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I found the following identity $\prod_{k=0}^{\infty}(1+\frac{1}{2^{2^k}-1})=\frac{1}{2}+\sum_{k=0}^{\infty}\frac{1}{\prod_{j=0}^{k-1}(2^{2^j}-1)}$

My first thought was to use eulers identity somehow $\prod_{k=1}^{\infty}(1+z^k)=\prod_{k=1}^{\infty}(1-z^{2k-1})^{-1}$ but it does not help me.

If you have an idea or know a helpful identity to prove this result I would really appreciate it.

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My first thought was to "expand the terms of the infinite product." – Lord Soth Apr 2 '13 at 19:47
But, then I said to myself "Well, that may not be a good idea, greed for getting the second term is not cool." – Lord Soth Apr 2 '13 at 20:00
Maybe using a recursive formula for $2^{2^n}-1$: $a(n)=(a(n-1)+1)^2-1, a(0)=1$ – Babla Apr 2 '13 at 20:04
My first reaction is to use the geometric series $1/(T-1) = 1/T + 1/T^2 + 1/T^3 + \cdots$ on both sides of the equation (with $T=2^{2^k}$ and $T=2^{2^j}$) and see if that helps. – Greg Martin Apr 2 '13 at 20:04
The funny thing is that while working on this problem, I "discovered" an extremely fast converging series that can approximate the constant $1$: $\sum_{k=0}^{\infty}\frac{2^{2^k}-1}{2^{2^{k+1}-1}} = 1$. – Lord Soth Apr 2 '13 at 20:29
up vote 4 down vote accepted

To prove this identity, observe that the left-hand side is \begin{eqnarray*} \prod_{k\ge 0} (1+\frac{1}{2^{2^k}-1}) &=& \prod_{k\ge 0} \frac{2^{2^k}}{2^{2^k}-1}\\ &=& \prod_{k\ge 0} \frac{1}{1-2^{-2^k}}\\ &=& \prod_{k\ge 0} (1 + 2^{-2^k} + 2^{-2\cdot 2^k} + 2^{-3\cdot 2^k} + \cdots) \end{eqnarray*} and then, partially expanding the infinite product, \begin{eqnarray*} &\ & \prod_{k\ge 0} (1 + 2^{-2^k} + 2^{-2\cdot 2^k} + 2^{-3\cdot 2^k} + \cdots)\\ &=& 1+\sum_{k\ge 0} (2^{-2^k} + 2^{-2\cdot 2^k}+ 2^{-3\cdot 2^k} + \cdots)\prod_{0\le j<k} (1 + 2^{-2^j} + 2^{-2\cdot 2^j} + 2^{-3\cdot 2^j} + \cdots)\\ &=& 1+\sum_{k\ge 0} \left(2^{-2^k} + \frac{2^{-2\cdot 2^k}}{1-2^{-2^k}}\right) \prod_{0\le j<k} (1-2^{-2^j})^{-1}\\ &=& 1+\sum_{k\ge 0} \left(2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}+ 2^{-2^{k+1}} \prod_{0\le j\le k} (1-2^{-2^j})^{-1}\right)\\ &=& 1+\sum_{k\ge 0} 2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1} +\sum_{\ell\ge 1} 2^{-2^\ell} \prod_{0\le j<\ell} (1-2^{-2^j})^{-1}, \ \ \text{setting } \ell=k+1\\ &=& 1-\frac{1}{2}+2\sum_{k\ge 0} 2^{-2^k} \prod_{0\le j<k} (1-2^{-2^j})^{-1}, \qquad \text{lumping the two sums into one}\\ &=& \frac12+\sum_{k\ge 0} 2^{-(2^0+\cdots+2^{k-1})} \prod_{0\le j<k} (1-2^{-2^j})^{-1}\\ &=& \frac12 + \sum_{k\ge 0} \prod_{0\le j\le k-1} \frac{1}{2^{2^j}-1}, \\ \end{eqnarray*} which is the right-hand side.

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