How to calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$ $$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
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!
 A: I use the Binomial expansion repeatedly without using logs, as in Tavish's answer. I just want to show that such a method works for this question. The method is long, but it is neither difficult to understand nor ugly.
Write $\ t = n^2 + n.\ $ Then we have
\begin{align}\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\ldots+\frac{1}{\sqrt{n^2+n+n}}\\ 
\\
= \frac{1}{\sqrt{t+1}}+\frac{1}{\sqrt{t+2}}+\ldots+\frac{1}{\sqrt{t+n}}\\
\\
=\frac{1}{\sqrt{t}}\left( \frac{1}{\sqrt{1+\frac{1}{t}}} + \frac{1}{\sqrt{1+\frac{2}{t}}} +\ldots + \frac{1}{\sqrt{1+\frac{n}{t}}} \right)\\
\\
=\frac{1}{\sqrt{t}}\left( \left(1+\frac{1}{t}\right)^{-\frac{1}{2}} + \left(1+\frac{2}{t}\right)^{-\frac{1}{2}} +\ldots + \left(1+\frac{n}{t}\right)^{-\frac{1}{2}} \right)\\
\\
=\frac{1}{\sqrt{t}}\left(\left[ 1+\left(-\frac{1}{2}\right)\left(\frac{1}{t}\right) + \ldots \right] + \left[ 1+\left(-\frac{1}{2}\right)\left(\frac{2}{t}\right) + \ldots \right] +\ldots + \left[ 1+\left(-\frac{1}{2}\right)\left(\frac{n}{t}\right) + \ldots \right] \right)\\
\\
=\frac{1}{\sqrt{t}}\left( 1\times n + \left(-\frac{1}{2}\right)\times\left(\frac{1}{t}\right)\times \left( \displaystyle\sum_{i=1}^n i \right) + O\left(\frac{1}{n}\right) \right)\\
\\
=\frac{1}{\sqrt{n^2+n}} \left( n + \frac{\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)n(n+1)}{n^2+n} + O\left(\frac{1}{n}\right) \right) \\
\\
=\frac{n-\frac{1}{4}+O\left(\frac{1}{n}\right)}{\sqrt{n^2+n}}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right)\left( \sqrt{n^2+n} \right)}{n^2+n}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right) \times n \times \left( \sqrt{1+\frac{1}{n}} \right)}{n^2+n}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right) \times \left( \left(1+\frac{1}{n}\right)^{\frac{1}{2}} \right)}{n+1}\\
\\
=\frac{\left(n-\frac{1}{4}+O\left(\frac{1}{n}\right)\right) \times \left(1+\left(\frac{1}{2}\right) \left(\frac{1}{n}\right) + O\left(\frac{1}{n^2}\right) \right)}{n+1}\\
\\
=\frac{n+\frac{1}{4}+O\left(\frac{1}{n}\right)}{n+1} \\
\\
=1 - \left(\frac{3}{4}\right)\frac{1}{n+1} + O\left(\frac{1}{n^2}\right). \\
\\
\end{align}
Therefore,
\begin{align}
I=\displaystyle\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\\
\\
=\displaystyle\lim_{n\to\infty}\left(\ \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right) + O\left(\frac{1}{n^2}\right)\ \right)^n\\
\\
=\displaystyle\lim_{n\to\infty}\left(\ \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^n + \underset{\Large{\to 0 \text{ as } n\to \infty}}{\underbrace{\displaystyle\sum_{k=1}^n \binom{n}{k} \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^{n-k}    O\left(\frac{1}{\left(n^2\right)^k}\right)}}\ \right)\\
\\
=\displaystyle\lim_{n\to\infty}\left(\ \left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^{n+1}\ \cdot\ \underset{\Large{\to 1 \text{ as } n\to \infty}}{\underbrace{\left(1 - \left(\frac{3}{4}\right)\frac{1}{n+1}\right)^{-1}}}    \right)\\
\\
=\displaystyle\lim_{(n+1)\to\infty}\left(\ \left(1 + \left(-\frac{3}{4}\right)\frac{1}{n+1}\right)^{n+1}  \right)\\
\\
=e^{-\frac{3}{4}}.\\
\end{align}
Within the proof, I actually left out a couple of details for the sake of brevity. For example, getting from line $5$ to line $6$ isn't immediately obvious, but is in fact not too difficult to prove.
For the final step, see here.
A: If you like generalized harmonic numbers
$$S_n=\sum_{k=1}^n \frac{1}{\sqrt{n^2+n+k}}=H_{n^2+2 n}^{\left(\frac{1}{2}\right)}-H_{n^2+n}^{\left(\frac{1}{2}\right)}$$
Using twice the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 p^{1/2}}-\frac{1}{24 p^{3/2}}+O\left(\frac{1}{p^{5/2}}\right)$$ and continuing with Taylor expansions
$$S_n=1-\frac{3}{4 n}+\frac{5}{8 n^2}+O\left(\frac{1}{n^3}\right)$$
$$n \log(S_n)=-\frac{3}{4}+\frac{11}{32 n}+O\left(\frac{1}{n^2}\right)$$
$$S_n^n=e^{n \log(S_n)}=e^{-3/4}\left(1+ \frac{11}{32 n}\right)+O\left(\frac{1}{n^2}\right)$$
Trying for $n=10$ the "exact" value is $0.487638$ while the truncated expansion gives $0.488604$.
A: 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 $$\Large{ \color{blue}{e^{-\frac 34 }} }$$
