find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem I'm trying to prove with squeeze theorem that the limit of the following series equals 1:
$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$
For the left side of the inequality I did:
$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} <  \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$
For the right side, at first I did the following:
$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} <  \frac{n\sqrt[n]{n}}{n}$$
But then I realized it wasn't true and that the direction of this inequality is the opposite.
Do you have any idea which series with limit 1 is bigger from the original series?
Thanks!

This is not a full answer to the question, but many answers are implying that the function $n\mapsto n^{1/n}$ is strictly increasing. This is not the case. To see this:
Let $y=x^{1/x}$. Then $\ln y=\frac 1x \ln x$ so $\frac{y'}{y}=\frac{1}{x^2}(1-\ln x)$. Since $y>0$, this implies that $y$ is increasing on $(0,e)$ and decreasing on $(e,\infty)$.
Hence, do not use upper bound of $n^{1/n}$.
