# Show that $\prod\limits_{n=1}^\infty \frac{(1-q^{6n})(1-q^n)^2}{(1-q^{3n})(1-q^{2n})}=\sum\limits_{n=-\infty}^\infty q^{2n^2+n}-3q^{9(2n^2+n)+1}$.

Show that $$\displaystyle \prod_{n=1}^\infty \frac{(1-q^{6n})(1-q^n)^2}{(1-q^{3n})(1-q^{2n})}=\sum_{n=-\infty}^\infty q^{2n^2+n}-3q^{9(2n^2+n)+1}$$.

I can't seem to be able to proceed with this question. I know that $$\displaystyle \prod_{n=1}^\infty \frac{(1-q^{2n})^2}{(1-q^n)}=\sum_{n=-\infty}^\infty q^{2n^2+n}$$. But the problem is the constant $$3$$, as I can't seem to get rid of it algebraically. I have used Maple and found that

$$\displaystyle \prod_{n=1}^\infty \frac{(1-q^{2n})^2}{(1-q^n)}-3q\prod_{n=1}^\infty \frac{(1-q^{18n})^2}{(1-q^{9n})}=\prod_{n=1}^\infty \frac{(1-q^{6n})(1-q^n)^2}{(1-q^{3n})(1-q^{2n})}$$. However, I am unsure on how I can prove this rigorously (since Maple is not considered as a proof) and unsure on how to combine the expressions into one single expression on the RHS.

I have learnt the Jacobi Triple Product Identity and the Quintuple Product Identity but unsure how I can use those identities to help me since those identities don't involve subtraction of terms.

• Show both side are modular, the quotient of the RHS by the LHS is weight $0$ modular and has no pole thus it is constant equal to the value at $q=0$ which is $1$. Commented Dec 9, 2019 at 15:41
• @reuns I don't think taking quotient is the right approach. Instead, after showing modularity of both sides, find their levels and cusps, and work with $q$-expansions to identify them. The space of modular forms of given level and given weight is finite dimensional, so comparing finitely many coefficients of the $q$-expansions will suffice for identifying them. Commented Dec 19, 2019 at 6:02

Consider the expression $$\sum_{n\in\mathbb {Z}}q^{2n^2+n}$$ which equals Ramanujan theta function $$f(q, q^3)=\psi(q)$$. Thus the RHS of the given identity equals $$\psi(q) - 3q\psi(q^9)$$ Ramanujan says in his notebooks that the following identity holds $$\psi(q) - 3q\psi(q^{9})=\frac{\phi(-q)} {\chi(-q^3)}$$ where $$\phi(-q)=\prod_{n=1}^{\infty} \frac{1-q^n}{1+q^n}$$ and by definition $$\chi(q) =\prod_{n=1}^{\infty}(1+q^{2n-1})$$ so that your identity holds.

The proof of Ramanujan's identity is provided by Bruce C. Berndt in his Ramanujan's Notebooks Vol. 3 in chapter 20, page 349, Entry 2 (ii).

Here is a complete proof based on Ramanujan's ideas presented in Berndt's book.

The following are definitions of Ramanujan theta functions: \begin{align} f(a, b) &=\sum_{n\in\mathbb {Z}} a^{n(n+1)/2}b^{n(n-1)/2}\notag\\ \phi(q)&=f(q,q)\notag\\ \psi(q) &=f(q, q^3)\notag\\ \chi(q)&=\prod_{n\geq 1}(1+q^{2n-1})\notag \end{align} and Jacobi Triple Product identity reads as $$f(a, b) =\prod_{n\geq 1}(1-(ab)^n)(1+a(ab)^{n-1})(1+b(ab)^{n-1})$$

