Per OP's request I made my comment into an answer. I am making this CW; if someone wants to add details of the solutions, feel free to do so. (Of course, if you prefer, you can post them in a separate post, so that you are rewarded by reputation for you effort.)
Such Examples are given in Problems 3.8.5 and 3.8.9 in Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series, p.113-114.
Problem 3.8.5 Set
$$a_{2n-1}=\frac1{\sqrt{n}}+\frac1n,\qquad a_{2n}=-\frac1{\sqrt{n}}, n\in\mathbb N.$$
Show that the product $\prod\limits_{n=1}^\infty(1+a_n)$ converges, although the series $\sum\limits_{n=1}^\infty a_n$ diverges.
A solution is given on p.364.
Problem 3.8.9. Prove that the product $\prod\limits_{n=1}^\infty \left(1+(-1)^{n+1}\frac1{\sqrt{n}}\right)$ diverges although the series $\sum\limits_{n=1}^\infty (-1)^{n+1}\frac1{\sqrt{n}}$ converges.
A solution is given on p.365.
The forward implication is false without the additional requirement that the limit $\,\prod_{k=1}^\infty b_k\,$ is strictly positive. Take for example $\,b_n = 1/2 \gt 0\,$, then $\,\prod_{n=1}^\infty b_n = 0\,$ but $\,\sum_{n=1}^\infty \ln b_n \to -\infty\,$.
With that caveat, let $\,p_n = \prod_{k=1}^n b_k\,$ and $\,s_n=\sum_{k=1}^n \ln b_k\,$, then what can be proved is that $\,p_n\,$ converges to a non-zero limit as $\,n \to \infty\,$ if and only if $\,s_n\,$ converges. To that end, note that:
$\;\displaystyle s_n=\ln p_n\,$ by the property of the logarithm that $\,\ln ab = \ln a + \ln b\,$;
$\;\displaystyle p_n=e^{s_n}\,$ since the natural logarithm and the exponential are inverses of each other.
$\log(x)$ is continuous on $(0, \infty)$, which I think might help in showing the forward implication
Intuition is right. Suppose that $\,p_n\,$ converges to $\,p \gt 0\,$ i.e. $\,\lim_{n \to \infty} p_n = p\,$, then $\,\lim_{n \to \infty} s_n\,$ $\,= \lim_{n \to \infty} \ln p_n\,$ $\, = \ln \big(\lim_{n \to \infty} p_n\big) = \ln p\,$ by continuity of $\ln$ on $\mathbb{R}^+$, so $\,s_n\,$ is convergent.
The reverse implication works much the same. Suppose that $\,\lim_{n \to \infty} s_n = s\,$, then $\,\lim_{n \to \infty} p_n\,$ $\,= \lim_{n \to \infty} e^{s_n}\,$ $\, = e^{\lim_{n \to \infty} s_n} = e^s\,$ by continuity of the exponential function, so $\,p_n\,$ is convergent.
Best Answer
As far as I know, the only formal result is $$\pi(F,m)=\prod_{k=m}^{\infty}(1-F^k)=\left(F^m;F\right){}_{\infty }$$ where appears the q-Pochhammer symbol
$$\sigma(F,m) = \prod_{k=1}^{m}(1-F^k)=(F;F)_m$$
You will find this function in Mathematica, Wolfram Alpha, Maple, Sage and most programming environments.