$X_1,X_2,\ldots$ is a sequence of independent random variables and
$$P\{X_n = n^2 - 1\} = 1 - P\{X_n = -1\} = n^{-2}$$
Clearly, $E[X_n] = 0$. However, $\frac{1}{n}S_n \to -1$ almost surely where $S_n = \sum_{i=1}^nX_i$.
My proof of this assertion is as follows and I would appreciate it if someone could check its correctness as well as the precision of my reasoning.
$(X_n)_n$ being a collection of independent random variables, the event $\left\{\omega: \lim_{n\to\infty}\frac{1}{n}S_n(\omega) = -1\right\}$ belongs to the tail $\sigma$-algebra and hence has a probability of either zero or one. If I can show that the probability of this event is strictly positive, then I am done.
I note that $$\{X_2 = -1, X_3 = -1, \ldots\} \subset \left\{\lim_{n\to\infty}\frac{1}{n}S_n = -1\right\}$$
By monotone convergence $$P\{X_2 = -1, X_3 = -1, \ldots\} = \lim_{n\to\infty}P\{X_2 = -1, \ldots, X_n = -1\}$$ This yields $P\{X_2 = -1, X_3 = -1, \ldots\} = \frac{1}{2}$. Therefore, $P\left\{\lim_{n\to\infty}\frac{1}{n}S_n = -1\right\} \geq \frac{1}{2}$, which implies $\frac{1}{n}S_n \to -1$ almost surely.
I would also appreciate seeing different proofs of this claim, in particular ones that do not invoke Kolmogorov's $0-1$ law.
