I have to calculate this sum in closed form
$$ \sum_{n=1}^\infty \prod_{k=1}^n \frac{x^{k-1}}{1 - x^k} $$
where $x < 1$.
Numerical evaluation shows that this converges. The product can be performed separately in the numerator and the denominator, but this doesn't seem to help. Any tip would be welcome.
Edit: Using the comment from TheBestMagician below, this becomes $ \sum_{n\geq1} \sum_{l_1 \geq 0} \sum_{l_2 \geq 0} \dots \sum_{l_n \geq 0} x^{\frac{n(n+1)}{2} + \sum_{k=1}^n \ k l_k} $. It is not obvious though how to sum it back.
I don't think there is a closed form in terms of ordinary functions. It is known that $$\sum_{n=1}^\infty \prod_{k=1}^n \frac{x^{k-1}}{1 - x^k}=2\prod_{n=1}^\infty (1+x^n)-1$$ where $\prod_{n=1}^\infty (1+x^n)$ is the generating function of the number of partitions of $n$ into distinct parts (see also the formula section in A000009 and A087135).