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I'm stuck in finding the sum: \begin{align} S = \sum_{k=2}^n \frac{1}{2^k-1} \end{align} It seems to be close to geometric without being geometric. I wanted to try to transform it to a geometric series but I don't seem to come up with a valid transformation to do it. Anyone has an idea?

Victor
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    This has answers here: https://math.stackexchange.com/q/1978310/431789 and here: https://math.stackexchange.com/q/2835362/431789 – David Diaz Apr 03 '19 at 03:30
  • There is a comment in the second post that "there is no transformation" because the sum is not multiplicative. It's not clear to me that this is true, only that any manipulation isnt obvious. – David Diaz Apr 03 '19 at 03:43

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$$S = \sum_{k=2}^n \frac{1}{2^k-1}$$ This sum cannot be expressed with a finite number of elementary functions. For a closed form a special function is required, namely the so called q-digamma function $\psi_{q}(x)$ :

http://mathworld.wolfram.com/q-PolygammaFunction.html

$$\sum_{k=2}^n \frac{1}{2^k-1}=\frac{1}{\ln(2)}\left(\psi_{1/2}(n+1)-\psi_{1/2}(2) \right)$$

JJacquelin
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