# Find Limit of Sum as a Riemann Sum : $\lim\limits_{n \to ∞} \frac{{\sqrt 1}+\cdots +{\sqrt n}}{n{\sqrt n}}$

Find $$\lim\limits_{n \to ∞} \frac{{\sqrt 1}+\cdots+{\sqrt n}}{n{\sqrt n}}$$

The series in the numerator is what throws me off.

I have tried just about everything, I'm stuck on how to write it in a form where I can find the limit.

$$\frac{\sum_{k=1}^n\sqrt k}{n\sqrt n}=\frac1n\sum_{k=1}^n\sqrt{\frac kn}\approx\int_0^1\sqrt x\,\mathrm dx$$
Using a Riemann sum: $$\dfrac{\sqrt{1}}{n\sqrt{n}}+\dfrac{\sqrt{2}}{n\sqrt{n}}+\cdots+\dfrac{\sqrt{n}}{n\sqrt{n}}=\frac1{n}\sum_{k=0}^{n}\frac{\sqrt{k}}{\sqrt{n}} \to \int_0^1\sqrt{x}\:dx=\color{red}{\frac23}.$$
$$\lim_{n\to \infty}\frac{1+\cdots+\sqrt{n}}{n\sqrt{n}}=\lim_{n\to \infty}\frac{\sqrt{n}}{n\sqrt{n}-(n-1)\sqrt{n-1}}=\lim_{n\to \infty}\frac{\sqrt{n}(n\sqrt{n}+(n-1)\sqrt{n-1})}{n^3-(n-1)^3}=\lim_{n\to \infty}\frac{n^2+n\sqrt{n(n-1)}-\sqrt{n(n-1)}}{3n^2-3n+1}=\frac{2}{3}$$ The last line is since the numerator behaves like $2n^2$ and denominator as $3n^2$