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I am trying to find

$$\lim_{n\to\infty}\frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{2n}}{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}$$

I believe I should use that

$$\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac1n \sum_{k=1}^n f\left(\frac{k}n\right)$$

to find the limit, but I am lost. Can anyone help?

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  • $\begingroup$ I have edited your question, as without proper formatting it was very hard to parse, but please check if that's what you intended to ask. $\endgroup$ – Fimpellizieri Mar 6 '17 at 7:07
  • $\begingroup$ thanks I was trying to find a way to fix it. $\endgroup$ – Christina Park Mar 6 '17 at 7:08
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I believe I should use that

$$\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac1n \sum_{k=1}^n f\left(\frac{k}n\right)$$ to find the limit, but I am lost. Can anyone help?

Hint. One may write, as $n\to \infty$, $$ \begin{align} \frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{2n}}{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}&=\frac{\frac1n\left(\sqrt{1+\frac1n}+\sqrt{1+\frac2n}+\dots+\sqrt{1+\frac nn}\right)}{\frac1n\left(\sqrt{\frac1n}+\sqrt{\frac2n}+\dots+\sqrt{\frac nn}\right)} \\\\&=\frac{\frac1n \sum_{k=1}^n \sqrt{1+\frac{k}n}}{\frac1n \sum_{k=1}^n \sqrt{\frac{k}n}}. \end{align} $$

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If we define $a_{n}=\sum_{k=1}^{n}\sqrt{n+k}$ and $b_{n}=\sum_{k=1}^{n}\sqrt{k}$ we have $$\lim_{n\rightarrow\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\lim_{n\rightarrow\infty}\frac{\sqrt{2n+1}+\sqrt{2n+2}-\sqrt{n+1}}{\sqrt{n+1}}=2\sqrt{2}-1$$ hence, by Stolz-Cesàro theorem, $$\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\color{red}{2\sqrt{2}-1}.$$

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