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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 $e$.

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

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