Given a sequence of positive numbers $(x_n)$ with $\lim x_n=\alpha$ show that $$\lim\sqrt[n]{x_1x_2\dots x_n}=\alpha$$
I'm not sure about how to proof it, but if each $x_n\to\alpha$ then $x_1 x_2\dots x_n\to a^n$ so $$\lim\sqrt[n]{x_1 x_2\dots x_n}=\sqrt[n]{a^n}=a$$
How I can proof it?
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.
