TITLE: How to simplify $\lim_{n\to \infty}\sum_{r=1}^n \tan^{-1} \dfrac{2r+1}{r^4+2r^3+r^2+1}$
QUESTION [5 upvotes]: $$\lim_{n\to \infty}\sum_{r=1}^n \tan^{-1} \dfrac{2r+1}{r^4+2r^3+r^2+1}$$
How am I supposed to do it? One thing I see here is 
$$\lim_{n\to \infty}\sum_{r=1}^n \tan^{-1} \dfrac{2r+1}{(r^2+r)^2+1}$$
Here derivative of $r^2+r$ is $2r+1$. (if it helps)
It's final answer is

 $\pi/4$

REPLY [2 votes]: You aren't too far from the answer. Credit on discovering what $(r+1)^2$ was. 
Write $r^2+r$ as $r(r+1)$. Squaring that would give you $r^2$ and $(r+1)^2$. And the difference of those two terms is the numerator. Now, it's in the form of $\tan(A-B)$. 
Now, telescope this sum which becomes 
$\arctan(r)- \arctan(1)$
$\pi/2 - \pi/4 = \pi/4$