$\Pi_{n=0}^\infty (1-a_n)>0$ if and only if $\sum a_n < \infty$. Let $a_n$ be sequence in (0,1).

$\Pi_{n=0}^\infty (1-a_n)>0$  if and only if $\sum_{n=0}^\infty a_n < \infty$. 

First I considered $\sum log(1-a_n)$
and tried to find sum inequality. 
I guess that This problem may be solved by taylor expansion. Could you help me?

First observe that $1-x\leq e^{-x}$ for all $x\in(0,1)$. This implies that if $\sum_{n=1}^{\infty}a_n=\infty$, then
$$ 0\leq\prod_{n=1}^N(1-a_n)\leq e^{-\sum_{n=1}^Na_n}\to 0$$
as $N\to\infty$. Therefore if the series diverges then the product is zero.
On the other hand, we have 
$$e^{-x}\leq 1-x+\frac{x^2}{2}\leq 1-\frac{x}{2}$$
for $x\in(0,1)$, hence
$$ e^{-2x}\leq 1-x$$ 
for $x\in(0,\frac{1}{2})$. If $\sum_{n=1}^{\infty}a_n<\infty$ then $a_n\to 0$, hence $0<a_n<\frac{1}{2}$ for all but finitely many terms. Assuming for simplicity that $0<a_n<\frac{1}{2}$ for all $n$, we have
$$ \prod_{n=1}^N(1-a_n)\geq e^{-2\sum_{n=1}^Na_n}\geq e^{-2\sum_{n=1}^{\infty}a_n}>0 $$
for all $N$, so the infinite product is non-zero.
