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

Question

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

Answer 1

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 $\sum_{r=1}^{\infty}\arctan(\frac{n}{n+1}(r+1))-\arctan(\frac{nr}{n+1})=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

Answer 2

$cot^{-1}(\frac {n^{2} (r^{2}+r+1)+2n+1} {n^{2}+n}) \to \frac {\pi} 4$ as $n \to \infty$ for each $r$ so the required limit is $\infty$. [ Look at the sum from $r=1$ to $N$ for fixed $N$ and then let $N \to \infty$].