Finding the closed form of the product $\prod_{i = 1}^{\infty} (1 - 1/q^{i}) $.

Question

I would like to know what is the closed form of the following product if it exists:

$$ \prod_{i = 1}^{\infty} \Big(1 - \frac{1}{q^{i}}\Big). $$

I gave it a little time but only managed to establish some pretty obviuos bounds:

$$ \frac{q - 2}{q - 1} < \prod_{i = 1}^{\infty} \Big(1 - \frac{1}{q^{i}}\Big) < \frac{q - 1}{q}. $$

If anyone is interested in a context - this is connected to full rank matrices over the finite field $F_{q}$. Particularly if we will pick square matrices of dimension $n$ over this field uniformly, then this product with $n$ in the place of infinity is the probability of finding a full rank matrix.

So if anyone has a suggestion related to how this probability can be calculated without resorting to finding this product, then I will also be glad to see it.

Answer 1

Depends on what you mean by "closed form". The infinite product $\;\prod_{i = 1}^\infty (1 - \frac{1}{q^{i}}) = f(-1/q)\;$ where $f()$ is a Ramanujan theta function. Other notation is $(1/q;1/q)_\infty\;$ which uses what is called q-Pochhammer symbol. As for calculating its value, in PARI/GP it is $\;\texttt{eta(log(1/q)/(2*Pi*I))}.$