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
 A: 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 $N \to \infty$ ,

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

A: $$
\begin{align}
\sum_{r=1}^\infty\arctan\left(\frac4{r^2+3}\right)
&=\sum_{r=1}^\infty\left[\arctan\left(\frac2{r-1}\right)-\arctan\left(\frac2{r+1}\right)\right]\\
&=\lim_{n\to\infty}\left[\sum_{r=1}^n\arctan\left(\frac2{r-1}\right)-\sum_{r=3}^{n+2}\arctan\left(\frac2{r-1}\right)\right]\\
&=\lim_{n\to\infty}\left[\frac\pi2+\arctan(2)-\arctan\left(\frac2n\right)-\arctan\left(\frac2{n+1}\right)\right]\\[6pt]
&=\frac\pi2+\arctan(2)
\end{align}
$$
A: $$\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r-1}2\right)=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=-1}^{N-2}\arctan\left(\frac{r+1}2\right)$$
$$=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\arctan\left(0\right)-\arctan\left(\frac{1}2\right)+\arctan\left(\frac{N}2\right)+\arctan\left(\frac{N+1}2\right)$$
Then let $N\rightarrow \infty$...
You then get $$\boxed {\pi -\arctan\left(\frac12\right)}$$
A: 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:
$$ \sum_{r\geq 1}\arctan\left(\frac{4}{r^2+3}\right) = \color{red}{\pi-\arctan\frac{1}{2}}.\tag{3}$$
