I need to show that if $a_k\in [0,1]$, and $\sum_{k=0}^\infty a_k=\infty$, then:
$$\prod_{k=0}^\infty(1-a_k)=0$$
I think the fact that I need to relate an infinite sum with an infinite product is throwing me off. I have the sense that I need to use the fact that $1-x\leq e^{-x}$ for $x\in [0,1)$, and I've tried looking at:
$$\prod_{k=0}^\infty \frac{1}{e^x}$$
to see what I could come up with, but regardless of what I look at, I'm not sure how to apply the assumed condition to this infinite product. I've tried separately to think about the geometric series:
$$\frac{1}{1-x}=\sum x^k$$
but then I'm not certain how to relate it to a product. I feel as though this is rather elementary and that I'm missing something very basic, but I really can't put my finger on it. Any suggestion as to where to look would be truly appreciated.
$0 \leq\prod_{k=1}^{n} (1-a_k) \leq \prod_{k=1}^{n} e^{-a_k}=e^{-\sum\limits_{k=1}^{n} a_k} \to 0$ since $e^{-x} \to 0$ as $x \to \infty$