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

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.

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.

• 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