Evaluating the series with arctangents: $\sum_{r=1}^\infty \tan^{-1}\frac{2r}{2+r^2+r^4}$

Question

If $$S=\sum\limits_{r=1}^\infty\tan^{-1}\left(\frac{2r}{2+r^2+r^4}\right)$$ then what is cot S?

Options: A) 1; B) 3; C) 1/3; D) 2

Does it converge? I don't really know how to find the convergence point. I tried using Abel's translation or make substitutions, regroup the terms, but didn't find it. How is it done?

Answer 1

I think your series should be $\text{arccot}\left(\dfrac{2r}{2+r^2+r^4} \right)$ or $\arctan\left(\dfrac{2r}{2+r^2+r^4} \right)$. The main trick is the following. Note that \begin{align} \arctan\left(\dfrac{2r}{2+r^2+r^4} \right) & = \arctan\left(\dfrac{(r^2+r+1)-(r^2-r+1)}{1+(r^2+r+1)(r^2-r+1)} \right)\\ & = \arctan(r^2+r+1)-\arctan(r^2-r+1)\\ & = \arctan(r(r+1)+1)-\arctan((r-1)r+1)\\ \end{align} Hence, $$\sum_{r=0}^n \arctan\left(\dfrac{2r}{2+r^2+r^4} \right) = \arctan(n(n+1)+1)-\arctan(1)$$ Adapt this to your problem at hand.