$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $ How do I find the following limit?
$$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $$
Can limit be find by Riemann sums?
$$\lim_{n\to \infty}\sum_{k=1}^{n}f(C_k)\Delta{x} = \int_{a}^{b}f(x)\,dx$$
I'm not sure what $f(C_k)$ is.

By Stolz we have $$ \lim\limits_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2}= \lim\limits_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n^\frac{3}{2}}= \lim\limits_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n^3}-\sqrt{(n-1)^3}}=$$
$$= \lim\limits_{n \to \infty} \frac{\sqrt{n}\left(\sqrt{n^3}+\sqrt{(n-1)^3}\right)}{n^3-(n-1)^3}=\lim\limits_{n \to \infty} \frac{\sqrt{n}\left(\sqrt{n^3}+\sqrt{(n-1)^3}\right)}{3n^2-3n+1}=\frac{2}{3}$$
