limit of a geometric mean Suppose that a sequence $a_n$ of positive numbers converges to $a$. Show that $$\lim_{n\rightarrow \infty}\left(\prod_{i=1}^{n}a_i\right)^{1/n}=a$$ This seems to be simple using that $x=e^{\log x}$, but I can't go any further after using that. 
PS. I'm almost sure that this post is a duplicate, but I wasn't able to find it here. I'm sorry if that's the case. 

Hint 1: $$\ln \left(\prod_{i=1}^{n}a_i\right)^{1/n} = \frac{\ln(a_1)+\ln(a_2)+...+\ln(a_n)}{n}$$
Hint 2: Suppose that $\lim_n b_n =l$ where $l$ is real number or $\pm \infty$. Prove that 
$$
\lim_n \frac{b_1+b_2+...+b_n}{n}=l
$$
This is a particular case of STolz-Cezaro, but easier to prove than the general version.
