If $(x_n) \to x$ then $(\sqrt[n]{x_1x_2\cdots x_n}) \to x$ This is not a duplicate of this question.
The linked question says that it suffices to show that if $(x_n)\to x$ then $(\frac{x_1+\cdots+x_n}{n})\to x$ to prove my question, but how so? I tried using the same strategy as how one proves that if $(x_n)\to x$ then $(\frac{x_1+\cdots+x_n}{n})\to x$, by "splitting" the product in th $N$th term:
$$\sqrt[n]{x_1x_2\cdots x_n}=\sqrt[n]{x_1x_2\cdots x_Nx_{N+1}\cdots x_n}=\sqrt[n]{x_1x_2\cdots x_N} \sqrt[n]{x_{N+1}\cdots x_n}$$
but it seems I can't use for now the definition of convergence of $(x_n)$ because of the $n$th root. I also tried to use a result: if $(x_n)\to x$ then $(\frac{x_n}{n})\to 1$ but don't know if this is true. Sadly I've had no real progress. Any help will be greatly appreciated, thanks in advance!
 A: I am going to assume $x_n\geq 0$ for any $n\geq 1$.
Cesàro mean theorem gives that if $\{\alpha_n\}_{n\geq 1}$ is a real sequence converging to $A$ and for every $n\geq 1$ we have $\beta_n=\frac{\alpha_1+\ldots+\alpha_n}{n}$, then $\{\beta_n\}_{n\geq 1}$ is a real sequence converging to $A$, too.
So you just have to consider $\alpha_n=\log x_n$ and derive from $\alpha_n\to \log\alpha$ that
$$ \beta_n = \log\sqrt[n]{x_1\cdot x_2\cdots x_n} \to \log\alpha, $$
hence $\sqrt[n]{x_1\cdot x_2\cdots x_n}\to \alpha$ by exponentiating back.
A: By A.M -G.M-H.M  inequality we have 
$$X_n=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}\le\{x_1x_2..x_n\}^{\frac{1}{n}} \le Y_n= \frac{x_1+x_2+..x_n}{n}$$..
Now show that both $X_n$ and $Y_n$ converge to $x$ and use Squeeze Principle. 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Stolz-Ces$\mrm{\grave{a}}$ro Theorem yields a straightforward answer:

\begin{align}
\lim_{n \to \infty}\root[n]{x_{1}\ldots x_{n}} & =
\exp\pars{\lim_{n \to \infty}{\ln\pars{x_{1}\ldots x_{n}} \over n}} =
\exp\pars{\lim_{n \to \infty}{\ln\pars{x_{1}\ldots x_{n}x_{n + 1}} - \ln\pars{x_{1}\ldots x_{n}} \over \bracks{n + 1} - n}}
\\[5mm] & = \exp\pars{\lim_{n \to \infty}\ln\pars{x_{n + 1}}} =
\exp\pars{\ln\pars{x}}\ =\ \bbox[10px,#ffe,border:1px dotted navy]{\ds{x}}
\end{align}
A: Here is a proof for the case $x>0$ and $x_n>0$ for all $n,$ using as little background material as I can manage:
We will need the following three facts:
(I). $\lim_{n\to \infty}y^{1/n}=1$ for any $y>0.$
(II).  If $0< s<1$ and  $0<k<1$ then $1-s< (1-s)^k< (1+s)^k< 1+s.$
(III). If  $0<r<1$ there exists $s\in (0,r)$ such that $1-r<yz<1+r$ whenever $\{y,z\}\subset [1-s,1+s].$
Let $x_n=xy_n$ for each $n.$ We have $y_n>0$ and  $\lim_{n\to \infty}y_n=1$.  And we have $(\prod_{j=1}^nx_j)^{1/n}=x\cdot (\prod_{j=1}^ny_j)^{1/n}.$ We show that $\lim_{n\to \infty}(\prod_{j=1}^ny_j)^{1/n}=1$.
Given $\epsilon >0,$ let $r=\min (1/2, \epsilon).$ Take $s \in (0,r)$ such that (III) applies to $s.$ Take $n_1>0$ such that $  n>n_1\implies  y_n\in (1-s,1+s).$
By (I) with $y=\prod_{j=1}^{n_1},$  take $n_2>n_1$ such that $$(\bullet) \quad n>n_2\implies 1-s<\left(\prod_{j=1}^{n_1}y_j\right)^{1/n}<1+s.$$ Now apply (II) with $k=(n-n_1)/n:$  Since $n_1<n_2$ and $1+n_1\leq j\implies 1-s<y_j<1+s,$ we have      $$(\bullet \bullet)\quad n>n_2\implies 1-s< (1-s)^k<\left(\prod_{j=1+n_1}^ny_j\right)^{1/n}<(1+s)^k< (1+s).$$  From $(\bullet)$ and $(\bullet \bullet)$  and by (III),  we have  $$n>n_2\implies 1-\epsilon\leq 1-r<(1-s)^2<\left(\prod_{j=1}^n y_j\right)^{1/n}<(1+s)^2<1+r\leq 1+\epsilon.$$
You can appreciate the ease of using some "technology", such as the Cesaro Mean Theorem applied to the sequence $(\log x_n)_n,$ compared to the above proof.
