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)

• Sorry due to low reputation , i cannot post image – user123733 Jul 25 '16 at 13:26
• Are you Ok with my edit? Let me know. – Olivier Oloa Jul 25 '16 at 13:36
• @OlivierOloa yes thank you – user123733 Jul 25 '16 at 13:37

\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}

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 $N \to \infty$ ,

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

• but how does other terms get cancelled – user123733 Jul 25 '16 at 13:40
• @user123733 Like this $(u_2-u_0)+(u_3-u_1)+(u_4-u_2)+\cdot+(u_N-u_{N-1})$, do you see the terms cancel? – Olivier Oloa Jul 25 '16 at 13:42
• @OlivierOloa are you sure? Shouldn't it be $\pi -\tan ^{-1}\left(\frac{1}{2}\right)$? – Pierpaolo Vivo Jul 25 '16 at 13:44
• @OlivierOloa then one term will left – user123733 Jul 25 '16 at 13:44
• Now this looks correct. (+1) – robjohn Jul 25 '16 at 14:13

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

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

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

• How r=-1 in your first step – user123733 Jul 25 '16 at 13:45
• @user123733 set for example $u+1=r-1$ then $u=r-2$ then for $r=1$ you have $u=-1$ – GeorgSaliba Jul 25 '16 at 13:46
• But everyone is getting another result – user123733 Jul 25 '16 at 13:54
• @user123733 it is the same result... – GeorgSaliba Jul 25 '16 at 13:54
• There it is pi/2 – user123733 Jul 25 '16 at 13:55

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