Basically the title. I'mo working on a proof right now and I'm not quite sure where to begin. I thought about writing out the $n$th partial product and then taking logarithms but that doesn't get rid of the $1$ in each of the terms. Just a nudge in the right direction would be very helpful. Basically I have to prove that if $a_k\leq 0$ or $a_k\geq 0$ then:
$$\prod_k^\infty c_k=\prod_k^\infty(1+a_k)=C\Leftrightarrow \sum_k^\infty a_k=S$$
Where $C$ is the limit of the $n$th partial product and $S$ is the limit of the $n$th partial sum.
Hint: $\displaystyle\log\left(\prod_{k=1}^N(1+a_k)\right)=\sum_{k=1}^N\log(1+a_k)$. Besides, $\displaystyle\lim_{k\to\infty}a_k=0\implies\lim_{k\to\infty}\frac{\log(1+a_k)}{a_k}=1$.