What is the value of $\prod_{i=1}^\infty \left(1-\frac{1}{2^i}\right)$?

Question

Also, what about in general, for some value p, which has the value 2 in the given formula?

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MOTIVATION:

I was wondering the probability of never getting tails if one forever flipped a coin whose probability of landing tails decreased (in this case, geometrically) each flip.

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The probability of tails on flip i is $\frac{1}{2^i}$, and the probability of heads on flip i is $1-\frac{1}{2^i}$. So, first flip the coin is 50-50, next it is 75-25, etc. And the probability of never landing tails is equal to the probability of always landing heads, which is the infinite product of the heads probabilities, yielding $\prod_{i=1}^\infty \left(1-\frac{1}{2^i}\right)$. => ($\frac{1}{2} * \frac{3}{4} * \frac{7}{8} ...$)

Answer 1

You can use the $q$-Pochhammer symbol to represent the infinite product as

$$ \prod_{i=1}^{\infty}\left( 1-\frac{1}{2^i} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)_{\infty} . $$

Note:

$$ \left( a, q \right)_{\infty}= \prod_{i=1}^{\infty}\left( 1-a q^k \right) $$

Answer 2

$$\sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) = -\sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{n 2^{kn}}=-\sum_{m\geq 1}\frac{1}{2^m}\sum_{d\mid m}\frac{1}{d}=-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}$$ hence: $$\prod_{i\geq 1}\left(1-\frac{1}{2^i}\right)=\exp\left(-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}\right)\geq\exp\left(-\sum_{m\geq 1}\frac{m+1}{2^{m+1}}\right)=e^{-3/2}$$ but the bound: $$\sigma_1(m)=\sum_{d\mid m}d \leq \sum_{k=1}^{m}k = \frac{m(m+1)}{2}$$ is obviously very crude.