Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$

Question

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) https://i.stack.imgur.com/26hA4.jpg (2) https://i.stack.imgur.com/g2vBb.jpg

Answer 1

$$ \begin{align} \sum_{r=1}^\infty\arctan\left(\frac4{r^2+3}\right) &=\sum_{r=1}^\infty\left[\arctan\left(\frac2{r-1}\right)-\arctan\left(\frac2{r+1}\right)\right]\\ &=\lim_{n\to\infty}\left[\sum_{r=1}^n\arctan\left(\frac2{r-1}\right)-\sum_{r=3}^{n+2}\arctan\left(\frac2{r-1}\right)\right]\\ &=\lim_{n\to\infty}\left[\frac\pi2+\arctan(2)-\arctan\left(\frac2n\right)-\arctan\left(\frac2{n+1}\right)\right]\\[6pt] &=\frac\pi2+\arctan(2) \end{align} $$

Answer 2

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 $N \to \infty$ ,

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

Answer 3

$$\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r-1}2\right)=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=-1}^{N-2}\arctan\left(\frac{r+1}2\right)$$ $$=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\arctan\left(0\right)-\arctan\left(\frac{1}2\right)+\arctan\left(\frac{N}2\right)+\arctan\left(\frac{N+1}2\right)$$

Then let $N\rightarrow \infty$...

You then get $$\boxed {\pi -\arctan\left(\frac12\right)}$$

Answer 4

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: $$ \sum_{r\geq 1}\arctan\left(\frac{4}{r^2+3}\right) = \color{red}{\pi-\arctan\frac{1}{2}}.\tag{3}$$