# Almost sure convergence of infinite product

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?

• Kolmogorov 3-series theorem tells that $$\sum_{n=1}^{\infty}\frac{X_n}{n}$$ converges almost surely. Now notice that the infinite product in question converges to a positive value if and only if this series converges. May 28, 2019 at 19:21
• I can see that the series converges almost surely. I suppose you mean that $\prod_{n=1}^\infty (1+a_n)$ converges to a positive number if and only if $\sum_{n=1}^\infty a_n$ converges. This is a known theorem, but don't we need the terms of the sequence $a_n$ to be non negative to use it?
– Mark
May 28, 2019 at 19:58
• Good point. I was lazy enough to skip explaining that I am using the fact that $\log(1 + X_k/k) = (X_k/k) + \mathcal{O}(1/k^2)$. May 28, 2019 at 20:00
• Oh, so here it's just Taylor expansion of logarithm. I guess this is the way to work with infinite products in general. Thanks a lot for your help.
– Mark
May 28, 2019 at 20:05

$$\prod_{k=2}^{n} \left( 1 - \frac{X_k}{k} \right) = \prod_{k=2}^{n} \left( 1 + \frac{X_k}{k} \right)^{-1} e^{\mathcal{O}(1/k^2)},$$
which follows from $$1 - x = \frac{1-x^2}{1+x}$$. From this, we find that $$\left[ \prod_{k=1}^{n} \left( 1 + \frac{X_k}{k} \right) \right]^{-1}$$ converges almost surely in $$(0, \infty)$$, and so, its reciprocal cannot be zero.