The groupoid cardinality of $\mathbb{F}_q$-vector spaces $\sum_{n=0}^{\infty} \, \bigl(\prod_{i=0}^{n-1} q^n-q^i\bigr)^{-1}$

Question

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 $e$.

Answer 1

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.

Answer 2

We start with \begin{align*} \color{blue}{\sum_{n=0}^{\infty}\left(\prod_{i=0}^{n-1}\left(q^n-q^i\right)\right)^{-1}} &=\sum_{n=1}^{\infty}\frac{1}{\prod_{i=0}^{n-1}\left(q^n-q^i\right)}\\ &=\sum_{n=0}^{\infty}\frac{q^{-n^2}}{\prod_{i=0}^{n-1}\left(1-q^{i-n}\right)}\\ &=\sum_{n=0}^{\infty}\frac{q^{-n^2}}{\prod_{i=0}^{n-1}\left(1-q^{-i-1}\right)}\tag{$i\to n-1-i$}\\ &\color{blue}{=\sum_{n=0}^{\infty}\frac{q^{-n^2}}{\prod_{i=1}^{n}\left(1-q^{-i}\right)}}\tag{1} \end{align*}

Now it's time for some beautiful math from chapter 7, Identities of the Rogers-Ramanujan Type from The Theory of Partitions by G. E. Andrews.

We consider a function $F(x)$ analytic in $x$ at $0$, $F(0)=1$ such that the linear, second-order $q$-difference equation \begin{align*} F(x)=F(xq)+xqF(xq^2)\tag{2} \end{align*} holds. We set \begin{align*} c(x,q):=\frac{F(x)}{F(xq)} \end{align*} from which \begin{align*} c(x,q)&=1+\frac{xq}{c(xq,q)}\\ &=1+\frac{xq}{1+\frac{xq^2}{c(xq^2,q)}}\cdots\tag{3} \end{align*} follows. If we set \begin{align*} F(x)=\sum_{n\geq 0}A_n(q)x^n\tag{4} \end{align*} we obtain from (2) \begin{align*} \sum_{n\geq 0}A_n(q)x^n=\sum_{n\geq 0}A_n(q)(xq)^n+xq\sum_{n\geq 0}A_n(q)(xq^2)^n \end{align*} and comparing coefficients gives \begin{align*} A_n(q)&=q^nA_n(q)+q^{2n-1}A_{n-1}(q)\\ \color{blue}{A_n(q)}&=\frac{q^{2n-1}}{1-q^n}A_{n-1}(q)\\ &=\frac{q^{(2n-1)+(2n-3)}}{\left(1-q^n\right)\left(1-q^{n-1}\right)}A_{n-2}(q)\\ &=\cdots\\ &\,\,\color{blue}{=\frac{q^{1+3+\cdots+(2n-1)}}{\prod_{i=1}^n\left(1-q^i\right)}A_0(q)}\tag{5} \end{align*}

We obtain from (4) and (5) \begin{align*} \color{blue}{F(x)=\sum_{n=0}^{\infty}\frac{q^{n^2}}{\prod_{i=1}^n\left(1-q^i\right)}x^n} \end{align*} and verbatim from the book:

Surprisingly $F(1)$ and $F(q)$ may be evaluated in terms of infinite products (as we shall prove later in this chapter), namely, \begin{align*} \color{blue}{F(1)}&=1+\frac{q}{1-q}+\frac{q^4}{(1-q)\left(1-q^2\right)} +\frac{q^9}{(1-q)\left(1-q^2\right)\left(1-q^3\right)}+\cdots\\ &\,\,\color{blue}{=\prod_{n=0}^{\infty}\left(1-q^{5n+1}\right)^{-1}\left(1-q^{5n+4}\right)^{-1}}\tag{6} \end{align*} and \begin{align*} F(q)&=1+\frac{q^2}{1-q}+\frac{q^6}{(1-q)\left(1-q^2\right)} +\frac{q^{12}}{(1-q)\left(1-q^2\right)\left(1-q^3\right)}+\cdots\\ &=\prod_{n=0}^{\infty}\left(1-q^{5n+2}\right)^{-1}\left(1-q^{5n+3}\right)^{-1} \end{align*}

We conclude from (1) and (6) \begin{align*} \color{blue}{\sum_{n=0}^{\infty}\left(\prod_{i=0}^{n-1}\left(q^n-q^i\right)\right)^{-1} =\prod_{n=0}^{\infty}\left(1-q^{-(5n+1)}\right)^{-1}\left(1-q^{-(5n+4)}\right)^{-1}\qquad\qquad |q|>1} \end{align*}

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Note: This is just a side-note from the introductory section of chapter 7. S.Ramanujan sent the following beautiful formulas to G. H. Hardy on 16 January 1913: \begin{align*} \frac{1}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\frac{e^{-6\pi}}{\vdots}}}} =\left(\sqrt{\frac{5+\sqrt{5}}{2}}-\frac{\sqrt{5}+1}{2}\right)e^{2\pi/5}\\ 1-\frac{e^{-\pi}}{1+\frac{e^{-2\pi}}{1-\frac{e^{-3\pi}}{\vdots}}} =\left(\sqrt{\frac{5-\sqrt{5}}{2}}-\frac{\sqrt{5}-1}{2}\right)e^{\pi/5}\\ \end{align*} Note that these continued fraction theorems are evaluations of (3), namely \begin{align*} c\left(1,e^{-2\pi}\right)\qquad\mathrm{and}\qquad c\left(1,-e^{-\pi}\right) \end{align*}