# Seeking a combinatorial proof for the invariance of a $q$-series

Start with some notations: $$(a,q)_n=(1-a)(1-aq)\cdots(1-aq^{n-1})$$, shortened by $$(a)_n$$, and $$(a)_{\infty}=\prod_{k=0}^{\infty}(1-aq^k)$$.

It's easy to verify (using algebraic means) that, for each $$m\in\mathbb{Z_{\geq0}}$$, $$\sum_{n\geq0}\frac{q^{n^2}q^{mn}}{(q)_n(q)_{n+m}}=\frac1{(q)_{\infty}}. \tag1$$

QUESTION. What is a combinatorial (conceptual) reason for (1) to be independent of $$m$$?

• I'm actually not so sure how to interpret this equation when $m$ is negative. Can you clarify what $(q)_{-5}$ means? – Sam Hopkins Jun 3 '20 at 16:27
• Perhaps, I'll drop $m<0$. – T. Amdeberhan Jun 3 '20 at 17:37

Namely, for $$m\in\mathbb{N}$$, let us define for a partition $$\lambda$$ the $$m$$-Durfee square of $$\lambda$$ to be the $$n\times (n+m)$$ rectangle of boxes in the upper left corner of $$\lambda$$, where $$n$$ is as large as it can be. (Note we can have $$n=0$$, so every partition has such an $$m$$-Durfee square.) Maybe it should be called "Durfee rectangle."
Then $$\frac{q^{n^2}q^{mn}}{(q)_n(q)_{n+m}}$$ is easilly seen to be the generating function of partitions with an $$m$$-Durfee square of dimension $$n\times (n+m)$$. Summing over all $$n$$ gives the generating function of all partitions.
• Are you asking about the last paragraph? The generating function for partitions by size of Durfee square is very well known: we form any partition placing a $n\times n$ Durfee square, then putting a partition with at most $n$ columns below it, and a partition with at most $n$ rows to its right. For $m$-Durfee squares the reasoning is exactly the same. – Sam Hopkins Jun 3 '20 at 19:04