Proof of Polynomial Identity

Question

I would like to prove the following identity: $$\prod_{n=1}^\infty\frac{1+q^{3n}}{1-q^{2n}} = \frac{\prod_{k=1}^\infty(1-q^{6k-5}) (1-q^{6k-1})}{\prod_{m=1}^\infty(1-q^m)}.$$

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From the components in $LHS$, $$\frac{1+q^{3n}}{1-q^{2n}} = \frac{1+q^{3n}}{(1+q^{n})(1-q^{n})}\\=\frac{(1+q^{n})(1-q^{n}+q^{2n})}{(1+q^{n})(1-q^{n})}\\=\frac{1-q^{n}+q^{2n}}{1-q^{n}}$$

Thus now I need to compare $1-q^{n}+q^{2n}$ and $(1-q^{6k-5}) (1-q^{6k-1})$, which is equal to $1-q^{6k-5}$.

How could I prove that $\prod_{n=1}^\infty(1-q^{n}+q^{2n}) = \prod_{k=1}^\infty (1-q^{6k-5})$?

Answer 1

$$ \prod_{m=1}^\infty (1-q^m) = \prod_{k=1}^\infty (1-q^{6k-5})(1-q^{6k-4})(1-q^{6k-3})(1-q^{6k-2})(1-q^{6k-1})(1-q^{6k}). $$

$$ \begin{aligned} &\frac{(1-q^{6k-5})(1-q^{6k-1})}{(1-q^{6k-5})(1-q^{6k-4})(1-q^{6k-3})(1-q^{6k-2})(1-q^{6k-1})(1-q^{6k})}\\ &=\frac{1}{(1-q^{6k-4})(1-q^{6k-3})(1-q^{6k-2})(1-q^{6k})}\\ &=\frac{(1-q^{6k})}{(1-q^{6k-4})(1-q^{6k-2})(1-q^{6k})(1-q^{6k-3})(1-q^{6k})}\\ &=\frac{(1+q^{3k})(1-q^{3k})}{(1-q^{6k-4})(1-q^{6k-2})(1-q^{6k})(1-q^{6k-3})(1-q^{6k})}. \end{aligned} $$

Starting from this, you need to claim that the term $(1-q^{3k})$ in the numerator cancels out "suitable terms" in the denominator.

Answer 2

Since $\;(1+x^3)/(1-x^2) = (1-x^6)/((1-x^2)(1-x^3)),\;$ all you need to do to prove the identity is to show that the infinite products are equal to $\;\prod_{n=1}^\infty (1-q^n)^{a_n}\;$ where $\;\{a_1,a_2,\dots\}\;$ is a period $6$ sequence with $\;a_n=0\;$ if $\;n\equiv 1,5\pmod{6}\;$ and $\;a_n=-1\;$ otherwise. Another way of stating this is $\;a_n=-1\;$ if $n$ is a multiple of $2$ or $3$ and $\;a_n=0\;$ otherwise. By the way, the infinite product is the generating function of OEIS sequence A007690 which states the second way.