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

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.

Solution 1:

$$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} = \int_{0}^{1}\sqrt{x}\,dx = \color{red}{\frac{2}{3}}$$ by Riemann sums and the integrability of $\sqrt{x}$ over $[0,1]$.

For a more elementary approach, notice that $\sqrt{k}$ is pretty close to $\frac{2}{3}\left[\left(k+\frac{1}{2}\right)^{3/2}-\left(k-\frac{1}{2}\right)^{3/2}\right]$ and apply creative telescoping and squeezing.

Solution 2:

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

Solution 3:

How about Stolz-Cesaro? If $x_n \to \infty $, $y_n \to \infty $ and $$ \lim_{n\to\infty} \frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$ exists, then : $$ \lim_{n\to\infty} \frac{x_n}{y_n} = \lim_{n\to\infty} \frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$