Given $p$ positives values $a_1...a_p$, define the sequence $x_n$ such that:
$$x_n = \frac{\sqrt[n]{a_1}+...+\sqrt[n]{a_p}}{p}$$
And define $S_n = (x_n)^n$
Prove that $S_n \rightarrow \sqrt[p]{a_1...a_p}$
The best idea here should be to use squeeze theorem. Of course, we have that $(\sqrt[p]{\sqrt[n]{a_1}...\sqrt[n]{a_p}})^n \leq S_n$ (as GM < AM), but the upper bound to squeeze it is what we haven't found yet. Any idea?
On
By using induction (and associativity of the various means involved here), it is enough to prove this for $p = 2$. By scaling, we may assume that $a_1 = 1$, i.e. it is enough to prove this for one pair $(a_1 = 1, a_2 = e^x)$. The formula now becomes: $$ \left(\frac{1+e^{x/n}}{2}\right)^n \rightarrow e^{x/2}.$$ To prove this, we note that we may write $\frac{1+e^{x/n}}{2} = 1+\frac{x}{2n}+O(n^{-2})$ and use the classic fact that $(1+t/n)^n \rightarrow e^t$.
On
You have already shown that $\liminf S_n \geq (a_1\cdots a_p)^{1/p}$, so we will complete the proof by showing that $\limsup S_n \leq (a_1\cdots a_p)^{1/p}$.
Remark 1: In order to prove this we will use the fact that for any $x>0$, we have that $\lim_{\alpha \to 0^{+}} \frac{x^{\alpha}-1}{\alpha} = \log x$, which can be derived via L'hopital's rule.
Returning to the proof, let $\epsilon >0$. Then from Remark 1, there exists some $N \in \mathbb{N}$ such that $\frac{a_i^{1/n}-1}{(1/n)} \leq \log a_i + \epsilon$ for all $1 \leq i \leq p$ whenever $n \geq N$. In particular, $n \geq N$ implies that $a_i^{1/n}-1 \leq \frac{1}{n}\log a_i + \frac{\epsilon}{n}$, so that $$S_n = \bigg(\frac{a_1^{1/n}+\dots +a_p^{1/n}}{p}\bigg)^n = \bigg(1+\frac{(a_1^{1/n}-1)+\dots+(a_1^{1/n}-1)}{p}\bigg)^n$$ $$ \leq \bigg(1+\frac{(\frac{1}{n}\log a_1+\frac{\epsilon}{n})+\dots+(\frac{1}{n}\log a_p+\frac{\epsilon}{n})}{p}\bigg)^n$$ $$=\bigg(1+\frac{\frac{1}{p}\log(a_1\cdots a_p)+\epsilon}{n}\bigg)^n$$
This last expression is of the form $(1+\frac{\beta}{n})^n$, and so it converges to $$e^{\beta} = e^{\frac{1}{p}\log(a_1\cdots a_p)+\epsilon}=e^{\epsilon}\cdot(a_1 \cdots a_p)^{1/p}$$
Thus we have shown that $\limsup S_n \leq e^{\epsilon} \cdot (a_1 \cdots a_p)^{1/p}$, for every $\epsilon>0$, which in turn implies that $\limsup S_n \leq (a_1 \cdots a_p)^{1/p}$, which is the desired result.
On
Suppose $\limsup\limits_{n\to\infty}x_n^n,\limsup\limits_{n\to\infty}y_n^n\le M$, then $$ \begin{align} \limsup_{n\to\infty}|x_n^n-y_n^n| &\le\limsup_{n\to\infty}\left|\frac{x_n^{n-1}+x_n^{n-2}y_n+\cdots+y_n^{n-1}}n\right|\limsup_{n\to\infty}n|x_n-y_n|\\ &\le M\limsup_{n\to\infty}n|x_n-y_n|\tag{1} \end{align} $$ Thus, $(1)$ implies
Lemma: If $x_n^n$ and $y_n^n$ are bounded, and $$ \lim_{n\to\infty}n(x_n-y_n)=0 $$ then $$ \lim_{n\to\infty}(x_n^n-y_n^n)=0 $$
We have the bound $$ x_n^n=\left(\frac{a_1^{1/n}+a_2^{1/n}+\cdots+a_p^{1/n}}p\right)^n\le\max_{1\le k\le p}(a_k)\tag{2} $$ and the limit $$ \lim_{n\to\infty}y_n^n=\lim_{n\to\infty}\left(1+\frac{\log(a_1a_2\cdots a_p)}{np}\right)^n=(a_1a_2\cdots a_p)^{1/p}\tag{3} $$ Since $$ \lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)\tag{4} $$ we have $$ \lim_{n\to\infty}n\left(\frac{a_1^{1/n}+a_2^{1/n}+\cdots+a_p^{1/n}}p-1\right)=\frac1p\log(a_1a_2\cdots a_p)\tag{5} $$ and therefore $$ \lim_{n\to\infty}n\left(\frac{a_1^{1/n}+a_2^{1/n}+\cdots+a_p^{1/n}}p-1-\frac{\log(a_1a_2\cdots a_p)}{pn}\right)=0\tag{6} $$
Applying the Lemma to $(2)$, $(3)$, and $(6)$, we get $$ \begin{align} \lim_{n\to\infty}\left(\frac{a_1^{1/n}+a_2^{1/n}+\cdots+a_p^{1/n}}p\right)^n &=\lim_{n\to\infty}\left(1+\frac{\log(a_1a_2\cdots a_p)}{np}\right)^n\\[6pt] &=(a_1a_2\cdots a_p)^{1/p}\tag{7} \end{align} $$
We have that
$$S_n = \left(\frac{\sqrt[n]{a_1}+...+\sqrt[n]{a_p}}{p}\right)^n$$
Which is the $\frac{1}{n}$-power mean. As $n$ goes to infinity, The above power mean will come closer to the $0$-power mean, which is the geometric mean.
It is a well known fact that if $n$ goes to infinity, the $\frac{1}{n}$-power mean goes to the geometric mean. See for a proof Wikipedia. It is in the first section.