$\sum|a_n|^2<\infty$ and $\sum a_n<\infty$ implies $\sum\log(1+a_n)$ converges

Note that all elements are complex numbers. Suppose we know $$\sum_{n=1}^\infty|a_n|^2\quad\text{and}\quad\sum_{n=1}^\infty a_n$$converge. How can we conclude $\sum_{n=1}^\infty\log(1+a_n)$ converges? This is an effort to prove that $$\prod_{n=1}^\infty(1+a_n)$$converges.

If I knew that $\sum a_n$ was absolutely convergent, then I think I'd be able to use the power series expansion of $\log(1+a_n)$ and interchange summations. However, I don't think this is the intended method of solution. Will someone help me with this homework problem?

Update: Other work I've done so far:

Disregarding finitely many terms if necessary, we can assume $|a_n|<\frac{1}{2}$ for all $n$. Thus, we know $0\leq|a_n-\log(1+a_n)|<|a_n|^2$, so by comparison, we know $\sum_{n=1}^\infty |a_n-\log(1+a_n)|$ converges. Since it is absolutely convergent, it is convergent, so now we have $$\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty\left( a_n-\log(1+a_n)\right)=\sum_{n=1}^\infty\log(1+a_n),$$therefore the series is convergent. Is this correct?

Use the fact that, for small $x$, $$\log(1+x)=x+\epsilon(x) |x|^2, \qquad |\epsilon(x)|\le 1,$$ to express $\sum_n \log(1+a_n)$ as a sum of two convergent series.

Comment re updated question: Yes, that's correct.