# Find the infinite product series [duplicate]

$$\lim_{n\to\infty} (1-\frac{1}{2})(1-\frac{1}{4}) \cdots (1-\frac{1}{2^n})=\prod_{n=1}^{\infty} (1-\frac{1}{2^n})$$

I've tried this techniques:

1) Using squeeze theorem, but i had not find right sequences to compare with

2) Taking log of both sides and trying to find the sum of the series, but the person that gived me this task has not been learning this topic yet(series). Although i`ve tried this:

$$x-x^2\leqslant\log(1+x)\leqslant x$$ at least for $$x \in [-0.5;0]$$

and summing this gives not that kind of approximation that we needed.

3) And that:

$$\prod_{n=1}^{\infty} (1-\frac{1}{2^n})=\prod_{n=1}^{\infty} (\frac{2^n-1}{2^n})= \prod_{n=1}^{\infty} (\frac{\sum_{k=0}^{n-1} 2^k}{2^n})$$

I thought that product of sums in numerator will be just the $$(\sum_{k=1}^{\frac{n(n-1)}{2}}2^k) - 1$$ using that $$\sum_{k=1}^{n-1}k = \frac{n(n-1)}{2}$$ (just trying to make a guess) But all that is not working for me, i need some help, using standard methods of limits topic preferably.

## marked as duplicate by smcc, ncmathsadist, user10354138, Leucippus, Claude Leibovici limits 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(); } ); }); }); Nov 10 '18 at 4:43

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