Product of absolute values nonzero implies sum of one minus absolute value is bounded

Question

I'm reading a proof that claims a well known equivalence: $\Pi_{k=1}^\infty |\alpha|>M>0$ implies $\sum_{k=1}^\infty 1-|\alpha|<\infty$ for $|\alpha_k|<1$ in the complex numbers. It seems to suggest writing $1-(1-|\alpha|)=\alpha$, but I don't really see how this helps. Expanding the product, many more terms appear than just our desired sum.

Answer 1

Hints: $\frac {\ln x} {x-1} \to 1$ as $ x \to 1$ by L'Hopital's Rule. The hypothesis implies that $|\alpha_k| \to 1$ as $ k \to \infty$ and $\sum \ln |\alpha_k|$ is absolutely convergent. By Comparison test $\sum |(|\alpha_k| -1)|$ is convergent.