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I am trying to evaluate $\lim_{n\to \infty} \frac{1}{\sqrt{n}} \left(\frac{1}{\sqrt{n+1}}+\dotsm +\frac{1}{\sqrt{n+n}} \right)$. I suspect identifying an appropriate Riemann sum is the trick. However after some toying with it I gave up on this suspicion and stumbled across the Stolz-Cesàro theorem, which I then used to calculate the limit as $\sqrt{2}$.

Does anybody see a way to do this as Riemann sum?

I tried putting it in this form $\frac{1}{n} \sum_{k=1}^n \sqrt{\frac{n}{n+k}}$ but then I don't see how carry on to identify the function from the partition to integrate.

Thank you for suggestions or comments.

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3 Answers 3

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This is a Riemann sum. Rewrite as

$$\frac1{n} \sum_{k=1}^n \frac1{\sqrt{1+\frac{k}{n}}} $$

which, as $n \to \infty$, becomes

$$\int_0^1 dx \frac1{\sqrt{1+x}} = 2 (\sqrt{2}-1)$$

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$ \sum_{k=1}^n \frac{1}{n}\sqrt{\frac{n}{n+k}}= \sum_{k=1}^n \frac{1}{n}\sqrt{\frac{1}{1+\frac{k}{n}}}$ and $\int_{0}^{1} \sqrt{\frac{1}{1+x}}$

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HINT:

$$\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{n+k}}=\frac1n\sum_{k=1}^n\frac{1}{\sqrt{1+k/n}}$$

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