Let $(X_i)_{i \in \mathbb{N}}$ be sequence of independent, identically distributed and integrable random variables with $\mathbb{P}(X_i < \varepsilon)=0$ for $\varepsilon >0$.
How to determine the almost sure limit of the random variable $Y_n:=(\prod_{i=1}^{n}{X_i})^{\frac{1}{n}}$ for $n \to \infty$?
My idea was:
Since $(X_i)_{i \in \mathbb{N}}$ are integrable, it's $\mathbb{E}[|X|]<\infty$.
$Y_n=(X_1X_2 \cdot \cdot \cdot X_n)^{\frac{1}{n}}=\sqrt[n]{X_1X_2\cdot \cdot \cdot X_n}$
Now I'm not sure how to continue.
I also tried: $Z:=e^Y$ is log-normal distributed with $\mathbb{E}[Z]=\mathbb{E}[e^Y]=e^{\mu + \frac{\sigma^2}{2}}$, but I don't know what to do next.
I know that $Y_n$ converges almost sure to $Y$ if $\mathbb{P}[\lbrace \omega : \lim\limits_{n\to\infty} Y_n(\omega)=Y(\omega)\rbrace]=1$, but how can it be applied here?
First hint: You mention $\log$-normal distributions (which is a mislead in your case). But it seems that you had the intuition that applying logarithms could be a good idea. Why don't you try it ?
Second hint: which very-well known theorem of probability gives you an almost-sure convergence? (sub-hint: it's not the one that gives you the normal distribution...)