TITLE: Limit $\lim_{n\to\infty} n^{-3/2}(1+\sqrt{2}+\ldots+\sqrt{n})=\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n\sqrt{n}}$
QUESTION [5 upvotes]: How do I find the following limit?
$$
\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n\sqrt{n}}
$$
The answer (from Wolfram) is $\frac{2}{3}$, but I'm not sure how to proceed.
Is this an application of the Squeeze theorem? I'm not quite sure.

REPLY [4 votes]: All the other answers are by Riemann sums, but one can also use the Stolz theorem. The limit becomes 
$$\frac{\sqrt{n}}{n\sqrt{n}-(n-1)\sqrt{n-1}}=\frac{\sqrt{n}(n\sqrt{n}+(n-1)\sqrt{n-1})}{n^3+(n-1)^3}=$$ 
$$\frac{n^2+(n-1)\sqrt{n^2-n})}{n^2+n(n-1)+(n-1)^2}\to\frac{2}{3}$$