Let $(X_n)$ be a sequence of independent uniform random variables in $[-1,1]$. Prove of disprove: $\lim_{n\to\infty} (1+X_1)(1+\frac{X_2}{2})...(1+\frac{X_n}{n})$ exists and belongs to $(0,\infty)$ almost surely.
Attempt: If we denote $Y_n=(1+X_1)(1+\frac{X_2}{2})...(1+\frac{X_n}{n})$ then it is well known that $(Y_n)$ is a martingale. Since it is non negative we conclude from the martingale convergence theorem that it must converge almost surely to a finite limit. So the only question is: is the limit almost surely in $(0,\infty)$ or it might be $0$ with a positive probability?
The limit is zero if and only if (we can forget about the case when $X_1=-1$ because it happens with probability $0$) for each $n$ we have $\prod_{k=n}^\infty (1+\frac{X_k}{k})=0$. So the event $\{Y_n\to 0\}$ belongs to the tail sigma algebra of $(X_n)$ and by Kolmogorov's $0-1$ law its probability is either $0$ or $1$. So now I just need to check if the probability of the limit of $Y_n$ being zero is positive or not. I had an idea to show that if at a point $\omega$ we have $(1+X_1(\omega))...(1+\frac{X_n(\omega)}{n})\to 0$ then at this same point we have $(1-X_1(\omega))...(1-\frac{X_n(\omega)}{n})\to\infty$. Since $Z_n:=(1-X_1(\omega))...(1-\frac{X_n(\omega)}{n})$ is a non negative martingale it must converge to a finite limit almost surely as well. So if my idea was correct then we would get that the set of such points $\omega$ must have probability zero, hence that would imply that almost surely the limit of $Y_n$ is in $(0,\infty)$. However, I failed to prove that my idea was correct. Any ideas?