# Convergence of $\sum\limits_{n=0}^{\infty} (1-|a_n|)$

So I must prove that if $$(a_n)$$ is a sequence of points in $$\mathbb{C}$$ with $$0< |a_n| < 1 \; \forall n \in \mathbb{N}$$ and verifying that $$|b| \leq \prod\limits_{n=1}^{\infty} |a_n|$$ with $$0<|b| < 1$$, then the series

$$\sum\limits_{n=0}^{\infty} \left(1-|a_n|\right)$$ converges.

My closest attempt:

Naming the partial sums as $$S_n = \sum\limits_{k=1}^n \left(1-|a_n|\right)$$ using the general Bernouilli inequality for $$-\frac{|a_k|}{n}$$ I get that

$$\prod\limits_{k=1}^n \left(1-\frac{|a_n|}{n}\right) \geq 1 - \sum\limits_{k=1}^n \frac{|a_n|}{n}$$

Multiplying by $$n$$ in each side give us $$n \prod\limits_{k=1}^n \left(1-\frac{|a_n|}{n}\right) \geq n - \sum\limits_{k=1}^n |a_n| = \sum\limits_{k=1}^n (1-|a_n|)=S_n$$

Therefore we get that $$S_n \leq n \prod\limits_{k=1}^n \left(1-\frac{|a_n|}{n}\right)$$

Taking $$\log$$ both sides give us $$\log{S_n} \leq \log{(n)} + \sum\limits_{k=1}^n \log{\left(1-\frac{|a_n|}{n}\right)}$$

Now using that $$\log{(1-x)} \leq -x$$ we get $$\log{S_n} \leq \log(n) - \sum\limits_{k=1}^n \frac{|a_n|}{n}$$

And now I can't continue, the problem is I need to use $$|b| \leq \prod\limits_{n=1}^{\infty} |a_n|$$ somewhere but I couldn't manipulate the series to use that. Any hint on how can I show convergence?

Using the “well-known” estimate $$\log x \le x -1$$ for $$x > 0$$: $$S_n = \sum_{k=1}^n (1-|a_n|) \le - \sum_{k=1}^n \log |a_k| = - \log \prod_{k=1}^n |a_k| \le -\log b$$ so the partial sums are bounded, which implies that $$\sum_{n=0}^{\infty} (1-|a_n|)$$ is convergent.