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!

One may combine the following two facts:
1. If $a_n\to a,\,$ then $\,\frac{1}{n}(a_1+\cdots+a_n)\to a$.
2. $\sqrt[n]{n}\to 1$.
Another way to show it is the following:
$$
\sqrt[2k]{k}=1+a_k\Longrightarrow \sqrt{k}=(1+a_k)^{k}\ge 1+ka_k
\Longrightarrow 0\le a_k<\frac{1}{\sqrt{k}}
$$
and hence
$$
1<\sqrt[n]{n}=(1+a_n)^2=1+2a_n+a_n^2<1+\frac{2}{\sqrt{n}}+\frac{1}{n}\le 
1+\frac{3}{\sqrt{n}}
$$
and thus
$$
1<\frac{1}{n}(1+\sqrt{2}+\cdots+\sqrt[n]{n})<1+\frac{3}{n}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) \\ <1+\frac{3}{n}\cdot (2\sqrt{n}+1)\to 1.
$$
It remains to show that
$$
1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}<2\sqrt{n}+1
$$
which can be easily done inductively.
