Approximate result for $\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)$?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

$$\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)=\Bigg[\prod_{n=1}^p\left(1-\frac{1}{2^n}\right)\Bigg]\Bigg[\prod_{n=p+1}^\infty\left(1-\frac{1}{2^n}\right)\Bigg]$$

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 $$\left( \begin{array}{cc} p & \text{result} \\ 1 & \color{red}{0.288}8615143 \\ 2 & \color{red}{0.2887}901234 \\ 3 & \color{red}{0.288788}1550 \\ 4 & \color{red}{0.28878809}69 \\ 5 & \color{red}{0.2887880951 } \end{array} \right)$$