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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.

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  • $\begingroup$ Do you know about Riemann sums? $\endgroup$ Jan 25 '17 at 21:46
  • $\begingroup$ @gobucksmath Not that much. Would appreciate a nudge. $\endgroup$ Jan 25 '17 at 21:47
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    $\begingroup$ Related: math.stackexchange.com/questions/5676/… $\endgroup$
    – Aryabhata
    Jan 25 '17 at 22:12
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    $\begingroup$ Another solution would be to compare $k^{1/2}$ with $\int_{k}^{k+1} t^{1/2} dt$ to get the behaviour of $\displaystyle\sum_{k=0}^n \sqrt{k}$ and then you can conclude. $\endgroup$ Jan 25 '17 at 22:42
  • $\begingroup$ @HungryBlueDev, I added an equivalent expression $\displaystyle\lim_{n\to\infty}n^{-\frac{3}{2}}(1+\sqrt{2}+\ldots+\sqrt{n})$ in the title to prevent future possible duplicates because I've seen this task in both forms. Sincerely $\endgroup$
    – Invisible
    Jan 15 '20 at 4:08
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$$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} = \int_{0}^{1}\sqrt{x}\,dx = \color{red}{\frac{2}{3}}$$ 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.

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

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  • $\begingroup$ @JackD'Aurizio No I think its the other way around, like l'hospitals rule, after all, its really the discrete version. You need also that $y_n\to\infty$. $\endgroup$ Jan 25 '17 at 22:21
  • $\begingroup$ All right, my bad. The (+1) stays :) $\endgroup$ Jan 25 '17 at 22:24
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    $\begingroup$ @JackD'Aurizio Phew, that was a close one.... I just looked it up in my notes $y_n$ also has to be increasing. $\endgroup$ Jan 25 '17 at 22:47
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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 : $$ \lim_{n\to\infty} \frac{x_n}{y_n} = \lim_{n\to\infty} \frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$

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  • $\begingroup$ Okay, that's interesting... But what will be $x_n$? I think $\sqrt{n}$? And what about $y_n$? $\endgroup$ Jan 25 '17 at 21:59
  • $\begingroup$ $x_n$ will be whole numerator of your fraction $\endgroup$ Jan 25 '17 at 22:00
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Fill in details: $$\frac1{n\sqrt n}\sum_{k=1}^n\sqrt k=\frac1{n}\sum_{k=1}^n\sqrt\frac kn\xrightarrow[n\to\infty]{}\int_0^1\sqrt x\,dx$$

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Using $\;\;\displaystyle\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)=\int_0^1f(x)\;dx,$ we get $$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} +\ldots + \sqrt{n}}{n\sqrt{n}}=\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{k}{n}}=\int_0^1\sqrt{x}\;dx=\frac{2}{3}.$$

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