6
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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 $.

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Could you please explain the second equality? $\endgroup$ – Itay4 Feb 18 '17 at 9:56
  • $\begingroup$ @Itay4 It's Stolz. See here: en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem $\endgroup$ – Michael Rozenberg Feb 18 '17 at 9:58
  • $\begingroup$ 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? $\endgroup$ – Itay4 Feb 18 '17 at 10:03
  • $\begingroup$ @Itay4Yes of course! We have $n\rightarrow+\infty$. $\endgroup$ – Michael Rozenberg Feb 18 '17 at 10:06
  • $\begingroup$ Got you, thanks ! $\endgroup$ – Itay4 Feb 18 '17 at 10:07
0
$\begingroup$

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 $\frac{403}{600}\approx 0.671667$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.