$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $

Hint: $$ \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{\sqrt{n}}\frac1n= \lim_{n \to \infty} \left(\sqrt{\frac1n} + \sqrt{\frac2n} + \sqrt{\frac3n} +\cdots+\sqrt{\frac{n}{ n}} \right) \frac1n $$ $f(C_k)=\sqrt{\dfrac{k}{n}}$ and $\Delta x=\dfrac1n $.

The sum $1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}$ is well approximated by the integral

$\int_1^n \sqrt{x} dx = \frac{2}{3} (n \sqrt{n} - 1),$

which can be seen by writing out the Riemann sum for this integral. Plugging this in, the limit of your series is

$\lim_{n \to \infty} \frac{2}{3} \frac{\sqrt{n}(n \sqrt{n} - 1)}{n^2} = 2/3.$

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