I've been studying Bressoud's paper "An easy proof of the Rogers-Ramanujan Identities" where he proves the R.R. identity:
$$\sum_{n \geq0}\frac{q^{n^2}}{(1-q) \cdots(1-q^n)}= \prod_{n \geq0}\frac{1}{(1-q^{5n+1})(1-q^{5n+4})}$$
Apparently the right hand side becomes modular when corrected by the term $q^{{-1}/{60}}$.
This seems rather arbitrary to me so I'm seeking an explanation where this may come from. Is there something more behind it? A greater context for instance?
The R-R identity you mention in your question is one of a pair and together they are part of a fascinating story told by William Duke, Continued Fractions and Modular Functions, Bull. Amer. Math. Soc. 42 (2005), 137-162. On page 143 he writes (NOTE: $q = e(\tau) := e^{2\pi i\tau})$
The theta constant with characteristic $[^{\,\epsilon}_{\epsilon'}]\in\mathbb{R}^2$ is defined by
$ \quad\quad\quad\quad\theta[^{\,\epsilon}_{\epsilon'}](\tau)= \sum_{n\in\mathbb{Z}}e(\frac12(n+\frac{\epsilon}2)^2\tau +\frac{\epsilon'}2(n+\frac{\epsilon}2))$
for $\tau\in\mathcal{H}.$ It satisfies the following basic properties for $\ell,m,N\in\mathbb{Z}$ with $N$ positive $\textrm{[FK pp. 72-77]}$:
$ (4.3)\quad\quad\quad \theta[^{\,\epsilon}_{\epsilon'}](\tau) = e(\mp\frac{\epsilon m}2) \theta[^{\,\pm\epsilon+2\ell}_{\pm\epsilon'+2m}](\tau) \\ (4.4)\quad\quad\quad \theta[^{\,\epsilon}_{\epsilon'}](\tau) = \sum_{k=0}^{N-1} \theta[^{\frac{\epsilon+2k}N}_{N\epsilon'}] (N^2\tau).$
It also satisfies for $(^{a\, b}_{c\, d})\in \textrm{SL}(2,\mathbb{Z})$ the fundamental transformation law $\textrm{[FK Thm 1.11, p. 81]}$:
$ (4.5)\quad\quad\quad \theta[^{\,\epsilon}_{\epsilon'}](\frac{a\tau+b}{c\tau+d}) = \kappa\sqrt{c\tau+d}\,\theta [^{a\epsilon+c\epsilon'-ac}_{b\epsilon+d\epsilon'+bd}](\tau)$
where
$\quad\quad\quad\kappa = e(-\frac14(a\epsilon+c\epsilon')bd-\frac18( ab\epsilon^2+cd\epsilon'^2+2bc\epsilon\epsilon'))\kappa_0,$
with $\kappa_0$ and eighth root of unity depending only on the matrix $(^{a\, b}_{c\, d}).$ The value of $\kappa_0$ is fixed by choosing the argument of the square root to be in $[0,\pi).\,$ In particular we have $\textrm{[FK, p. 86]}$
$(4.6)\quad\quad\quad \theta[^{\,\epsilon}_{\epsilon'}](\tau+1)= e(-\frac{\epsilon}4(1+\frac{\epsilon}2))\theta[^{\quad\epsilon} _{\epsilon+\epsilon'+1}](\tau)$
and
$(4.7)\quad\quad\quad \theta[^{\,\epsilon}_{\epsilon'}](\frac{-1}\tau)= e(-\frac18)\sqrt{\tau}e(\frac{\epsilon\epsilon'}4)\theta [^{\,\,\epsilon'}_{-\epsilon}](\tau).$
Additional transformation formulas invoked later in the text are given in Appendix A. we also have the product formula $\textrm{[FK, p. 141]}$:
$ (4.8)\quad \theta[^{\,\epsilon}_{\epsilon'}](\tau) \!=\! e(\frac{\epsilon\epsilon'}4) q^{\frac{\epsilon^2}8} \!\prod_{n\ge 1} (1\!-\!q^n) (1\!+\!e(\frac{\epsilon'}2)q^{n-\frac{1+\epsilon}2}) (1\!+\!e(\frac{-\epsilon'}2)q^{n-\frac{1-\epsilon}2}),$
which follows from the Jacobi triple product identity.
