If $\sum\limits_{n=1}^∞|a_n|^2$ converges, then $\prod\limits_{n=1}^∞(1+a_n)$ converges if and only if $\sum\limits_{n=1}^∞a_n$ converges. 
If $\sum\limits_{n=1}^{\infty}|a_n|^2$ converges, then $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges if and only if $\sum\limits_{n=1}^{\infty} a_n$ converges. 

My attempt is to use the Cauchy citerion, given $m>n$, $$\left|\prod_{i=1}^{m}(1+a_i)-\prod_{i=1}^{n}(1+a_i)\right|=\left|\prod_{i=1}^{n}(1+a_i)\left[\prod_{i=n+1}^{m}(1+a_i)-1\right]\right|\\
=\left|\prod_{i=1}^{n}(1+a_i)\right|\left|\prod_{i=n+1}^{m}(1+a_i)-1\right|,\\
\left|\prod_{i=1}^{n}(1+a_i)\right| \leqslant \prod_{i=1}^{n}(1+|a_i|),$$ then how to proceed?
 A: We first assume that $a_n\in \mathbb{R}$.
Note that since $\sum_{n=1}^\infty a_n^2$ converges, then $\displaystyle \lim_{n\to \infty}a_n=0$.  There exists, therefore, a number $N$ so that $1+a_n>0$ whenever $n>N$.
Applying the Mean Value Theorem, it is straightforward to show that
$$x-\frac12x^2+\frac13x^3\le \log(1+x)\le x \tag 1$$
Applying $(1)$, we have for $n>N$
$$a_n-\frac12 a_n^2+\frac13 a_n^3 \le \log(1+a_n)\le a_n\le \frac12 a_n^2-\frac13a_n^3+\log(1+a_n) \tag 2$$
Since $\sum_{n=1}^\infty a_n^2$ converges, then so does $\sum_{n=1}^\infty a_n^3$.  From $(2)$, it is easy to see that $\sum_{n=1}^\infty \log(1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n$ converges.
Finally, we note that for $n>N$
$$\prod_{n=N}^\infty (1+a_n)=e^{\sum_{n=N}^\infty \log(1+a_n)} \tag 3$$
whence we arrive at the coveted result!

If $a_n\in \mathbb{C}$, and if $n$ is so large that $\text{Re}(1+a_n)> 0$, then $\log(1+a_n)$ can be expanded as
$$\log(1+a_n)=a_n-\frac12 a_n^2+O(a_n^3)$$
where we are free to choose to use the principal branch of the logarithm function to define $\log(1+a_n)$.
Since $\sum_{n=1}^\infty a_n^2$ converges, then so does $\sum_{n=1}^\infty a_n^3$.
Now we can proceed along similar lines as to the proof for real-valued $a_n$.  The details are left as an exercise.
A: For $x\gt-1$, writing $x-\log(1+x)=\int_0^x\frac{t}{1+t}\,\mathrm{d}t$, it follows that
$$
0\le x-\log(1+x)\le\frac{\frac12x^2}{\min(1,1+x)}\tag1
$$
Thus, for $|x|\le\frac12$, we have
$$
|\,x-\log(1+x)\,|\le x^2\tag2
$$
If either $\sum\limits_{n=1}^\infty a_n$ or $\sum\limits_{n=1}^\infty\log(1+a_n)$ converges, then $\lim\limits_{n\to\infty}a_n=0$. Thus, there is an $N$ so that $n\ge N\implies|a_n|\le\frac12$. Therefore,
$$
\begin{align}
&\left|\,\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty\log(1+a_n)\,\right|\\
&\le\left|\,\sum_{n=1}^{N-1}a_n-\sum_{n=1}^{N-1}\log(1+a_n)\,\right|+\left|\,\sum_{n=N}^\infty a_n-\sum_{n=N}^\infty\log(1+a_n)\,\right|\tag{3a}\\
&\le\left|\,\sum_{n=1}^{N-1}a_n-\sum_{n=1}^{N-1}\log(1+a_n)\,\right|+\sum_{n=N}^\infty a_n^2\tag{3b}\\[9pt]
&\lt\infty\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: triangle inequality
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: the sum of two finite sums is finite
