Characterize the type of sequence that satisfies $\prod (1-a_i) \leq c$ Consider a product $\prod_{i=1}^{n} (1-a_i)$ where $n\leq \infty$ and $a_i\in [0,1)$ for all $i$. I'm hoping to see if there exist conditions on the sequence $\{a_i\}$ so that
$$\prod_{i=1}^{n} (1-a_i)\leq c.$$
The product can be reduced to $\sum_{B\subseteq \{1,2,\dots,n\}} (-1)^{|B|} \prod_{b\in B} a_b$, but from there it's not clear where to go. Simple limiting convergence results aren't strong enough to address the bound $c$. When $c=0$ and $n=\infty$, this will require that $\sum_{i=1}^\infty a_i = \infty$ if I am not mistaken.
I realize this question is very general, so feel free to impose additional assumptions on $\{a_i\}$ if necessary.
 A: We can take logarithms, then the inequality becomes
$$\sum_{i=1}^n \log (1-a_i) \leqslant \log c.\tag{1}$$
Expanding the logarithms in Taylor series,
$$\log (1-x) = -\sum_{k=1}^\infty \frac{x^k}{k}$$
and multiplying with $-1$, it becomes
$$\sum_{i=1}^n \sum_{k=1}^\infty \frac{a_i^k}{k} \geqslant -\log c.\tag{2}$$
Since all terms on the left are non-negative, we can change the order of summation and get
$$\sum_{k=1}^\infty \frac{1}{k}\sum_{i=1}^n a_i^k \geqslant -\log c.\tag{3}$$
In all this, we can allow $c = 0$ with the small abuse of notation that $\log 0 = -\infty$ and $-\log 0 = +\infty$.
Then we get a chain of sufficient conditions


*

*$\sum\limits_{i=1}^n a_i \geqslant -\log c$,

*$\sum\limits_{i=1}^n a_i + \frac{1}{2}\sum\limits_{i=1}^n a_i^2 \geqslant -\log c$


and so on.
For $c = 0$ the conditions $n = \infty$ and
$$\sum_{i=1}^\infty a_i = +\infty$$
are necessary and sufficient, for if $\sum a_i$ were finite, then also $\sum \log (1-a_i)$ would be finite, and the product strictly positive.
For $c \geqslant 1$, the inequality is trivially satisfied, and for $0 < c < 1$, $(1),(2),(3)$ are somewhat tautological necessary and sufficient conditions, but the sufficient conditions involving only the finitely many of the power sums may be of the kind you are interested in.
A: Let $\{a_i\}_{i=1,\ldots,n-1}$ be any sequence in $[0,1)$. Then you can just define the $n$th term to be large enough so that full product is small enough: $$a_n\geq1-c\prod\limits_{i=1}^{n-1}(1-a_i)^{-1}$$
So if there is no constraint on $n-1$ out of $n$ terms, then is this the right question to ask? Perhaps if the sequence $\{a_i\}$ were required to be decreasing? But then there would be no constraints on the last $n-1$ terms, and you could make the first term as large as needed. Putting a constraint on only one term of a sequence but letting the rest of the sequence do whatever it wants doesn't feel like characterizing the sequence in a meaningful way.
If $n=\infty$, these considerations don't quite apply.
