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.

 A: 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 $0$).
A: 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.
A: Suppose $\{a_n\}$ is a sequence of complex numbers such that convergence/divergence of $\prod_{n=1}^\infty (1 + a_n)$ is equivalent to convergence/divergence of $\sum_{n=1}^\infty a_n$. (For simplicity's sake, let us assume $a_n \neq -1$ for all $n$, so that we don't trivially get a product of zero.) It is self-evident that
$$\sum_{n=1}^\infty a_n \text{ and } \prod_{n=1}^\infty e^{a_n} = e^{\sum_{n=1}^\infty a_n}$$
either both converge, or both diverge. So we can rephrase OP's question as:
"Suppose $\{a_n\}$ is a sequence of complex numbers such that convergence/divergence of $\prod_{n=1}^\infty (1 + a_n)$ is equivalent to convergence/divergence of $\prod_{n=1}^\infty e^{a_n}$. Must $\sum_{n=1}^\infty a_n^2$ converge absolutely?"


Lemma: In order for convergence/divergence of $\prod_{n=1}^\infty (1 + a_n)$ to be equivalent to convergence/divergence of $\prod_{n=1}^\infty e^{a_n}$, then
$$\prod_{n=1}^\infty (1 + a_n) e^{-a_n}$$
must converge.
Proof of lemma: We need to show that if one of these products diverges but the other one converges, then $\prod_{n=1}^\infty (1 + a_n) e^{-a_n}$ diverges.

*

*Case 1: $\prod_{n=1}^\infty (1 + a_n)$ diverges but $\prod_{n=1}^\infty e^{a_n} \rightarrow P \neq 0$. Then

$$\lim_{N \rightarrow \infty} \prod_{n=1}^N (1 + a_n) e^{-a_n} = \frac{1}{P} \lim_{N \rightarrow \infty} \prod_{n=1}^N (1 + a_n),$$
and by hypothesis $\lim_{N \rightarrow \infty} \prod_{n=1}^N (1 + a_n)$ DNE, so neither does $\lim_{N \rightarrow \infty} \prod_{n=1}^N (1 + a_n) e^{-a_n}$.

*

*Case 2: $\prod_{n=1}^\infty e^{a_n}$ diverges but $\prod_{n=1}^\infty (1 + a_n)$ converges to a nonzero limit. The same idea as in Case 1 works to show that $\lim_{N \rightarrow \infty} \prod_{n=1}^N (1 + a_n) e^{-a_n}$ DNE.



Armed with our lemma, it is now easy to find examples of sequences $\{ a_n \}$ for which $\prod_{n=1}^\infty (1 + a_n) e^{-a_n}$ converges, but $\sum_{n=1}^\infty |a_n|^2$ is divergent.
For instance, if we pick $a_n = \frac{i^n}{\sqrt{n}}$, where $i = \sqrt{-1}$, then $\sum_{n=1}^\infty |a_n|^2 = \sum_{n=1}^\infty n^{-1}$ diverges. However:

*

*Since

$$(1 + a_n) e^{-a_n} = 1 - a_n^2/2 + \Theta(a_n^3),$$
and:

*

*For this sequence, $\sum_{n=1}^\infty a_n^2 = \sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges, while $\sum_{n=1}^\infty a_n^3$ converges absolutely because $|a_n^3| = n^{-3/2}$,

this is enough to show that
$$\prod_{n=1}^\infty (1 + a_n) e^{-a_n} = \prod_{n=1}^\infty [1 - a_n^2/2 + \Theta(a_n^3)]$$
converges, since the sum
$$\sum_{n=1}^\infty -a_n^2/2 + \Theta(a_n^3)$$
converges and the sum
$$\sum_{n=1}^\infty [-a_n^2/2 + \Theta(a_n^3)]^2$$
converges absolutely: $[a_n^2/2 + \Theta(a_n^3)]^2 = \Theta(n^{-2})$.


However: if the $\{ a_n \}$ are restricted to be real numbers, then it does become true that "convergence/divergence of $\prod_{n=1}^\infty (1 + a_n)$ is equivalent to convergence/divergence of $\sum_{n=1}^\infty a_n$ only if $\sum_{n=1}^\infty |a_n|^2 = \sum_{n=1}^\infty a_n^2$ converges".
The only interesting case is when $a_n \rightarrow 0$, so making the assumption $a_n \rightarrow 0$, we will eventually have, for any $\epsilon > 0$,
$$1 - (1 + \epsilon)a_n^2/2 \leq (1 + a_n) e^{-a_n} \leq 1 - (1-\epsilon)a_n^2/2.$$
But $(1 \pm \epsilon)a_n^2/2$ is always positive for sufficiently small $\epsilon$, since $a_n \in \mathbb{R}$, and we know that for a positive sequence $p_n \geq 0$,
$$\prod_{n=1}^\infty (1 - p_n) \text{ converges iff } \sum_{n=1}^\infty p_n \text{ does.}$$
Fix $\epsilon$ and let $N$ be large enough so that for all $n \geq N$,
$$1 - (1+\epsilon)a_n^2/2 \leq (1 + a_n) e^{-a_n} \leq 1 - (1-\epsilon)a_n^2/2.$$
Then
$$\prod_{n=N}^\infty [1 - (1+\epsilon)a_n^2/2] \leq \prod_{n=N}^\infty (1 + a_n) e^{-a_n} \leq \prod_{n=N}^\infty [1 - (1-\epsilon)a_n^2/2],$$
so the middle product converges iff the left-hand product converges iff the right-hand product converges iff $\sum_{n=1}^\infty a_n^2$ converges. This proves OP's hypothesis for sequences $\{ a_n \}$ of real numbers.
