finding $\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\arctan\left(\frac{2r}{1-r^2+r^4}\right)$

Question

I want to find $$L=\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\arctan\left(\frac{2r}{1-r^2+r^4}\right)$$ I already know that I need to split the expression $\frac{2r}{1-r^2+r^4}$ of the form $\frac{a_r+a_{r-1}}{1-a_ra_{r-1}}$ or $\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}$ so that the main expression gives $$\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\arctan\left(\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}\right)=\lim_{n\rightarrow\infty}\sum_{r=1}^{n}(\arctan(a_r)-\arctan(a_{r-1}))=\arctan(a_\infty)-\arctan(a_0)$$ But I am not able to split up that thing. Any hints on how to start? Thanks in advance.

Answer 1

Hint $$a_r=r(r+1)$$ $$\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}=\frac{2r}{1-r^2+r^4}$$