Let $q > 1$. What can we say about the value of $$\sum_{n=0}^{\infty} \, \bigl(\prod\limits_{i=0}^{n-1} q^n-q^i\bigr)^{-1} ~~?$$ The series clearly converges. Is there a closed form or something like that?

Background: If $q$ is a prime power, then this is the cardinality of the groupoid of finite-dimensional $\mathbb{F}_q$-vector spaces. Since $\mathbb{F}_1$-vector spaces are pointed sets and the cardinality of the category of finite pointed sets is $\sum_{n=0}^{\infty} \frac{1}{n!} = e$, the above series may be seen as a $q$-analog of $e$.

  • $\begingroup$ @Normal Human: Thank you for editing. $\endgroup$ – Martin Brandenburg Sep 30 '15 at 13:28

The function you asked is related to the function

$$ G(x) = \sum_{n=0}^{\infty} \frac{x^{n^2}}{(1-x)\cdots(1-x^n)} $$

appearing in Rogers-Ramanujan identity by

$$ \sum_{n=0}^{\infty} \Bigg( \prod\limits_{i=0}^{n-1} (q^n-q^i) \Bigg)^{-1} = G(1/q). $$

I learned that this has been studied in combinatorics since the coefficients in the series expansion admit combinatorial interpretation in terms of partitions (see OEIS A003114 for the coefficient list of $G(x)$), but I am not sure if this will help you and I also do not know its number-theoretic properties.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.