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I have fo find: $$\displaystyle\lim_{n\to\infty}\frac{1}{n}\Bigg(1+\frac{2}{1+\sqrt{2}}+\frac{3}{1+\sqrt{2}+\sqrt{3}}+\cdots+\frac{n}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}\Bigg)$$

using the Cesaro-Stolz Theorem.

Applying what the theorem states once i get: $\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}$

My question is: can i apply it again to get: $\displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n+1}}=0$?

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Yes, why not? If this new exercise was a completely new exercise, wouldn't you had solved it with S-C? Then, why would the fact that this is not a new exercise, but part of a larger one change your technique? – N. S. Jan 4 '13 at 21:33
Well in school they never tell us what freedom we have while doing math, so i was unsure. I thought logically its possible, i just wanted to be sure. – phi Jan 4 '13 at 21:36
BTW if you use SC, the new limit should be $\lim_n \frac{1}{\sqrt{n+1}}$. – N. S. Jan 4 '13 at 21:38
up vote 1 down vote accepted

S-C is the easiest way to calculate

$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}$$

If you want an alternate solution, note that

$$1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}=\sqrt{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} \,,$$


$$\lim_n \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} =\int_0^1 \sqrt{x} dx \,.$$


$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}=\displaystyle\lim_{n\to\infty} \frac{n+1}{n\sqrt{n}}\frac{1}{ \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} }=0$$

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