Could you please tell me how to compute $\prod_{1\le i\le n} \left(1-\frac{1}{2^i}\right)$? Especially, I would like to know its limit when $n\to \infty$.
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The last sentence doesn't make sense. Do you want the product, or do you want the limit? There is no in-between. – Arnaud Mortier Jul 02 '18 at 13:09
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Thanks for pointing out. I would like to know the limit as $n$ tends to infinity. – Xiuping Jul 02 '18 at 13:11
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https://www.wolframalpha.com/input/?i=prod+(1-2%5E(-i-1)) – anonymus Jul 02 '18 at 13:27
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2See https://math.stackexchange.com/questions/3776/limit-of-a-particular-variety-of-infinite-product-series – lhf Jul 02 '18 at 13:28
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@anonymus Thank you very much – Xiuping Jul 02 '18 at 14:18
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@lhf Thank you very much – Xiuping Jul 02 '18 at 14:19
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The limit is $Q$ in Tree searching, which has many different forms. – Simply Beautiful Art Jul 02 '18 at 14:36
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@SimplyBeautifulArt Okay, many thanks – Xiuping Jul 03 '18 at 19:54
1 Answers
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$$\prod_{n\geq 1}\left(1-\frac{1}{2^n}\right) = \exp\left[-\sum_{n,m\geq 1}\frac{1}{m 2^{mn}}\right]=\exp\left[-\sum_{k\geq 1}\frac{\sigma(k)}{k 2^k}\right] $$
where $\sum_{k\geq 1}\frac{\sigma(k)}{k 2^k}$ converges pretty fast to $\approx 1.2420620948124$. It follows that
$$ \prod_{n\geq 1}\left(1-\frac{1}{2^n}\right) \approx 0.288788095. $$
For the acceleration of similar Lambert-like series, have a look at this question, too.
The approach outlined by Marko Riedel (Mellin transform and approximation through the trivial zeroes of the Riemann $\zeta$ function) here is also extremely interesting.
Jack D'Aurizio
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