An equivalent of a central limit theorem for a geometric mean

Question

I'm taking a probability theory actuary exam next week and came accross an interesting problem - consider a sequence $(X_n)_{n=1}^{\infty}$ of i.i.d. random variables with density $$f(x) = \frac{3}{x^4}\mathbb{1}_{(1, \infty)}(x)$$ and let $$U_n = (X_1 \cdot \ldots \cdot X_n)^{\frac{1}{n}}$$ I have to prove that $$P(3(U_n - e^{\frac{1}{3}})\sqrt{n} > e^{\frac{1}{3}}) \approx 0.1587$$ but I don't know how to approach it. My first thought was to attain some kind of an equivalent of a central limit theorem for the geometric mean. Taking the random variable $\ln U_n$ and proving $\mathbb{E} \ln U_n = \frac{1}{3}$, therefore $\ln U_n \rightarrow \frac{1}{3}$ almost surely and $U_n \rightarrow e^{\frac{1}{3}}$ a.s., alas I don't know what to do next.