I have a problem with the calculation of the following limit. \begin{equation} \lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n} \end{equation} I do not know where to start! Thank you very much
Computing $\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}$.
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sequences-and-series
limits
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0See also: [Evaluating Limit Question $\lim\limits_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$?](https://math.stackexchange.com/q/130442) – 2018-04-17
2 Answers
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There are at least two possibilities. In all of them you use that $\lim_{n\to\infty}=\sqrt[n]{n}=1$.
The first one uses the following result: if $a_n$ is a convergent sequence, then $\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n}=\lim_{n\to\infty}a_n.$
The second is to use Stolz's criterion.
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0@user49685 It is (a version of) Stolz-Cesaro theorem, see e.g. [this answer](http://math.stackexchange.com/questions/100338/limit-of-quotient-of-two-series/100542#100542). – 2012-11-19
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Hint: Note that $\sqrt[n]{n} \rightarrow 1.$
You'll need to know and use that fact.
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0@DonAntonio I second that. – 2012-11-19