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 $m$?
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> 
> 



Seeking a combinatorial proof for the invariance of a $q$-series
The RHS is the (size) generating function for all integer partitions.


The LHS is a modification of the idea of keeping track of the [Durfee square](https://en.wikipedia.org/wiki/Durfee_square) of a partition.


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.


