Computing $\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}$.

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

Solution 1:

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.

Solution 2:

Hint: Note that $$\sqrt[n]{n} \rightarrow 1.$$

You'll need to know and use that fact.