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This question already has an answer here:

I have the product $\prod_{k=1}^\infty \frac{2^k-1}{2^k}$. I know that every successive partial product will necessarily be smaller than the last, as we are multiplying always by a number smaller than 1. But, as the expression keeps getting bigger as k does, and it does so very fast, does this product actually converge on a constant different than $0$? I tried calculating the first partial products but just saw that the terms kept approaching $0.28$. It looks like a very random number, is this just a case of slow convergence to $0$? I have no idea of how to calulate this, any help would be appreciated.

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marked as duplicate by Martin Sleziak, saz, kingW3, b00n heT, Stella Biderman Feb 7 '17 at 20:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Take the logarithm of the partial product: $$A_n = \ln\left(\prod_{k=1}^n \frac{2^k - 1}{2^k}\right) = \sum_{k=1}^n \ln \left( 1 - \frac{1}{2^k}\right)$$ Since $-x - x^2 \le \ln (1 - x) \le -x$, we can bound the sum above as $$ - \sum_{k=1}^n \frac{1}{2^k} - \sum_{k=1}^n \frac{1}{4^k} \le A_n \le - \sum_{k=1}^n \frac{1}{2^k} $$ So $$ -1-\frac13 \le \ln\left(\prod_{k=1}^\infty \frac{2^k - 1}{2^k}\right) \le -1$$ $$ 0.2636 \approx e^{-\frac43} \le \prod_{k=1}^\infty \frac{2^k - 1}{2^k} \le e^{-1} \approx 0.3679$$
So the product is not zero.

If you want to compute the exact value. A double summation change leads to this: $$ \ln\left(\prod_{k=1}^\infty \frac{2^k - 1}{2^k}\right) = - \sum_{k=1}^\infty \frac{1}{k(2^k-1)}.$$

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In general an infinite product $\prod_{k=1}^{\infty}(1+a_k)$ only has a chance to converge (to something nonzero) if $a_k\rightarrow 0$; and if $a_k\rightarrow 0$ and $\sum_k |a_k|^2 < \infty$, then the product converges iff $\sum a_k$ converges. In your case, $a_k = \frac{2^k-1}{2^k}-1=-2^{-k}$, which goes to zero rapidly enough that $\sum_k |a_k|^2 < \infty$ and $\sum a_k$ converges, and hence the product also converges. Now, as to what it converges to, $$ \prod_{k=1}^{\infty} \left(1 - 2^{-k}\right)=\sum_{m=0}^{\infty}2^{-m}(n_m - p_m), $$ where $n_m$ (resp. $p_m$) is the number of partitions of $m$ into an odd (resp. even) number of distinct parts. This is known to equal $$ 1 + \sum_{n=1}^{\infty} (-1)^n \left((1/2)^{n(3n+1)/2} + (1/2)^{n(3n-1)/2}\right)=1 + \sum_{n=1}^{\infty} \frac{(-1)^n(2^n+1)}{2^{n(3n+1)/2}}\\=1-\frac{3}{4}+\frac{5}{2^7}-\frac{9}{2^{15}}+\frac{17}{2^{26}}-\ldots \approx 0.28879, $$ with superlinear convergence.

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Recalling that the $q $-Pochhammer symbol is defined as $$\left(a,q\right)_{\infty}=\prod_{k\geq0}\left(1-aq^{k}\right) $$ we have $$\prod_{k\geq1}\left(1-\frac{1}{2^{k}}\right)=\left(\frac{1}{2},\frac{1}{2}\right)_{\infty} $$ and it's value is $\approx0.289 $.

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