Folland 1.32; Relation between infinite product of sequence and infinite sum of sequence I found this question asked here
Folland exercise 1.32
but I am still having issues solving it.

Suppose $\{a_j\}\subset(0,1)$.
a. $\prod(1-a_j)>0$ iff $\sum a_j<\infty$. (Compare $\sum\log(1-a_j)$ to $\sum a_j$).
b. Given $B\in(0,1)$, exhibit a sequence $a_j$ such that $\prod(1-a_j)=B$.

In particular, Curious says I should easily check that $\log(1-a_j)<-a_j$ when $0<a_j<1$, but for whatever reason I'm having issues seeing this. If I raise each side by e, I get $1-a_j<\frac{1}{e^{a_j}}$ and I don't see what to do from there.
Even if I get that step, the only thing I see from the hint in the problem is that $\log(1-a_j)<-a_j<a_j$ so $\sum\log(1-a_j)<\sum a_j$. So if $\sum a_j$ converges then $\sum\log(1-a_j)$ converges. Am I forgetting some basic notions about series and convergence?
 A: What you are forgetting is that the comparison test for convergence applies only to series with non-negative terms.  If $x_j < 0 < y_j$ then convergence of $\sum y_j$ does not necessarily imply convergence of $\sum x_j$.
It is true for $0 < a_j < 1$ that $\log(1 - a_j) < - a_j$ which implies
$$\log \prod_{j=1}^n (1 - a_j) = \sum_{j=1}^n \log(1-a_j) < - \sum_{j=1}^n a_j, $$
and, hence,
$$0 \leqslant P_n = \prod_{j=1}^n (1 - a_j) < \exp(- \sum_{j=1}^na_j).$$
All this gives us  is that if $\sum a_j$ diverges (to $+\infty$) then $\lim_{n \to \infty} P_n = 0$.  In this case we say that the infinite product "diverges to $0$."
It can be shown that convergence of $\sum a_j$ implies the convergence of the product but this requires a different approach.
Specifically if the series converges, then for any $1 > \epsilon > 0$ there is a positive integer $N$ such that $0 \leqslant \sum_{j=N}^\infty a_j < \epsilon$.  We also have
$$ \frac{P_n}{P_{N-1}}= \prod_{j=N}^n (1 - a_j) \geqslant 1 - \sum_{j=N}^n a_j > 1 - \epsilon > 0.$$
Hence, the LHS is bounded below. It is straightforward to show that $P_n/P_{N-1}$ is monotonically decreasing and , therefore, $P_n$ is convergent to a non-zero limit.
