# How to evaluate this infinite product: $\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}$ [duplicate]

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

## marked as duplicate by Martin Sleziak, Did, Lucian, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 4 '17 at 11:25

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.
• 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
• @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