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.


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 $0$).
