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

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.

## 4 Answers

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

• Could you please explain the second equality? – Itay4 Feb 18 '17 at 9:56
• @Itay4 It's Stolz. See here: en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – Michael Rozenberg Feb 18 '17 at 9:58
• Yes, I understand it's Stolz, but how did you get that expression? Did you use it on $n-1$? Is it ok to do that? – Itay4 Feb 18 '17 at 10:03
• @Itay4Yes of course! We have $n\rightarrow+\infty$. – Michael Rozenberg Feb 18 '17 at 10:06
• Got you, thanks ! – Itay4 Feb 18 '17 at 10:07

You could also do it using harmonic numbers since $$S_n=\sum_{i=1}^n \sqrt i=H_n^{\left(-\frac{1}{2}\right)}$$ For large values of $n$, Taylor expansion would be $$S_n=\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta \left(-\frac{1}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ which makes $$\frac {\sqrt n\,S_n}{n^2}=\frac{2}{3}+\frac{1}{2 n}+O\left(\frac{1}{n^{3/2}}\right)$$ showing the limit and how it is approached.

Using Excel for $n=100$, you would find $S_{100}\approx 0.671463$ while the above formula would give $\frac{403}{600}\approx 0.671667$.