how can I show this equations? $\lim \left( x_{1}x_{2}...x_{n}\right) ^{\frac {1} {n}}=x_{0}$ Suppose $x_{n}>0$ for all $n\in {\bf N}$ and $\lim_n x_{n}=x_{0}>0$. How can I show that 
$$
\lim_n \left( x_{1}x_{2}...x_{n}\right)^{\frac {1} {n}}=x_{0}\quad ?
$$
 A: We have 
$$
\log \left( ( x_1 x_2 ... x_n)^{1/n} \right) = \frac{1}{n} \log(x_1 x_2 ... x_n) = \frac{1}{n} ( \log(x_1) + \log(x_2) + ... + \log(x_n) ) = \frac{\log(x_1) + \log(x_2) + ... + \log(x_n)}{n}
$$
As $ \lim_{n \rightarrow \infty } \log(x_n) = \log( \lim_{ n \rightarrow \infty } x_n ) = \log(x_0) $, we deduce that 
$$ 
\lim_{n \rightarrow \infty } \frac{\log(x_1) + \log(x_2) + ... + \log(x_n)}{n} =
 \lim_{ n \rightarrow \infty } \log(x_n) = \log(x_0)
$$
And finally
$$
\lim_{ n \rightarrow \infty } \log \left( ( x_1 x_2 ... x_n)^{1/n} \right) =\log \left( \lim_{ n \rightarrow \infty } ( x_1 x_2 ... x_n)^{1/n} \right) = \log(x_0) \longrightarrow \lim_{ n \rightarrow \infty } ( x_1 x_2 ... x_n)^{1/n} = x_0
$$
A: Let $f(n)$ be defined by 
$$f(n)=\left(\prod_{m=1}^n x_m\right)^{1/n}$$
Clearly the logarithm of $f(n)$ is given by
$$\log(f(n))=\frac1n \sum_{m=1}^n\log(x_m)$$
From the Stolz-Cesaro Theorem, we have
$$\begin{align}
\lim_{n\to \infty}\log(f(n)) &=\lim_{n\to \infty} \left(\frac1n \sum_{m=1}^n\log(x_m)\right)\\\\
&=\lim_{n\to \infty}\left( \frac{\sum_{m=1}^{n+1}\log(x_m)-\sum_{m=1}^{n}\log(x_m)}{(n+1)-n}\right)\\\\
&=\lim_{n\to \infty}\log(x_{n+1})\\\\
&=\log(x_0)
\end{align}$$
