Finding $\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\cot^{-1}\bigg(\frac{n^2(r^2+r+1)+2n+1}{n^2+n}\bigg)$ 
Finding $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\cot^{-1}\bigg(\frac{n^2(r^2+r+1)+2n+1}{n^2+n}\bigg)$$

Plan
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\tan^{-1}\bigg(\frac{n^2+n}{1+n^2r^2+nr+n^2+2n}\bigg)$$
I am trying to convert it into $$\tan^{-1}\bigg(\frac{x-y}{1+xy}\bigg)=\tan^{-1}(x)-\tan^{-1}(y)$$
But not find any way
How do i solve it Help me please

Doing some basic manipulation we get the following terms $cot^{-1}(\frac{n(r^2+r)}{n+1}+\frac{(n+1)}{n})$ . Now I let $\frac{n}{n+1}=a$ so we have $cot^{-1}(a(r^2+r)+\frac{1}{a})=\arctan(\frac{a}{a^2(r^2+r)+1})$ we have $y=ar+a,x=ar$ thus it becomes $\arctan(ar+a)-\arctan(ar)$ thus sum is $\sum_{r=1}^{\infty}\arctan(\frac{n}{n+1}(r+1))-\arctan(\frac{nr}{n+1})=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$
