Limit $\lim\limits_{n \rightarrow +\infty} \frac{\sum\limits_{k=1}^{n} \sqrt[k] {k} }{n}= 1$

Question

I would like to prove that:

$$\lim\limits_{n \rightarrow +\infty} \frac{\sum\limits_{k=1}^{n} \sqrt[k] {k} }{n}= 1$$

I thought to write $\sqrt[k] {k} = e^{\frac{\ln({k})}{k}}$ but I don't know how to continue.

Answer 1

Use Stolz–Cesàro theorem or a version of it here.

$$\lim\limits_{n \rightarrow +\infty} \frac{\sum\limits_{k=1}^{n+1} \sqrt[k] {k} -\sum\limits_{k=1}^{n} \sqrt[k] {k} }{n+1 - n}= \lim\limits_{n \rightarrow +\infty}\sqrt[n+1] {n+1} =1$$