I'm trying to understand the relationship between series convergence and product convergence. Suppose we have a sequence $a = (a_1, a_2, \ldots)$, and let $S(a) = \sum a_n$ and $P(a) = \prod (1 + a_n)$. When we restrict our $a_n$ to be positive, then we can show that $S(a) \leq P(a) \leq e^{S(a)}$, and since the sum (and product) is monotonically increasing as we introduce new terms (or factors), $S(a)$ converges if and only if $P(a)$ does. However, if we relax the condition that $a_n$ is positive, I don't know to what extent this relationship still holds.
For instance, let $a$ be the alternating harmonic series. Then $S(a)$ converges to $\ln(2)$ (Wikipedia), and $P(a)$ converges to $1$, since:
$$P(a) = \prod_{n=1} (1 + \frac{(-1)^{n+1}}{n}) = 2 \cdot \frac{1}{2} \cdot \frac{4}{3} \cdot \frac{3}{4} \cdot \ldots$$
And the $\limsup$ and the $\liminf$ both agree on $1$.
In order to keep my question simple, let's look at this particular sequence
$$b = 2a_{n \geq 3} = \Big(\frac{2}{3}, - \frac{1}{2}, \frac{2}{5}, - \frac{1}{3}, \ldots \Big)$$
That is, the alternating harmonic sequence where every term is doubled, started at $n = 3$. Then we should have $S(b) = 2(S(a) - (1 - \frac{1}{2})) = 2\ln(2) - 1$.
But then what is $P(b)$?
$$P(b) = \prod (1 + b_n) = \Big(1+ \frac{2}{3}\Big)\Big(1 - \frac{1}{2}\Big)\Big(1 + \frac{2}{5}\Big)\Big( 1 - \frac{1}{3}\Big) \ldots \\ = \frac{5}{3} \cdot \frac{1}{2} \cdot \frac{7}{5} \cdot \frac{2}{3} \cdot \ldots$$
Because $1 + x \leq e^x$ for all real $x$, we know that $P(b) \leq e^{S(b)}$. But how do I prove that $P(b)$ converges, and what does it converge to?