Prove that $\prod_1^{\infty}(1-a_i) > 0$ iff $\sum_1^{\infty}{a_i} < \infty$ 

Suppose $\{a_i\}_1^{\infty} \subset (0,1)$
    a) $\prod_1^{\infty}(1-a_i) > 0$ iff $\sum_1^{\infty}{a_i} < \infty$
    b) Given $\beta \in (0,1)$, exhibit a sequence $\{a_i\}$ such that $\prod_1^{\infty}(1-a_i) = \beta$


This is not my homework, but I'm learning measure theory from Real Analysis of Folland, and I get stuck on this problem. My idea is to prove that $\sum_1^{\infty}{\ln(1-a_i)} > -\infty$ (for sure this sum is smaller than 0). At the first glance, I try to prove that $\ln(1-x) + x > 0$, but finally, the inequality should be reversed. Using maclaurine expansion, I can expand:
$$\ln(1-x) = -(x + x^2/2 + x^3/3 +...)$$
So seem like I can't find a function $f(x)$ such that $\ln(1-x) +f(x) > 0$. Can anyone give me some hint to solve this? For the second problem, I got no idea. Thanks so much. I really appreciate!
 A: I think if you use $\ln(1-x)<-x$, then you can show that $%
%TCIMACRO{\dprod }%
%BeginExpansion
{\displaystyle\prod}
%EndExpansion
\left(  1-a_{i}\right)  >0\rightarrow%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
\ln\left(  1-a_{i}\right)  >-\infty\rightarrow%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
-a_{i}>-\infty\rightarrow%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
a_{i}<\infty.$
Now, if $%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
a_{i}<\infty$, then $a_{i}\rightarrow0$. Since $\frac{\ln\left(  1-x\right)
}{x}\rightarrow-1$ as $x\rightarrow0$, we have for $i$ big enough (means
$a_{i}$ small enough): $\ln\left(  1-a_{i}\right)  >-2a_{i}\rightarrow%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
\ln\left(  1-a_{i}\right)  >-2%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
a_{i}>-\infty$.
For b) if you want $%
%TCIMACRO{\dprod }%
%BeginExpansion
{\displaystyle\prod}
%EndExpansion
\left(  1-a_{i}\right)  =\beta\rightarrow$ $%
%TCIMACRO{\dsum }%
%BeginExpansion
{\displaystyle\sum}
%EndExpansion
\ln\left(  1-a_{i}\right)  =\ln\beta$ then you can use the Taylor expansion to
expand $\ln\beta$ and then choosing $a_{i}$ accordingly.
