 Solution

If $S_n/n$ converged almost surely to a finite limit, thenBut $\mathbb{P}(X_n/n>1)=\mathbb{P}(X_1>n)\sim 1/(\pi n)$. These independent-event probabilities have divergent sum, so the second Borel-Cantelli lemmas says $X_n/n>1$ infinitely often almost surely, a contradiction.