Calculate $\prod\limits_{n=2}^\infty (1-q^n)$ for $0\leqslant q < 1$, or prove that $\prod\limits_{n=2}^\infty (1-q^n)>0$

Question

How to calculate $$\prod_{n=2}^\infty (1-q^n)$$ for $0\leqslant q < 1$?

I tried to apply $\ln$, giving this: $$\sum_{n\geqslant 2} \ln(1-q^n)$$ but I don't know how to calculate either.

Also, is it possible to show that $\prod\limits_{n=2}^\infty (1-q^n)$ is positive, without using known results about $q$-Pochhammer symbols?

Answer 1

You could e.g. first show that $\log (1-t) + \frac{t}{1-t}\geq 0$ for $t<1$ (minimum for $t=0$) so that $$ \sum_{n\geq 2} \log(1-q^n) \geq -\sum_{n\geq 2} \frac{q^n}{1-q^n} \geq -\sum_{n\geq 2} \frac{q^n}{1-q}= -\frac{q^2}{(1-q)^2}$$

Answer 2

For a completely elementary (and quantitative) solution to the answerable part of the question, note that, by convexity of the exponential, for every $x$ in $[0,q]$, $$1-x\geqslant c(q)^x$$ where $c(q)$ solves $$1-q=c(q)^q$$ Using this for $x=q^n\leqslant q$, for every $n\geqslant1$, one gets $$\prod_{n=1}^\infty(1-q^n)\geqslant c(q)^{q+q^2+q^3+\cdots}=c(q)^{q/(1-q)}=(1-q)^{1/(1-q)}>0$$ Likewise, $$\prod_{n=2}^\infty(1-q^n)\geqslant (1-q)^{q/(1-q)}>0$$