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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?

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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.

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  • $\begingroup$ oh my god I tried several ways and I didn´t notice that simple solution. Thank you very much!! $\endgroup$ – Ahlfkushevich Feb 14 at 21:08

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