Show that $S_n/n$ converges in probabilty but not almost surely - Borel Cantelli

Question

Let $X_n$ be independent random variables with the following distribution:

$$ P(X_n=\pm n)=\frac{1}{2(n+1)\log (n+1)}, \;\;\;\; P(X_n=0)=1-\frac{1}{(n+1)\log (n+1)} $$

and let $S_n=\sum_{k=1}^n X_k$. Show that $\frac{S_n}{n}\to 0$ in probability but not almost surely.

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I'm a bit doubtful on almost sure convergence part. This was what I thought:

set $A_n=\{X_n=n\}$, then clearly $A_n$ are independent events, and since

$$ \sum_{n=1}^{\infty} P(A_n)=\sum_{n=1}^{\infty} \frac{1}{2(n+1)\log (n+1)}=\infty $$

By BC lemma $P(A_n \;\;\text{i.o.})=1$. But then we know

$$ \frac{S_n}{n}=\frac{X_n}{n}+\frac{S_{n-1}}{n} \implies \frac{S_n}{n}=1+\frac{S_{n-1}}{n-1}+\delta_n\; \text{infinitely often} $$

where $\delta_n=S_{n-1}\left(\frac{1}{n}-\frac{1}{n-1}\right)\to 0$. So for all sufficiently large $n$, $S_n/n$ deviates from its previous value by approximately $1$ almost surely, so no convergence.

Couple of questions: Is my argument reasonably rigorous? Also, it seems that this sequence almost surely does not converge (to any limit, not just $0$) according to my reasoning. I'm also a bit curious about what $\limsup \frac{S_n}{n}$ would be - surely it would not be $0$, otherwise by symmetry we would have $\liminf=\limsup=0=\lim$.