For infinite products I know the following theorems:
Theorem I. If $a_n \geq 0$ for all $n$ then $\prod (1+a_n)$ converges if and only if $\sum a_n$ converges.
Theorem II. If $-1 < a_n \leq 0$ for all $n$ then $\prod (1+a_n)$ converges if and only if $\sum a_n$ converges.
My question is what are necessary and sufficient conditions (possibly including more restrictions than in I and II) for $\prod (1+a_n)$ convergence/divergence as related to $\sum a_n$ convergence/divergence if the sequence of terms $a_n$ can switch signs repeatedly but always stay greater than $-1$.
I already know the trivial equivalence of $\prod(1+a_n)$ and $\sum\ln(1+a_n)$.
An example to consider.
$$ a_n = \frac{(-1)^n}{\sqrt{n}}\;. $$ [Note, for $n \ge 2$, we have $-1 < a_n < 1$ so $0 < 1+a_n < 2$.]
Then of course the alternating series $\sum_{n=2}^\infty a_n$ converges. But what about infinite product $$ \prod_{n=2}^\infty\big(1+a_n\big)=\prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt{n}}\right)\;? $$ We have as $n \to \infty$:
$$ \log\left(1+\frac{(-1)^n}{\sqrt{n}}\right) = \frac{(-1)^n}{\sqrt{n}}-\frac{1}{2n} + O(n^{-3/2})\;. $$ Thus, $$ \sum_{n=2}^\infty\log\left(1+\frac{(-1)^n}{\sqrt{n}}\right) $$ is the sum of a series $\sum\frac{(-1)^n}{\sqrt{n}}$that converges, a series $-\sum\frac{1}{2n}$ that diverges to $-\infty$, and a series that converges by limit comparison with $\sum n^{-3/2}$.
Therefore $$ \prod_{n=2}^\infty\big(1+a_n\big) $$ diverges to $0$.
The reciprocal series
$$ \prod_{n=2}^\infty\frac{1}{1+a_n}= \prod_{n=2}^\infty\big({1+b_n}\big) $$ diverges to $+\infty$ while the series $\sum b_n$ still converges. And $-1 <b_n< 1$ for $n \ge 4$.