Let $X_1, \ldots, X_n$ i.i.d random variables and consider the random walk $X_1 + \ldots X_n$.
I consider the event $\{ S_n = 0 \text{ i.o}\}$. I've been told at lectures that this event lies in the exchangeable $\sigma$-algebra, but not in the tail $\sigma$-algebra $\mathcal{T}= \cap_n \sigma(X_n,X_{n+1}...)$, I'd like to convince myself.
Since $\{ S_n = 0 i.o\} = \{ \limsup_n S_n =0 \}$, if I permute a finite number of indices , the limit remains the same, and hence it should belong to the exchangeable sigma algebra.
BUT it's not a tail event since if, for instance, I could change the value of $X_1$ with a real large number as to make the limsup stricly positive. Then the event is influenced by a changed of a finite number of $X_n$, and hence would not be in $\mathcal{T}$.
Of course, this is just something intuitive, and now I want to formalize the second one
I note that $$S_{n+k} = S_n + X_{n+1} + \ldots X_{n+k}$$
Hence $$\limsup_n S_n = \limsup_k S_n + X_{n+1} + \ldots + X_{n+k}$$ can I conclude now that this event is not $\sigma(X_{n+1}, X_{n+2})$-measuralbe, and hence it's not a terminal event?
$\{S_n=0 \, i.o.\}$ and $\{lim \sup S_n=0\}$ are very different. If $S_n>0 $ for all $n$ and $S_n \to 0$ almost surely then the first set is empty and the second one has probability $1$.
It is not possible to prove that the first event never belongs to tail sigma filed. For example if $X_i=0$ with probability $1$ then the event becomes $\Omega$ which does belong to the tail sigma filed. What you can say is that this event need not be in the tail sigma field.