The convergence of the arithmetic mean into the geometric mean 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?
 A: 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 $(1+t/n)^n \rightarrow e^t$.
A: 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.
A: 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.
A: 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
$$
\begin{align}
\lim_{n\to\infty}\left(\frac{a_1^{1/n}+a_2^{1/n}+\cdots+a_p^{1/n}}p\right)^n
&=\lim_{n\to\infty}\left(1+\frac{\log(a_1a_2\cdots a_p)}{np}\right)^n\\[6pt]
&=(a_1a_2\cdots a_p)^{1/p}\tag{7}
\end{align}
$$
