We have $$|V_{n+1}-V^\prime_{n+1}|=\prod_{m=1}^n|W_m||V_1-V^\prime_1| \tag{1}\label{1}$$ where $P(|W|=1)\ne 1$ (with $W\in[-1,1]$) and $W_n$s are iid and independent of $V$s and $V^\prime$s.
Show that \eqref{1} converges to $0$, using Borel-Cantelli lemma.
I have the proof in the paper I am reading, but I do not understand it. The proof states:
\eqref{1} converges to $0$ almost surely, because $W_n$s are iid and $P(W=1)<1$.
Attempt
$$E_{n+1}(\epsilon)=\{\omega\in\Omega: |V_{n+1}-V^\prime_{n+1}| > \epsilon\}$$ Than I would have to show that $$\sum_{n=1}^\infty P(E_n(\epsilon)) < \infty$$
But I do not know how to proceed.
Edit
This is the proof of Lemma 3.3 in the paper at the link. It seems simple, but I just don't get it... Please help me understand this proof; I am a self learner, mostly relying on stackExchange when I get stuck (like now...)
I really need an answer to this, it's really bothering me... If there is something wrong with my question, please tell me.
Here is a proof using the law of large numbers. The random variables $\{\log|W_m|\}$ are iid and non-positive. By the strong law of large numbers, we have $$ \frac1n\sum_{m=1}^n\log|W_m|\to E(\log|W|) $$ almost surely. But by assumption, $E|W|<1$, so by Jensen's inequality $$ E(\log|W|)\le \log E|W|<0. $$ Recall that for any sequence $\{a_n\}$ of non-positive reals, $$\lim_n\frac1n\sum_{m=1}^na_m<0 \quad\Longrightarrow\quad \sum _{m=1}^na_m\to-\infty.$$ Applying this result pointwise with $a_m:=\log|W_m(\omega)|$, it follows that with probability one, $\sum_{m=1}^n\log|W_m|\to -\infty$. By exponentiating, this last event is the same as the event $\prod_{m=1}^n|W_m|\to0$.