Characterize the type of sequence that satisfies $\prod (1-a_i) \leq c$

Question

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.

Answer 1

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.

Answer 2

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.