<s>[INST] 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!
 [/INST] First of all we have, for any $ n\in\mathbb{N}^{*} $, the following : $$ \sqrt[n]{n}=1+\frac{\ln{n}}{n}\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x} $$
Since : \begin{aligned}0\leq\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x}&\leq n^{\frac{1}{n}}\\ &\leq 2\end{aligned}
We have : \begin{aligned} 1\leq \sqrt[n]{n}\leq 1+\frac{2\ln{n}}{n}&=1+\frac{4\ln{\sqrt{n}}}{n}\\ &\leq 1+\frac{4\sqrt{n}}{n}= 1+\frac{8}{2\sqrt{n}}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq1+\frac{8}{\sqrt{n}+\sqrt{n-1}} \end{aligned}
That remains true for any $ n\in\mathbb{N}^{*} $, which means given $ n\in\mathbb{N}^{*} $, we have : \begin{aligned} 1\leq\frac{1}{n}\sum_{k=1}^{n}{\sqrt[k]{k}}&\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\frac{1}{\sqrt{k}+\sqrt{k-1}}} \\ &\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\left(\sqrt{k}-\sqrt{k-1}\right)}\\ &\leq 1+\frac{8}{\sqrt{n}} \end{aligned}
Thus, using the squeezing theorem, the limit would be $ 1 \cdot$
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