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How to evaluate this one $$\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}$$

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I'm pretty sure there's no closed form for this, and that it has come up here before (though finding it on this site may not be so easy). – Gerry Myerson Feb 8 '13 at 12:04
Closest I've found to it here:… – Gerry Myerson Feb 8 '13 at 12:18
This is closer... – Neves Feb 8 '13 at 12:25
More here and here. – Neves Feb 8 '13 at 12:39
up vote 1 down vote accepted

Hint(From Complex Analysis by Lars V. Ahlfors, $3$rd Edition, Page-$192$, Theorem-$5$):The infinite product $\Pi (1+a_n)$ with $1+a_n\neq 0$ converges simultaneously with the series $\sum_{1}^{\infty} \log(1+a_n)$ whose terms represent the values of the principal branch of the logarithm.

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Just a small side comment, your statement of simultaneous convergence is not true. consider $a_n = -\frac{1}{2}$. Then the product converges to $0$, and the sum of logarithms diverges towards $-\infty$. Still, it is true in OP's case. – Arthur Feb 8 '13 at 10:24
He just "took" log on either side. – Gautam Shenoy Feb 8 '13 at 10:25
@GautamShenoy Yes, he did, and I'm saying that if the product converges to $0$, then the log diverges, and thus there is not simultaneous convergence, as proposed. He needs a stronger assumption than $1+a_n \neq 0$, like "The product does not converge to a non-positive number, and we have $1+a_n > 0$ for all $n$." Also, these are positive, real numbers, so there is no need to mention the principal branch. – Arthur Feb 8 '13 at 10:26
@Arthur It's not unheard of to define such a product as "diverging to 0", so as to eliminate this minor technicality. The expression sounds strange at first but it makes the correspondence between sums and products very natural. – Erick Wong Feb 8 '13 at 10:31
The question wasn't whether it converged, but what does it converge to, and I don't think this gives an answer. – Gerry Myerson Feb 8 '13 at 12:01

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