How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $\frac{2}{3}$. But how to show?

A solution based on Stolz-Cesàro theorem. Let $f(x)=x^{\frac{3}{2}}$
\begin{align*}
\lim_{n \to \infty} \frac{\sum_{i=1}^{n-1}\sqrt{i}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{(n+1)\sqrt{n+1} - n\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{(1 + \frac{1}{n})^{\frac{3}{2}} - 1} = \frac{1}{f'(1)} = \frac{2}{3}
\end{align*}
