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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.

Martin Sleziak
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asv
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  • Hint: for any $\varepsilon>0$, you have $\sqrt[k]{k}<1+\varepsilon$ for $k\geq n_0$ for some $n_0$. – Wojowu Mar 03 '19 at 19:57
  • The key is that $\sqrt[k]{k}\to1$ as $k\to\infty$ – saulspatz Mar 03 '19 at 19:57
  • for any $\epsilon >0$ we have $\lim_{n \rightarrow +\infty} \frac{\sum_{k=1}^{n} \sqrt[k] {k} }{n} \le \lim_{n \rightarrow +\infty} \frac{ n(1+\epsilon) }{n}= 1+ \epsilon$ – asv Mar 03 '19 at 20:01

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

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

rtybase
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