Convergence of random series $\sum_{k=1}^{n}\prod_{i=k}^{n} X_i$ If $X_i$ are i.i.d. random variable such that $|X_i|\leq 1$ almost surely for any $i$.
Define
$$S_1 = X_1$$
$$S_2 = X_2(1+S_1) = X_2 +X_1X_2$$
$$S_3 = X_3(1+S_2) = X_3 + X_2X_3 + X_1 X_2 X_3$$
$$S_3 = X_4(1+S_3) = X_4 + X_3X_4 + X_2 X_3 X_4+ X_1X_2 X_3 X_4$$
......
$$S_n = X_n(1+S_{n-1}) = \sum_{k=1}^{n}\prod_{i=k}^{n} X_i$$
It seems to be a random version of the geometric sum (if every $X_i=c$ then it is just a geometric sum). But in this case, the summands $\prod_{i=k}^{n} X_i$ are not independent, so Kolmogorov's three-series theorem does not apply.
I wonder if there is any chance we can prove the convergence of $S_n$ to some constant or to some limiting random variable? Thank you!
 A: Under the assumption, we find that

Claim. $(S_n)$ converges a.s. if and only if $X_n = c$ a.e. for some constant $c \in (-1, 1)$.

Since the implication $(\impliedby)$ is trivial, let's prove the direction $(\implies)$:
Suppose $(S_n)$ converges a.s. and let $S_{\infty}$ denote the a.s. limit of $(S_n)$.  From the recurrence relation and the condition $|X_i| \leq 1$, we have $|S_{\infty}| \leq |S_{\infty} + 1|$ and hence $S_{\infty} \geq -\frac{1}{2}$ a.s. Then
$$ X_n = \frac{S_n}{1 + S_{n-1}} $$
also converges a.s. Now using the lemma below, we find that $(X_n)$ converges a.s. to a constant:

Lemma. Let $(Y_n)$ be a sequence of independent random variables that converges a.s. Then the limit is constant a.s.
Proof. Let $Y_{\infty}$ denote the a.s.-limit of $(Y_n)$. Then $Y_{\infty}$ is measurable w.r.t. the tail $\sigma$-algebra of $(Y_n)$, hence $Y_{\infty}$ is constant a.s. by the Kolmogorov $0$-$1$ theorem. $\square$

However, since $X_n$'s are identically distributed, this implies that each $X_n$ is constant. Then it is easy to check that this constant value must lie in $(-1, 1)$ for $(S_n)$ to converge.
