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Find $$\lim\limits_{n \to ∞} \frac{{\sqrt 1}+\cdots+{\sqrt n}}{n{\sqrt n}}$$

The series in the numerator is what throws me off.

I have tried just about everything, I'm stuck on how to write it in a form where I can find the limit.

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$$ \frac{\sum_{k=1}^n\sqrt k}{n\sqrt n}=\frac1n\sum_{k=1}^n\sqrt{\frac kn}\approx\int_0^1\sqrt x\,\mathrm dx$$

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  • $\begingroup$ So it has to be expressed as a definite integral because finding the limit itself would be complicated. Correct? $\endgroup$ – User 210 Nov 12 '15 at 22:03
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Using a Riemann sum: $$ \dfrac{\sqrt{1}}{n\sqrt{n}}+\dfrac{\sqrt{2}}{n\sqrt{n}}+\cdots+\dfrac{\sqrt{n}}{n\sqrt{n}}=\frac1{n}\sum_{k=0}^{n}\frac{\sqrt{k}}{\sqrt{n}} \to \int_0^1\sqrt{x}\:dx=\color{red}{\frac23}. $$

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  • $\begingroup$ is there any way to see that without a riemann sum...? $\endgroup$ – tired Nov 12 '15 at 22:24
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Here is a approach using Stolz theorem(since op asked for another approach in comments).

$$\lim_{n\to \infty}\frac{1+\cdots+\sqrt{n}}{n\sqrt{n}}=\lim_{n\to \infty}\frac{\sqrt{n}}{n\sqrt{n}-(n-1)\sqrt{n-1}}=\lim_{n\to \infty}\frac{\sqrt{n}(n\sqrt{n}+(n-1)\sqrt{n-1})}{n^3-(n-1)^3}=\lim_{n\to \infty}\frac{n^2+n\sqrt{n(n-1)}-\sqrt{n(n-1)}}{3n^2-3n+1}=\frac{2}{3}$$ The last line is since the numerator behaves like $2n^2$ and denominator as $3n^2$

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