The Wikipedia article on the Rogers-Ramanujan identities remarks that $$q^{-1/60} G(q) = q^{-1/60} \sum_{n=0}^{\infty} \frac{q^{n^2}}{(1-q)...(1-q^n)} = q^{-1/60} \prod_{n=0}^{\infty} \frac{1}{(1 - q^{5n+1})(1 - q^{5n+4})}$$ is a modular function; that is, a meromorphic modular form of weight zero whose only possible poles are at cusps. This is also discussed on mathoverflow here.
Taken at face value this cannot be true because it implies that $(q^{-1/60} G(q))^{-1}$ is a holomorphic modular form of weight $0$ and therefore constant (right?) Edit: this logic is flawed; $q^{-1/60} G(q)$ can have zeros at cusps other than $\infty$.
I would like to know in a little more detail how to interpret this as an identity of modular forms.