Convergence of $\prod_{n=1}^\infty(1+a_n)$

The question is motivated by the following exercise in complex analysis:

Let $\{a_n\}\subset{\Bbb C}$ such that $a_n\neq-1$ for all $n$. Show that if $\sum_{n=1}^\infty |a_n|^2$ converges, then the product $\prod_{n=1}^\infty(1+a_n)$ converges to a non-zero limit if and only if $\sum_{n=1}^\infty a_n$ converges.

One can get a proof by using $|a_n|^2$ to bound $|\log(1+a_n)-a_n|$.

Here is my question: is the converse of this statement also true?

If "the product $\prod_{n=1}^\infty(1+a_n)$ converges to a non-zero limit if and only if $\sum_{n=1}^\infty a_n$ converges", then $\sum_{n=1}^\infty |a_n|^2$ converges.

Solution 1:

I shall try to give examples where $\sum|a_n|^2$ is divergent and all possible combinations of convergence/divergence for $\prod(1+a_n)$ and $\sum a_n$.

Let $a_{2n}=\frac1{\sqrt n}$ and $a_{2n+1}=\frac1{1+a_{2n}}-1=-\frac{1}{1+\sqrt n}$. Then $(1+a_{2n})(1+a_{2n+1})= 1$, hence the product converges. But $a_{2n}+a_{2n+1}=\frac1{n+\sqrt n}>\frac1{2n}$, hence $\sum a_n$ diverges.

Let $a_{2n}=\frac1{\sqrt n}$ and $a_{2n+1}=-\frac1{\sqrt n}$. Then $a_{2n}+a_{2n+1}=0$, hence $\sum a_n$ converges. But $(1+a_{2n})(1+a_{2n+1})=1-\frac1n$; the $\log$ of this is $\sim -\frac1n$, hence $\sum \log(1+a_n)$ and also $\prod(1+a_n)$ diverges.

Let $a_n=\frac1{\sqrt n}$. Then $\prod(1+a_n)$ and $\sum a_n$ diverge.

It almost looks as if it is not possible to have both $\prod(1+a_n)$ and $\sum a_n$ convergent if $\sum |a_n|^2$ diverges because $\ln(1+a_n) = a_n-\frac12a_n^2\pm\ldots$, but here we go: If $n=4k+r$ with $r\in\{0,1,2,3\}$, let $a_n = \frac{i^r}{\sqrt k}$. Then the product of four such consecutive terms is $(1+\frac1{\sqrt k})(1+\frac i{\sqrt k})(1-\frac1{\sqrt k})(1-\frac i{\sqrt k})=1-\frac1{k^2}$, hence the log of these is $\sim -\frac1{k^2}$ and the product converges. The sum also converges (to $0$).

Solution 2:

The convergence of $\Pi_{n=0}^\infty(1+a_n)$ is equivalent to that of $\sum_{n=0}^\infty\ln(1+a_n)$. Note $$ \ln(1+x)=x-\frac{1}{2}x^2+O(|x|^3). $$ So $$\sum_{n=0}^\infty\ln(1+a_n)=\sum_{n=0}^\infty (a_n-\frac{1}{2}a_n^2+O(|a_n|^3))=\sum_{n=0}^\infty a_n-\frac{1}{2}\sum_{n=0}^\infty a_n^2+\sum_{n=0}^\infty O(|a_n|^3).$$ Note the convergence of $\sum_{n=0}^\infty|a_n|^2$ implies the convergence of $\sum_{n=0}^\infty a_n^2$ and $\sum_{n=0}^\infty O(|a_n|^3)$. Thus the product $\Pi_{n=0}^\infty(1+a_n)$ converges to a non-zero limit if and only if $\sum_{n=0}^\infty a_n$ converges.