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I know that

$$\prod_1^\infty \left(1-\frac{1}{2^n}\right)$$

converges to a positive number because the series $\sum 2^{-n}$ is convergent. Do we know the limit? If so, how?

Aside: I am interested in this product because it describes asymptotically the fraction of $n\times n$ matrices with entries in $\mathbb{F}_2$ that are nonsingular.

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You are using $n$ as a bound variable in the sum and product. So what is it doing as a matrix dimension? – John Bentin Sep 12 '13 at 20:22
up vote 7 down vote accepted

This is a tautology but, your product is a $q$-Pochhammer symbol $(\frac{1}{2}; \frac{1}{2})_\infty$. One connection to Jacobi-Theta functions is:

$$\left(\frac{1}{2}; \frac{1}{2}\right)_\infty=\frac{2^{1/24}}{\sqrt{3}}\vartheta_2\left(\frac{1}{6}\pi,\frac{1}{2^{1/6}}\right)$$

where $\vartheta_n$ is the Jacobi theta function.

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@O.L.: good point. I was mostly just thinking of gamma functions. I've adjusted the answer. – Alex R. Sep 12 '13 at 20:28

Let $$P = \prod_{n\ge 1} \left(1 - \frac{1}{2^n} \right).$$ Introduce $$S = \log P = \sum_{n\ge 1} \log\left(1 - \frac{1}{2^n} \right).$$

Now recall the harmonic sum identity for Mellin transforms: $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

Introducing $$S(x)= \sum_{n\ge 1} \log\left(1 - \frac{1}{2^{nx}} \right)$$ so that $S=S(1)$ we see that $S(x)$ is harmonic with parameters $$\lambda_k = 1, \quad \mu_k = k \quad\text{and}\quad g(x) = \log\left(1 - \frac{1}{2^x} \right)$$ and may be evaluated (approximated) by inverting its Mellin transform.

Now $$\mathfrak{M}\left(\log\left(1 - \frac{1}{2^x} \right);s\right) = \int_0^\infty \log\left(1 - \frac{1}{2^x} \right) x^{s-1} dx = - \int_0^\infty \sum_{q\ge 1} \frac{2^{-qx}}{q} x^{s-1} dx.$$ This is $$-\frac{\Gamma(s)}{(\log 2)^s} \sum_{q\ge 1} \frac{1}{q} \frac{1}{q^s} = -\frac{\Gamma(s)}{(\log 2)^s} \zeta(s+1).$$ It follows that the Mellin transform of $S(x)$ is $$Q(s) = \mathfrak{M}(S(x); s) = -\frac{1}{(\log 2)^s}\Gamma(s)\zeta(s)\zeta(s+1).$$ Now perform Mellin inversion with the inversion integral being $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ and shifting the integral to the left for an expansion about zero. There are only three singularities to consider because the trivial zeros of the two zeta function terms cancel the poles of the gamma function.

We have $$\operatorname{Res}(Q(s)/x^s; s=1) = -1/6\,{\frac {{\pi }^{2}}{\log 2\times x}},$$ $$\operatorname{Res}(Q(s)/x^s; s=0) = 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \log 2 -1/2\,\log x$$ and $$\operatorname{Res}(Q(s)/x^s; s=-1) = 1/24\,\log 2 \times x.$$ Putting $x=1$ we obtain that $$S(1) = S \approx -1/6\,{\frac {{\pi }^{2}}{\log 2}} + 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \log 2 + 1/24\,\log 2.$$ This approximation is excellent (good to $24$ digits) but not quite exact. E.g. setting $x=1/2$ which is closer to zero we get $50$ good digits, setting $x=1/5$ we get $123$ good digits and so on.

The conclusion is that $$ P \approx e^{-\pi^2/6/\log 2} \times \sqrt{2\pi} \times \frac{2^{1/24}}{\sqrt{\log 2}} \approx 0.2887880950866024212788997.$$

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+1 Thank you, this is fascinating! – Ben Blum-Smith Jan 3 '15 at 14:40

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