Let $(X_n)_{n\geq 1}$ be a sequence of independent and identically distributed standard normal random variables and define for $n\geq 1$,
$$ W_n = \exp\left\{\sum_{i=1}^nX_i-\frac{n}{2}\right\} $$ Show that $W_n$ converges almost-surely and check if the sequence is a uniformly integrable collection.
$W_n$ is clearly a log-normal random variable with the following moments, \begin{align} \mathbb{E}[W_n] &= 1\\ \operatorname{\mathbb{V}ar}[W_n] &= \exp(n)-1 \end{align} that are easily obtainable applying the fact that the moment generating function for the normal distribution is $$ M_X(t)=\exp\left\{\mu t + \frac{t^2\sigma^2}{2} \right\}. $$
I tried to apply the Kolmogorov 3-series theorem and the Levy equivalence, but failed as the variance explodes. I also tried working with $S_n -\frac{n}{2}$ and applying the continuous mapping theorem, but it also didn't work, I would appreciate some guidance.
By taking logarithms, dividing by $n$, and re-arranging, we obtain: $$\frac{1}{n} \sum_{i=1}^n X_i = \frac{1}{n} \log W_n + \frac{1}{2}$$ By the strong law of large numbers, $$\frac{1}{n} \sum_{i=1}^n X_i \to 0$$ almost surely. Thus, $\frac{1}{n} \log W_n \to - \frac{1}{2}$ almost surely, which implies $\log W_n \to - \infty$ almost surely. The continuous mapping theorem allows us to conclude that $W_n \to 0$ almost surely.
The sequence is not uniformly integrable since $E(W_n) = 1$ for all $n$, yet $E(W_\infty) = 0 \neq 1$.