Limit of a particular variety of infinite product/series

Question

I was musing about a particular limit,

$L = \prod\limits_{n > 0} \bigl(1 - 2^{-n}\bigr)$:

we may bound 0.288 < L < 0.308, which we may show by taking the logarithm:

$\ln(L) = \ln \bigl( \frac{315}{1024}\bigr) + \sum\limits_{n > 4} \ln\bigl(1 - 2^{-n}\bigr) > \ln\bigl(\frac{315}{1024}\bigr) - \sum\limits_{n > 4} 2^{-n} =\; \ln\bigl(\frac{315}{1024} \cdot \mathrm e^{-1/16}\bigr)$.

I was wondering if this type of infinite product (or the corresponding sum of logarithms) has a name, and whether there are techniques for obtaining a closed form expression for the limit.

Answer 1

The product $$\phi (x) = \prod_{n > 0} (1 - x^n)$$ is called the Euler function, and is well studied. I don't know of any way to compute values at special points. The Euler identity can be used to compute numerical values.

Answer 2

If you want to evaluate $$\prod_{n=1}^\infty (1-1/2^n)$$ numerically you will find that it's VERY close to

$$2^{1/24}\sqrt{\frac{2\pi}{\log 2}}\exp{(-\pi^2/(6\log 2))}.$$

Which is about 0.2887880950866... And that's because $$\exp{(-4\pi^2/\log 2)}$$ is pretty small, approximately 1.839x10^(-25).

This can be seen from the fact that $$\Delta(-1/z) = z^{12}\Delta(z),$$ where $$\Delta(z)$$ is the cusp form of weight 12 defined by

$$\Delta(z)=q\prod_{n=1}^\infty (1-q^n)^{24},$$

where $$q=e^{2\pi iz}$$ for Im(z)>0.

Just put $$z=\frac{2\pi i}{\log 2}.$$

You can use this technique to greatly accelerate the convergence of your product for terms other than (1/2)^n.