$$ \lim_{n\to\infty}\sum_{r=0}^n \tan^{-1}\left(\frac{2r}{1+2r^4}\right)$$
So I really couldn't get started. I know that it is probably of form $$ \tan^{-1}(y) - \tan^{-1}(x) = \tan^{-1}\left(\frac{y-x}{1+xy}\right)$$ , but I couldn't bring it in that form. The denominator won't let me . Help please...
I'll leave it to you to prove by induction that the partial sum is $\arctan\left(1-\frac{1}{n^2+n+1}\right)$, so the limit is $\pi/4$. One approach to obtaining this partial sum is that of @achillehui's telescope, viz. $$\left[\arctan(2r^2+2r+1)\right]_{-1}^n=\arctan\frac{n^2+n}{n^2+n+1}.$$
Edit: just to spell it out, the definition $f(r):=\arctan(2r^2+2r+1)$ gives $$\sum_{r=0}^n\arctan\frac{2r}{1+2r^4}=\sum_{r=0}^n\left(f(r)-f(r-1)\right)=f(n)-f(-1),$$where the second $=$ can be proven by induction; this is the basis of any proof with telescoping series. In particular $$\lim_{n\to\infty}(f(n)-f(-1))=\arctan\infty-\arctan1=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{2}.$$