Calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$

Question

$$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
I tried $$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\leq\frac{n}{\sqrt{n^2+n+1}}$$
But$$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+n}}\right)^n=\lim_{n\to\infty}\left(1+\frac2n\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac2n)}=\frac1e$$
And in the same way,i got $$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+1}}\right)^n=\lim_{n\to\infty}\left(1+\frac1n+\frac{1}{n^2}\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac1n+\frac{1}{n^2})}=\frac{1}{\sqrt e}$$
So i only got $$\frac1e\leq I\leq\frac{1}{\sqrt e}$$
Could someone help me get the value of $I$. Thanks!

Answer 1

The natural log of this expression is $$ K= n\ln( 1+ (x-1) )$$ where $$x=\sum_{r=1}^n \frac{1}{\sqrt{n^2+n+r}} $$ Due to the bounds you have shown, $K$ must tend to a finite non-zero number, which can only happen if $x\to 1$. Now, $$K = n(x-1) +n O((x-1)^2)$$ The first term is $$n \sum_1^n \left( \frac{1}{\sqrt{n^2+n+r}} -\frac 1n \right) \\ = \sum_1^n \left( \frac{1}{\sqrt{ 1+\frac 1n +\frac{r}{n^2}}} -1 \right) \\ = \sum_1^n \left( -\frac 12 \left( \frac 1n +\frac{r}{n^2} \right) +O(\frac{1}{n^2}) \right) \\ = -\frac 12 -\frac{n(n+1)}{4n^2} +O(\frac 1n ) \\ \to -\frac 12 -\frac 14 \\ =-\frac 34$$ The rest of the terms in $K$ can be shown to go to zero in a similar way. Hence, your limit is $$\Large{ \color{blue}{e^{-\frac 34 }} }$$