How to find the value of the above product up to infinity?

Asked Jul 21 '23 at 13:57

Active Jul 21 '23 at 14:47

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$$P=\prod_{r=1}^{\infty} \left(1-\frac{1}{2^r}\right)$$

How to find the value of the above product up to infinity? Though I think we can find the minimum and maximum values of the above product using Weierstrass Inequality. Let $S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\cdots$. Then $(1-S)<P<\frac{1}{(1-S)}$; where P is the above product. But how to find the actual value of the above product?

edited Jul 21 '23 at 14:44

J. W. Tanner

60,406

asked Jul 21 '23 at 13:57

Subha Sankar Roy

1

I doubt there exists a closed form; it is actually equal to the Euler function $\phi(q)$ evaluated at $q = -1/2$ (cf. https://en.wikipedia.org/wiki/Euler_function). – Abezhiko Jul 21 '23 at 14:13
But is that true for any value of q where |q|<1? – Subha Sankar Roy Jul 21 '23 at 14:17
You will not get an answer in terms of transcendental functions. Read https://en.wikipedia.org/wiki/Q-Pochhammer_symbol and https://en.wikipedia.org/wiki/Euler%27s_totient_function. Your product P is $\left(\frac{1}{2};\frac{1}{2}\right)_n$ – NadiKeUssPar Jul 21 '23 at 14:18
related: https://mathoverflow.net/q/104685/454 – GEdgar Jul 21 '23 at 14:32
check out this question – vallev Jul 21 '23 at 15:52

How to find the value of the above product up to infinity?

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