More Limits

The product generalizes to

$P(q) = \prod_{k=1}^{\infty} \left( 1 - q^{k} \right)$

The product above is a special case of the q-Pochhammer symbol known as Euler function denoted by

$\phi (q) = (q ; q)_{\infty} = P(q)$

The value is asymptotically equal to

$(q;q)_{\infty} \sim \sqrt{\dfrac{2 \pi}{\ln \left( q^{-1} \right)}} \exp \left( - \dfrac{\pi^2}{6 \ln \left( q^{-1} \right)} + \dfrac{\ln \left( q^{-1} \right)}{24} \right)$

Setting $q= \dfrac 12$ gives our desired result which is approximately equal to

$A = P \left( \dfrac 12 \right) = \left( \dfrac 12 ; \dfrac 12 \right)_{\infty} \approx \sqrt{\dfrac{2 \pi}{\ln 2}} \exp \left( - \dfrac{\pi^2}{6 \ln 2} + \dfrac{\ln 2}{24} \right) = 0.2887880950866$

correct to 13 places of decimals.

Thus, $\left\lfloor 10^9 A \right\rfloor = \boxed{288788095}$ .