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Use Riemann integral to evaluate the limit $$\lim_{n\to\infty}\frac{\sum_{k=1}^n \sqrt k}{\sum_{k=1}^n\sqrt{n+k}}$$

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closed as off-topic by heropup, ILoveMath, Claude Leibovici, mrp, астон вілла олоф мэллбэрг Nov 14 '16 at 10:22

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    $\begingroup$ By Cesaro-Stolz, this limit is equal to the limit $$\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n-1}-\sqrt{n}}=\frac1{2\sqrt2-1}.$$ $\endgroup$ – LutzL Nov 14 '16 at 8:16
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Hint. We have that $$\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^n \sqrt{k}}{\sum_{k=1}^n \sqrt{n+k}}= \lim_{n \rightarrow \infty}\frac{\displaystyle\frac{1}{n}\sum_{k=1}^n \sqrt{\frac{k}{n}}}{\displaystyle\frac{1}{n}\sum_{k=1}^n \sqrt{1+\frac{k}{n}}} = \frac{\displaystyle\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^n \sqrt{\frac{k}{n}}}{\displaystyle\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^n \sqrt{1+\frac{k}{n}}}=\frac{\int_0^1 f(x)\,dx}{\int_0^1 g(x)\, dx}.$$ What are the integrand functions $f$ and $g$? What is the final result?

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  • $\begingroup$ Please edit your answer. In the last expression you wrote, somehow two question marks have appeared instead of algebraic quantities. $\endgroup$ – Landon Carter Nov 14 '16 at 7:30
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    $\begingroup$ @Landon Carter The two question marks are intentional. My post is a hint... $\endgroup$ – Robert Z Nov 14 '16 at 7:35
  • $\begingroup$ @LandonCarter your comment is for real? The question marks are intented because RobertZ provide a hint, not a solution. $\endgroup$ – Masacroso Nov 14 '16 at 7:35
  • $\begingroup$ Oh I was thinking it was a problem with my browser. $\endgroup$ – Landon Carter Nov 14 '16 at 7:50
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    $\begingroup$ @DonAntonio Is it better now? $\endgroup$ – Robert Z Nov 14 '16 at 8:44

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