So I have a sequence of positive random variables, and I know that $E(|\log X_k|) < \infty$ for all $k \in \mathbb{N}$.
I'm trying to compute $\lim_{n \rightarrow \infty} \left(\prod_{k=1}^n X_k\right)^{\frac{1}{n}}$. So the most obvious thing I thought of to do was:
that $$\log\left( \left(\prod_{k=1}^n X_k\right)^{\frac{1}{n}}\right) = \sum_{k=1}^n \frac{\log(X_k)}{n} \implies E\left( \log\left(\left( \prod_{k=1}^n X_k\right)^{\frac{1}{n}}\right)\right) = \sum_{k=1}^n \frac{E(\log X_k)}{n}.$$ Since $\log X_k \in L^1((\Omega,\mathcal{F},P))$, it follows that the summation of the righthand side makes sense. Now since the $X_k$'s are i.i.d, it follows that $E(\log X_k) = \mu < \infty$ for all $k \in \mathbb{N}$. Hence, we conclude: $$\sum_{k=1}^n \frac{E(\log X_k)}{n} = \mu\sum_{k=1}^n \frac{1}{n} = \mu.$$ Thus, since $E$ is linear (and hence continuous) and $\log$ is continuous on $(0,\infty)$, we conclude that: $$\lim_{n \rightarrow \infty} E\left( \log\left(\left( \prod_{k=1}^n X_k\right)^{\frac{1}{n}}\right)\right) = E\left( \log\left(\lim_{n \rightarrow \infty}\left( \prod_{k=1}^n X_k\right)^{\frac{1}{n}}\right)\right) = \lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{E(\log X_k)}{n} = \mu.$$
So now I have $E(\log(\boldsymbol{X})) = \mu =E(\log(X_k))$ for $\boldsymbol{X} = \lim_{n \rightarrow \infty} \left(\prod_{k=1}^n X_k \right)^{\frac{1}{n}}$. I'm not sure what to do from here.
I know that what I'm looking for is $e^{\mu} = e^{E(\log(X_k))}$ for any $k \in \mathbb{N}$, but I don't see how I can remove the expected value to conclude that $$\log(\boldsymbol{X}) = \mu \implies \boldsymbol{X} = e^\mu.$$