Value of $\lim_{n \to \infty}\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}$ 
Evaluate
  $$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}.$$

I am trying to use the Sandwich principle here..
$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}=\lim_{n \to \infty}\dfrac{\sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+\cdots+\sqrt{\dfrac{n-1}{n}}}{n}\ge \lim_{n \to \infty}\bigg(\sqrt{\dfrac{1}{n}}.\sqrt{\dfrac{2}{n}}.\sqrt{\dfrac{3}{n}}\cdots\sqrt{\dfrac{n-1}{n}}\bigg )^\dfrac{1}{n-1}=\lim_{n \to \infty}\bigg(\dfrac{(n-1)!}{n^n}\bigg )^\dfrac{1}{2(n-1)}$
But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)$ but I do not understand how it would help my problem..
In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link 
$\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)=\lim\limits_{n \to \infty }\sum_{k=1}^{n}{\dfrac{1}{\sqrt{kn}}}=\int_{0}^{1}\dfrac{1}{\sqrt{x}}dx+\lim\limits_{n \to \infty }\dfrac{1}{n}=2$
 A: Just for your curiosity since you already received good answers.
We can approximate the value of
$$a_n=\frac{\sum_{i=1}^{n-1} \sqrt i }{n \sqrt n}$$
$$\sum_{i=1}^{n-1} \sqrt i =\sum_{i=1}^{n} \sqrt i -\sqrt n=H_n^{\left(-\frac{1}{2}\right)}-\sqrt n$$ where appear generalized harmonic numbers.
Now, using the asymptotics
$$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n\sqrt n}{3}+\frac{\sqrt{n}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt n}+O\left(\frac{1}{n^{5/2}}
   \right)$$ making, for large $n$
$$a_n=\frac{\frac{2 n\sqrt n}{3}-\frac{\sqrt{n}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt n}+O\left(\frac{1}{n^{5/2}}
   \right)} {n \sqrt n}=\frac{2}{3}-\frac{1}{2 n}+\frac{\zeta
   \left(-\frac{1}{2}\right) } {n \sqrt n }+O\left(\frac{1}{n^{2}}
   \right)$$ which, for sure, shows the limit and also how it is approached.
But it also gives a good approximation even for small values of $n$. Using $\zeta
   \left(-\frac{1}{2}\right)\approx -0.207886$ and $n=10$, this would give $a_{10}\approx 0.610093$ while the "exact"  value is $a_{10}\approx 0.610509$.
A: One more approach. By Stolz-Cesaro Theorem
$$\begin{align}\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n-1}}{n\sqrt{n}} &= \lim_{n \to \infty} \frac{\sqrt{n}}{(n+1)\sqrt{n+1}-n\sqrt{n}}
\\&=\lim_{n \to \infty} \frac{1}{(n+1)\sqrt{1+\frac{1}{n}}-n}\\
&=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\end{align}$$
where we used the fact that $\sqrt{1+\frac{1}{n}}=1+\frac{1}{2n}+o(1/n)$.
A: Hint :
$$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n-1}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} \sqrt{\frac{k}{n}} = \int_0^1\sqrt{x}\mathrm{d}x$$
A: It is sufficient to obtain integral of $\sqrt{x}$, when $0\leq x\leq 1$.
