For any real number $x$, $\tan^{-1}(x)$ denote unique real number $\theta$ in (-$\pi\over2$, $\pi\over2$) such that $\tan\theta=x$. Then evaluate
$\lim_{n\to\infty}$ $\sum_{m=1}^n \tan^{-1}$$1\over(1+m+m^2)$
My approach: I separated the above to $\tan^{-1}(m+1)-\tan^{-1}(m)$
But I could not proceed further. I sense that it needs to be converted into a Riemann integral, however I am not being able to do that.