By crude estimations we have $\sum_{r\geq 3}\frac{4}{r^2+3}\leq\frac{\pi}{2}$, hence:
$$ \sum_{r\geq 3}\arctan\left(\frac{4}{r^2+3}\right) = \text{Arg}\prod_{r\geq 3}\frac{r^2+3+4i}{r^2}\tag{1}=\text{Arg}\prod_{r\geq 3}\left(1+\frac{(2+i)^2}{r^2}\right)$$
but due to the Weierstrass product for the $\sinh$ function:
$$\prod_{r\geq 1}\left(1+\frac{(2+i)^2}{r^2}\right)=\frac{\sinh(2\pi+\pi i)}{\pi(2+i)}=-\frac{\sinh(2\pi)}{\pi(2+i)}\tag{2}$$
and that leads to:
$$ \sum_{r\geq 1}\arctan\left(\frac{4}{r^2+3}\right) = \color{red}{\pi-\arctan\frac{1}{2}}.\tag{3}$$
