How to evaluate this infinite sum? $\sum_{n=1}^{\infty}\frac{1}{2^n-1}$
Find the infinite sum $\sum_{n=1}^{\infty}\frac{1}{2^n-1}$
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0I made a comment as an answer, see below... – 2012-12-29
4 Answers
(This is not meant as an answer but it is too long for the comment.box)
Since you mentioned interest in variations of the problem: here is a text, in which L. Euler discussed that sum:
"Consideratio quarumdam serierum quae singularibus proprietatibus sunt praeditae" (“Consideration of some series which are distinguished by special properties”)
L. Euler Eneström-index E190.
You can find it online.
A further discussion of this by Prof. Ed Sandifer, where he sheds light on a very interesting discussion about a "false series for the logarithm" at which that constant pops up (and which actually had pointed me originally to L.Euler's article):
A false logarithm series (Discussion of E190)
Ed. Sandifer in: "How Euler did it" Dec 2007
http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2050%20false%20log%20series.pdf
I've fiddled then with it myself a bit further, maybe you find that amateurish explorations interesting too. The constant is part of a consideration on page 5.
Yes. I found it. It is called the Erdős-Borwein Constant.
$E=\sum_{n\in Z^+}\frac{1}{2^n-1}$
Check http://mathworld.wolfram.com/Erdos-BorweinConstant.html
According to the page, Erdős showed that it is irrational.
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0@ZevChonoles Ok, I will in the future. – 2012-12-28
I think you wanna see this:
Ramanujan’s Notebooks Part I
Click me and try Entry $14$ (ii) / pag 146 where you set $x=\ln2$
Chris.
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0The following [MSE post](http://math.stackexchange.com/questions/157040/a-new-formula-for-aperys-constant-and-other-zetas) shows how to evaluate this type of sum by inverting Mellin transforms. – 2014-04-08
$ \displaystyle \sum _{k=1}^n \frac{1}{\left(\frac{1}{q}\right)^k-\frac{1}{r}}=\frac{r}{\log (q)} \left(\psi _q^{(0)}\left(1-\frac{\log (r)}{\log (q)}\right)-\psi _q^{(0)}\left(n+1-\frac{\log (r)}{\log (q)}\right)\right) $
In trying to get Mathematica to solve the series, I eventually found the preceding form which assumes $0. If we take $q=1/2$, $r=1$ and let n approach infinity, we get the same solution that Amr references. The partial sum solution utilizes the function, $\psi _q^{(n)}(z)$.
$ \displaystyle \lim_{n\to \infty } \, \frac{1}{\log (1/2)}\left(\psi _{\frac{1}{2}}^{(0)}(1)-\psi _{\frac{1}{2}}^{(0)}(n+1)\right)=1+\frac{\psi _{\frac{1}{2}}^{(0)}(1)}{\log (1/2)}=E $