# Products of $|q|<1$ : $\prod_{n=1}^{\infty}(1-q^n) = \prod_{n=1}^{\infty} (1-q^{2n})(1-q^{2n-1})$

I want to show the following infinity products of $q$ $|q|<1$.

\begin{align} & \prod_{n=1}^{\infty} (1-q^n) = \prod_{n=1}^{\infty} ( 1- q^{2n-1})(1-q^{2n}) \\ & \prod_{n=1}^{\infty} (1-q^n) = \prod_{n=1}^{\infty} ( 1- q^{3n-1})(1-q^{3n-2}) (1-q^{3n}) \\ & \prod_{n=1}^{\infty} (1-q^n) = \prod_{n=1}^{\infty} (1-q^{4n-1})(1-q^{4n-2})(1-q^{4n-3})(1-q^{4n}) \\ & \cdots \end{align}

It seems for decomposition of $kn$, the products can be expressed as a $k$ sequence products of $q$ $q^{kn-1}, \cdots$ $q^{kn-(k-1)n}$, $q^{kn}$

I have some trouble understanding this. Any comment will be helpful.

• It's a joke ? ${}{}{}$ – reuns Jan 2 '17 at 12:09

## 1 Answer

Given a positive integer $k$, break the product into products extended over congruence classes modulo $k$. Thus

$$\prod_{n \,=\, 1}^\infty (1 - q^n) = \prod_{r\, =\, 0}^{k-1} \prod_{n\, \equiv \,r\pmod{k}} (1 - q^n) = \prod_{r\, =\, 0}^{k-1}\prod_{j = 1}^\infty (1 - q^{kj-r}) = \prod_{j\, =\, 1}^\infty \prod_{r\, =\, 0}^{k-1}(1 - q^{kj-r})$$

Since $j$ is a dummy variable, we can write

$$\prod_{n\, =\, 1}^\infty (1 - q^n) = \prod_{n\, =\, 1}^\infty \prod_{r\, =\, 0}^{k-1}(1 - q^{kj-r}) = \prod_{n\, =\, 1}^\infty (1 - q^{kn-k+1})\cdots(1-q^{kn-1})(1-q^{kn})$$