$\quad$ A short calculation using $(4.1)$ and $(4.8)$ shows that
$ (4.9) \quad\quad\quad r(\tau) = e(-\frac1{10})\frac {\theta[^{3/5}_{\,\,1}](5\tau)}{\theta[^{1/5}_{\,\,1}](5\tau)}. $
N.B.: In the infinite product in $(4.8)$ the $\frac{\epsilon'}2$ and $-\frac{\epsilon'}2$ should be swapped.
The reciprocal of the right side of the R-R equation corresponds to the numerator of equation $(4.9)$ of the Duke article but not exactly. What I mean is that the theta constant has both an infinite sum and an infinite product expansion. As a special case of this result we have the equation
$$ \sum_{n=-\infty}^{\infty}\!(-1)^n q^{(10n+3)^2} \!=\! q^9 \prod_{n=1}^{\infty} (1\!-\!q^{200n}) (1\!-\!q^{200n-160})(1\!-\!q^{200n-40}) \tag{1}$$
The Wikipedia article Rogers-Ramanujan identities states that
If $\,q=e^{2\pi i\tau},\,$ then $\,q^{-1/60}G(q)\,$ and $\,q^{11/60}H(q)\,$ are modular functions of $\,\tau.$
Now, by definition of $\,G(q)\,$ we have $$ 1/G(q) = \prod_{n=1}^\infty (1-q^{5n-4})(1-q^{5n-1}) \tag{2}$$ and therefore we immediately have $$ 1/G(q^{40}) = \prod_{n=1}^\infty (1-q^{200n-160})(1-q^{200n-40}). \tag{3}$$ Comparing this with equation $(1)$ we now have $$ q^9/G(q^{40})\prod_{n=1}^\infty (1-q^{200n}) = \sum_{n=-\infty}^{\infty} (-1)^n q^{(10n+3)^2}. \tag{4}$$ The infinite product on the left side is the Dedekind $\eta$ function missing a factor of $q^{200/24}=q^{25/3}.$ When we supply the missing factor, the power of $q$ on the left side is now $\,q^{9-25/3}=q^{2/3}.\,$ Now $\,q^{2/3}/G(q^{40})\,$ is a modular function and therefore so is $\,q^{1/60}/G(q) \,$ and $\,q^{-1/60}G(q)\,$ which answers your question.
P.S. I have found a very brief reference that may answer your question. In Bruce C. Berndt, Ramanujan's Notebooks, Part IV, page 285 is written:
McIntosh [1] has improved $(8.3)$ by proving that, as $q$ tends to $1-,$ $$ \log\left(\sum_{n=0}^\infty \frac{q^{n^2}}{(q)_n}\right) = -\frac1{\log q} \frac{\pi^2}{15} -\frac12\log\left(\frac{5-\sqrt{5}}2\right) + \frac{\log q}{60} + E_1(q), $$ where $E_1(q)$ is exponentially small. The small error term arises from $(0.1)$ and the transformation formula for the infinite product.
A little further on is this
By setting $\,b=c=1\,$ in Entry 7 and using McIntosh's further work, we find that, as $\,q\,$ tends to $1-$, $$ \log\left(\sum_{n=0}^\infty \frac{q^{n(n+1)}}{(q)_n}\right) = -\frac1{\log q} \frac{\pi^2}{15} -\frac12\log\left(\frac{5+\sqrt{5}}2\right) - \frac{11\log q}{60}+ E_2(q), $$ where $\,E_2(q)\,$ is exponentially small.
The coefficient of $\,\log q\,$ is why you need $\,q^{-1/60}\,$ in the 1st case and $\,q^{11/60}\,$ in the 2nd.
