For each $\beta \in(0,1)$, give a sequence $\{a_j\}\subset(0,1)$ with $\prod (1-\alpha_j) = \beta $

Question

Suppose $\{\alpha_j\} \subset (0,1)$. Give a sequence ${\alpha_j}$ such that $\prod (1-\alpha_j) = \beta $, where $\beta \in (0,1)$

I was able to prove part a. That is the infinite product of $(1-a_j)$ is positive iff $\sum \alpha_j < \infty $.

However, I have no idea how to exhibit a sequence ${a_j}$ so that the infinite product of $1-\alpha_j$ is some $\beta$ that is in $(0,1)$. Would I need to construct a specific sequence or a more general case?

Any help would be much appreciated as I am really stuck and don't have much of an idea even where to begin.

Answer 1

Let $\alpha_j=1-\exp(-t\,/\,2^{\,j})$ where $t =-\log \, \beta$. Can you verify that this sequence has the desired properties?