Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that for complex sequence $\{a_n\}_{n\in\mathbb{N}}$ :

if $\displaystyle \sum_{n=1}^\infty |a_n|^2<\infty$, Then : $\displaystyle \sum_{n=1}^\infty a_n$ Converges $\Leftrightarrow \prod_{n=1}^\infty(1+a_n)$ Converges

#problem from Stein, Complex Analysis#

As we've done in proof of $\displaystyle \sum_{n=1}^\infty |a_n|<\infty\Longrightarrow \prod_{n=1}^\infty(1+a_n)<\infty$ we can write :

\begin{align*} \forall \,|z|<\frac12\,:\,|\log(1+z)|&\leq|z|+\frac{|z|^2}{2}+\frac{|z|^3}{3}+\cdots\\ &\leq|z|+|z|^2(\frac12+\frac14+\cdots)\\ &\leq|z|+|z|^2\tag{*} \end{align*} Now we may have : \begin{align*} \left|\prod_{n=1}^\infty(1+a_n)\right|&\leq\exp \left|\displaystyle \sum_{n=1}^\infty\log(1+a_n)\right|\\ &\leq \exp \left(\displaystyle \sum_{n=1}^\infty|\log(1+a_n)|\right)\\ &\leq\exp\left(\sum_{n=N}^\infty |a_n|+|a_n|^2\right)\qquad\text{by Ineq. ($*$)}\tag{**} \end{align*} But it seems it doesn't help us. Now can we change $(**)$ to check Cauchy-ness of the series !

For the converse I've no idea yet and let's share our ideas.

Any help is appreciated.

share|cite|improve this question
up vote 2 down vote accepted

I suppose that where you wrote $\sum a_n < \infty$ resp $\prod (1+a_n) < \infty$, you meant "$\sum a_n$ converges" resp. "$\prod (1+a_n)$ converges". The condition $<\infty$ doesn't make sense unless all $a_n$ are real, and that would be an undue limitation.

You know that $\prod (1+a_n)$ converges if and only if $\sum \log (1+a_n)$ converges (where $a_n \neq -1$ for all $n$, and $\log$ denotes the principal branch for all large enough $n$).

So the exercise is to show that if $\sum \lvert a_n\rvert^2 < \infty$, then $\sum \log (1+a_n)$ converges if and only if $\sum a_n$ converges.

Instead of your $(\ast)$, look at

$$\lvert a_n - \log (1+a_n)\rvert$$

to obtain the conclusion.

share|cite|improve this answer
Thank you It was such a useful idea to remove the term $a_n$ from direct calculation – Fardad Pouran Jun 7 '14 at 12:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.