Let's observe that $$\psi(q^{1/3})=\sum_{n\geq 0}q^{n(n+1)/6}=1+q^{1/3}+q+q^{2}+q^3q^{1/3}+q^5+q^{7}+q^{9}q^{1/3}+\dots$$ The terms corresponding to $$n=1\pmod 3$$ contain a factor $$q^{1/3}$$ and other terms don't contain any fractional powers of $$q$$. If $$n=(3m+1)$$ then $$\frac{n(n+1)}{6}=\frac{3m(m+1)}{2}+\frac{1}{3}$$ and hence such terms combined can be written as $$q^{1/3}\psi(q^3)$$ and thus we have $$\psi(q^{1/3})=q^{1/3}\psi(q^3)+\sum_{n=0\pmod 3}q^{n(n+1)/2}+\sum_{n=2\pmod 3}q^{n(n+1)/2}$$ The remaining two sums can be combined to form $$\sum_{n\in\mathbb{Z}} q^{n(3n+1)/2}=f(q,q^2)$$ and thus we arrive at $$\psi(q^{1/3})=q^{1/3}\psi(q^3)+f(q,q^2)\tag{1}$$ We next apply the same technique with $$\phi(-q^{1/3})=\sum_{n\in\mathbb {Z}} (-1)^nq^{n^2/3}$$ The sum on the right can be expressed as $$\sum_{n\in\mathbb {Z}}(-1)^nq^{3n^2}+\sum_{n=1\pmod 3}(-1)^nq^{n^2/3}+\sum_{n=-1\pmod 3}(-1)^nq^{n^2/3}$$ which is the same as $$\phi(-q^3)-q^{1/3}\sum_{n\in\mathbb {Z}} (-1)^nq^{3n^2+2n}-q^{1/3}\sum_{n\in\mathbb {Z}} (-1)^nq^{3n^2-2n}$$ The last two sums are same as is evident by changing index $$n$$ into $$-n$$ and therefore we have \begin{align} \phi(-q^{1/3})&=\phi(-q^3)-2q^{1/3}\sum_{n\in\mathbb {Z}} (-1)^nq^{3n^2+2n}\notag\\ &=\phi(-q^{3})-2q^{1/3}f(-q,-q^5)\tag{2} \end{align} Next note that \begin{align} f(-q, - q^5)&=\prod_{n\geq 1}(1-q^{6n})(1-q^{6n-5})(1-q^{6n-1})\notag\\ &=\prod_{n\geq 1}\frac{1-q^{6n}}{1-q^{6n-3}}\prod_{n\geq 1}(1-q^{6n-1})(1-q^{6n-3})(1-q^{6n-5})\notag\\ &=\psi(q^3)\chi(-q)\tag{3} \end{align} Therefore $$(2)$$ can be recast as $$\phi(-q^{1/3})=\phi(-q^3)-2q^{1/3}\chi(-q)\psi(q^3)\tag{4}$$ Using similar manipulation we can write $$f(q, q^2)f(-q,-q^5)=\prod_{n\geq 1}(1-q^{3n})(1-q^{6n})=\psi(q^3)\phi(-q^3)\tag{5}$$ We can combine equations $$(1)-(5)$$ and prove Ramanujan's identity mentioned earlier and thus complete the proof of the identity in question.

Thus starting from $$(1)$$ we have \begin{align} \frac{\psi(q^{1/3})}{q^{1/3}\psi(q^3)}&=1+\frac{\phi(-q^3)}{q^{1/3}f(-q,-q^{5})}\text{ (via (5))}\notag\\ &=3+\frac{-2q^{1/3}f(-q,-q^5)+\phi(-q^{3})} {q^{1/3}f(-q,-q^5)}\notag\\ &=3+\frac{\phi(-q^{1/3})}{q^{1/3}f(-q,-q^5)}\text{ (via (2))}\notag\\ &=3+\frac{\phi(-q^{1/3})}{q^{1/3}\psi(q^3)\chi(-q)}\text{ (via (3))}\notag \end{align} Replacing $$q$$ by $$q^3$$ we get Ramanujan's identity.

• (+1) Really nicely done. I deeply envy your mastery of this topic. Commented Dec 20, 2019 at 2:07
• @clathratus: thanks for compliments, but just to clarify, Ramanujan was the master of this art and all mathematicians working in this area are like his students. Commented Dec 20, 2019 at 2:44