Let $ \{X_n\}_{n \ge 1} $ be a sequence of independent random variables satisfying $$ \mathbb{P}(X_n = -n^2) = 1 - \mathbb{P}(X_n = \frac{n^2}{n^2 - 1}) = \frac{1}{n^2}. $$ The question is to show that $S_n = X_1 + ... + X_n$ is a martingale such that $S_n / n \rightarrow 1$ a.s. and $S_n \rightarrow \infty$ a.s.
My attempt: $$ \mathbb{E}[X_m] = -m^2 \cdot \frac{1}{m^2} + (\frac{m^2}{m^2 - 1}) \cdot (1 - \frac{1}{m^2}) = -1 + 1=0, $$ and $\mathbb{E}[S_n] = \sum_{i = 1}^n \mathbb{E}[X_i] = 0.$ For $\mathcal{F}_{n} = \sigma(X_1, ..., X_n)$, we see that
$$ \begin{align*} \mathbb{E}[S_{n} | \mathcal{F}_{n - 1}] &= \mathbb{E}[S_{n - 1} + X_n |\mathcal{F}_{n - 1}] \\ &= \mathbb{E}[S_{n - 1} |\mathcal{F}_{n - 1}] + \mathbb{E}[X_n] \\ &= S_{n - 1} + 0 \\ &= S_{n - 1}. \end{align*} $$ Moreover, $\mathbb{E}|S_n| < +\infty$ since $$ \mathbb{E}|S_n| \le \mathbb{E}( \sum_{i = 1}^n |X_i|) = \sum_{i = 1}^n\mathbb{E}|X_i| = 2n. $$ If we let $A_n$ be the event that $X_n = \frac{n^2}{n^2 - 1}$, then $\sum_{n \ge 1} \mathbb{P}(A_n) = \sum_{n \ge 1} (1 - \frac{1}{n^2}) = +\infty $ and since $X_n$ independent for $n = 1, 2, ...$ we have by Borel-Cantelli lemma that $\mathbb{P}(A_n \text{ i.o}) = 1$. And since $\lim_n X_n = 1$, we can conclude that $S_n \rightarrow \infty$ almost surely.
Is this reasoning correct? If it is, how do I apply the same reasoning using Borel-Cantelli for $S_n/n \rightarrow 1$ a.s. ?