Approximate result for $\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)$? What would be a quick way to approximately determine the value of
$$\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)=\phi\left(\frac{1}{2}\right) , $$
where $\phi(q)$ is the Euler function? By approximating, I mean determine the first few digits of the result starting from the product representation without additional knowledge of other properties of $\phi(q)$.
For instance, my non-programmable CASIO fx-991ES calculator comes with a summation function $\sum_n$, but no multiplication function $\prod_n$, so assuming that I have access to this low complexity piece of equipment, I got an approximation as
$$\begin{align}\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right) &= \exp\left(\ln\left(\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)\right)\right)\\
&\approx\exp\left(\sum_{n=1}^{100}\ln\left(1-\frac{1}{2^n}\right)\right)\\
&=0.2887880951\end{align}$$
by doing a finite sum and exponentiating the result instead. Curiously, all of the digits are correct, so maybe summing even fewer terms would be sufficient to get just a few digits right.
But what if all I have is paper and pen? Is there a way to get a good approximation in a few lines?
 A: A quadratic convergence formula is given by the pentagonal number theorem $$\prod_{n=1}^\infty(1-x^n)=\sum_{n=-\infty}^\infty(-1)^n x^{n(3n-1)/2}=1+\sum_{n=1}^\infty(-1)^n(1+x^n)x^{n(3n-1)/2}.$$ The summation (at $x=1/2$) over $1\leqslant n\leqslant 10$ already gives over $50$ correct digits.
A: Put
$$
p_{\;n}  = \prod\limits_{k = 1}^n {\left( {1 - {1 \over {2^{\,k} }}} \right)} 
$$
Then
$$
p_{\;n + 1}  = p_{\;n} \left( {1 - {1 \over {2^{\,n + 1} }}} \right)
 = p_{\;n}  - {{p_{\;n} } \over {2^{\,n + 1} }}\quad \left| {\;p_{\;1}  = {1 \over 2}} \right.
$$
which means that the $\{ p_n \}$ sequence is steadily decreasing, while remain positive, and thus converges.
Being steadily decreasing, at a decreasing absolute rate, you can multiply the first terms untill the required number of digits
become stable.
A: $$\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)=\Bigg[\prod_{n=1}^p\left(1-\frac{1}{2^n}\right)\Bigg]\Bigg[\prod_{n=p+1}^\infty\left(1-\frac{1}{2^n}\right)\Bigg]$$
Fot the second product, take its logarithm
$$\log\Bigg[\prod_{n=p+1}^\infty\left(1-\frac{1}{2^n}\right)\Bigg]=-\sum_{n=p+1}^\infty \left(\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{3 x^3}+\frac{1}{4 x^4}+O\left(\frac{1}{x^5}\right)\right)$$ where $x=2^n$.
So, the approximation will write
$$\Pi_p=\frac {a_p}{b_p} \exp\left(- \frac {c_p}{d_p}\right) $$ Below are listed the various constants
$$\left(
\begin{array}{ccccc}
p & a_p & b_p & c_p & d_p \\
 1 & 1 & 2 & 1229 & 2240 \\
 2 & 3 & 8 & 28087 & 107520 \\
 3 & 21 & 64 & 73229 & 573440 \\
 4 & 315 & 1024 & 1738567 & 27525120 \\
 5 & 9765 & 32768 & 4611629 & 146800640 
\end{array}
\right)$$ and now the numerical values
$$\left(
\begin{array}{cc}
p & \text{result} \\
 1 & \color{red}{0.288}8615143 \\
 2 & \color{red}{0.2887}901234 \\
 3 & \color{red}{0.288788}1550 \\
 4 & \color{red}{0.28878809}69 \\
 5 & \color{red}{0.2887880951 }
\end{array}
\right)$$
