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?

Using right Riemann sums, \begin{align*}
\frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {k - 1} } & = \frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt k }  = \frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt {\frac{k}{n}} } \\ & = \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} } } \right) - \frac{1}{n} \to \int_0^1 {\sqrt x dx}  - 0 = \frac{2}{3}.
\end{align*}
