Suppose we have a sequence $0\leqslant a_n \leqslant 1$. Assume $$ 0<\lim _{n\to\infty} \prod _{k=n}^\infty (1-a_k) = p \leqslant 1 $$ Can we say anything about $p$? Does it have to be equal to $1$?
The question is inspired by the following. Given a sequence of independent events $A_n$, find the probability of $\limsup A_n$ (or $\liminf A_n$). The conjecture is that such events have probability of either $0$ or $1$. Assume $$ \mathbb P( \liminf A_n^c) = \lim _{n\to\infty} \mathbb P\left ( \bigcap _{k\geqslant n} A_k^c \right ) = \lim _{n\to\infty} \prod _{k\geqslant n}(1-\mathbb P(A_k)) >0 $$ Does it follow that this "remainder product" converges to $1$? Of course, we know the conjecture is true due to either the Borel Cantelli lemmas or Kolmogorov 0-1. But suppose we know nothing about them. Can we justify this purely by some infinite product analysis?
We may assume that eventually $a_k<1$. I know something like $$ \forall n,\quad \prod _{k\geqslant n} (1-a_k) >0 \Leftrightarrow \sum _{k\geqslant n} a_k <\infty $$ I think it is reasonable to assume $\prod _{k\geqslant n}(1-a_k)$ is eventually positive, so by the above it would have to be that $a_k\to 0$. So $1-a_k \to 1$ and intuitively speaking, if we keep removing terms then surely the product would have to converge to $1$? How do I rigorously fill this gap? Or is this all hogwash and the conjecture regarding $p$ is false?
From the hypothesis, it follows that $a_k=1$ for at most finitely many $n$ and those finitely many can be ignored for the final conclusion . Now $\prod_{k=n}^{\infty} (1-a_k)=\prod_{k=1}^{\infty} (1-a_k)/\prod_{k=1}^{n-1} (1-a_k) \to 1$.
This is essentially a restatement of the fact that if a series converges then the tail sums tend to $0$.