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

Hint :

$$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n-1}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} \sqrt{\frac{k}{n}} = \int_0^1\sqrt{x}\mathrm{d}x$$

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 $a_{10}\approx 0.610509$.