An identity related to $q$-series

Question

While studying Ramanujan's theta functions, I encountered a q-series $(q;q)_\infty^2\phi(q)$. I calculated the first few terms of $(q;q)_\infty^2\phi(q)$ and observed that it seems to have the following form:

$$(q;q)_\infty^2\phi(q)=\sum_{\genfrac{}{}{0pt}{}{n\geq1}{n\ \text{odd}}}\left(\frac n3\right)nq^{(n^2-1)/12},\tag1$$

where $(q;q)_\infty=\prod_{n=1}^\infty(1-q^n)$ and $\phi(q)=\sum_{n=-\infty}^\infty q^{n^2}$. I have attempted to find a reference for this series but have not made any progress.

My question is: How to prove (1)?

Any comments will be appreciated.

Answer 1

As Kevin Zheng says in his answer, you can use the Jacobi triple product identity on $\phi(q)$. Then, substituting $s=e^{2\pi i z}=q^2$ reduces the identity (1) to $$s^{-1/24}\frac{\eta(z)^5}{\eta(2z)^2}=\frac{(s;s)_\infty^5}{(s^2;s^2)_\infty^2}=\prod_{n\ge 1} (1-s^n)^3 (1-s^{2n-1})^2 = \sum_{ \text{odd}\ n\geq1}\left(\frac n3\right)ns^{(n^2-1)/24}.\qquad (2)$$ This follows from the quintuple product identity $$ \prod_{n\ge 1} (1-s^n)(1-s^nt)(1-s^{n-1}t^{-1})(1-s^{2n-1}t^2)(1-s^{2n-1}t^{-2}) = \sum_{n}s^{(3n^2+n)/2}(t^{3n}-t^{-3n-1}) $$ if you divide both sides by $1-t^{-1}$, expand the quotient $(t^{3n}-t^{-3n-1})/(1-t^{-1})$, and then set $t$ equal to 1. This proof of (2) was given by Basil Gordon in 1961 [1]. (2) was earlier found by Ramanujan [3]. Also, (2) is identity (8.19) in Theorem 8.5 in [2], and there is some more discussion of the history of it there.

1: "Some Identities in Combinatorial Analysis", B. Gordon, The Quarterly Journal of Mathematics, 12, #1 (1961), pp. 285-290.

2: Eta Products and Theta Series Identities, Günter Köhler, Springer, 2011.

3: "On certain arithmetical functions", Transactions of the Cambridge Philosophical Society 22 (1916), pp. 159-184, at p. 170.

Answer 2

I found an identity in the last one of Theorem 1.1 of [1], say

$$\frac{\eta^5(24z)}{\eta^2(48z)}=\sum_{n=1}^\infty\left(\frac{n}{12}\right)nq^{n^2}.$$

Moreover, we have

$$\sum_{n=-\infty}^\infty q^{n^2}=\frac{(q^2;q^2)_\infty^5}{(q;q)_\infty^2(q^4;q^4)_\infty^2},$$

by Jacobi's Triple Product Identity. Thus, the desired result come from if we substitude $q$ by $q^{12}$ of (1) and then multiply the both sides of (1) with $q$.

[1] Lemke Oliver, Robert J., Eta-quotients and theta functions, Adv. Math. 241, 1-17 (2013). ZBL1282.11030.