Convergence in distribution of the sample geometric mean for positive random variables

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I am stumped on the current problem regarding asymptotic distributions of random variables.

Suppose that $X_n$ are positive, iid random variables with finite mean $\mu$ and finite variance $\sigma^2$. Let $P_n=(\prod_{k=1}^{n}X_k)^{1/n}$ and let $Z_n=\frac{P_n - \mu}{\sigma/\sqrt{n}}$ for $n \geq 1$. Find the limit of $Z_n$ in distribution as $n \to \infty$.


Thus far, what I have noticed is that $\log P_n=\frac{1}{n}\sum_{k=1}^{n}\log(X_k)=\overline{\log(X_n)}$, so one could apply the Delta Method to $\frac{\overline{\log(X_n)}-E(\log (X_1))}{\sigma/\sqrt{n}}$ as this will converge to a normal distribution by the CLT. However, in the problem statement we have $\mu$ and not $e^{E(\log (X_1))}$, so I would not be sure on how to proceed. On top of that, we may not even know whether $E(\log (X_1))$ is finite or not. What would be some hints on how to proceed? I think it would be more straight forward if the $X_n$ followed a particular distribution, but in this case it is completely arbitrary (just that they are positive and have finite mean and variance).