I have seen that infinite products $\prod(1+a_n)$ where $a_n$ can change sign repeatedly is more complicated than in situations where $-1 < a_n <0$ or $a_n > 0$ for all $n$ where product and series $\sum a_n$ must both converge or diverge. From other questions I understand that if $\sum a_n^2$ converges then simultaneous convergence/divergence of $\prod (1+a_n)$ and $\sum a_n$ still holds. I also was given the example $a_n = \frac{(-1)^n}{\sqrt{n}}$ where $\sum a_n$ converges and $\prod(1+a_n)$ diverges. In this case we also have $\sum a_n^2 = \sum\frac{1}{n}$ divergent.
Summarizing, I know that $$\sum a_n^2 \text{ converges}, \quad \prod(1+a_n) \text{ converges} \implies \sum a_n \text{ converges}$$ and I have an example where $$\prod(1+a_n) \text{ diverges}, \sum a_n \text{ converges, and } \sum a_n^2 \text{ diverges}$$
My question is if $\prod(1+a_n)$ converges (where $a_n$ changes sign) must both $\sum a_n$ and $\sum a_n^2$ converge?
Counterexample: Let $a_n = e^{(-1)^n/\sqrt n}-1.$ Then $\prod (1+a_n)$ converges. But
$$\sum a_n = \sum \left (e^{(-1)^n/\sqrt n} -1\right) = \sum \left (\frac{(-1)^n}{\sqrt n} + \frac{1}{2n} + O(1/n^{3/2})\right ).$$
The series on the right is the sum of a convergent series, a divergent series, and a convergent series. Hence it diverges.