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How to calculate infinite product $\prod_{n=1}^\infty(2^n-1)/2^n$ i.e. $(1-1/2)(1-1/4)(1-1/8)(1-1/16) \cdots $

I know the value is around $0.2888 \cdots$ but I can't figure out how the number come.

If the answer is so complicated, anyone could explain me how to show that the value doesn't converge to $0$?

thanks.

balddraz
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이성진
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  • Now I see convergence value is strictly positive. – 이성진 Jul 31 '19 at 05:50
  • I have voted to close this as the primary question asked is an exact duplicate of https://math.stackexchange.com/questions/2128456/evaluate-prod-n-1-infty-frac2n-12n, which itself has been linked to 3 more exact duplicates https://math.stackexchange.com/questions/linked/1200575 – pre-kidney Jul 31 '19 at 06:25

2 Answers2

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Partial answer: To answer the last part, just note that $ \sum\limits_{k=1}^{\infty} \frac 1 {2^{n}} <\infty$. Hence $\log \prod_1^{N} \frac {2^{n}-1} {2^{n}}=\log \prod_1^{N} (1- \frac 1 {2^{n}})=\sum\limits_{k=1}^{N} \log (1-\frac 1 {2^{n}})$. Now use the fact that $\frac {log (1-x)} x \to -1$ as $x $ increases to $0$. Can you finish the proof now?

Hardik Kalra
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Kavi Rama Murthy
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This product is related to the Dedekind eta-function. Indeed for $|q|<1$, Euler's pentagonal number formula states that $$\prod_{n=1}^\infty(1-q^n)=1+\sum_{k=1}^\infty(-1)^k(q^{k(3k-1)/2}+q^{k(3k+1)/2}).$$ The sum has the advantage over the product of being more rapidly convergent.

A general result on infinite products is that if $(a_n)$ is a sequence with $0<a_n<1$ and $\sum_{n=1}^\infty a_n$ is convergent, then $\prod_{n=1}^\infty(1-a_n)$ converges to a strictly positive value. This proves your product must be positive.

Angina Seng
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  • Please my answer here https://math.stackexchange.com/questions/2128456/evaluate-prod-n-1-infty-frac2n-12n/2128460#2128460 presenting the same rapidly convergent series as well as proof of transcendence. – pre-kidney Jul 31 '19 at 06:29