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How to evaluate this infinite sum? $$\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

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Why the expression appears so small? How can I enlarge that? : ( –  Ryan Dec 28 '12 at 18:54
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use \displaystyle –  sxd Dec 28 '12 at 18:55
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If I'm not wrong, this one appears in one volume of Ramanujan's notebook. (Chris) –  Chris's sis Dec 28 '12 at 18:59
    
$1.606695152415291$ –  Henry Dec 28 '12 at 19:12
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Not an answer, but it is easy to convert the formula to: $$\sum_{m=1}^\infty \frac{\tau(m)}{2^m}$$ where $\tau(m)$ is the number of distinct divisors of $m$. –  Thomas Andrews Dec 28 '12 at 19:33

4 Answers 4

up vote 4 down vote accepted

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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I want to write out that particular equation just because it's so . . . out there: $$ \sum_{k \ge 1}\frac{1}{e^{kx}-1}=\frac{\gamma}{x}-\frac{\log x}{x}+\frac{1}{4}-\sum_{1 \le k \le n}\frac{B^2_{2k}x^{2k-1}}{(2k)(2k)!}+R_n, $$ where $\gamma$ is Euler's constant, $B_{2k}$ are the conventionally defined Bernoulli numbers, $x>0$, $n\ge 1$, and $R_n$ is a constant bounded by the inequality $$ |R_n|\le \frac{|B_{2n}B_{2n+2}|x^{2n}}{(2n)!}\left(\frac{x^2}{4\pi^2}+\frac{\pi^2}{6} \right). $$ For our case, let $x=\ln 2$ as Chris's sister stated. –  000 Dec 29 '12 at 18:00
    
@Limitless: thank you for the details provided! (+1) –  Chris's sis Dec 29 '12 at 18:03
    
The following MSE post shows how to evaluate this type of sum by inverting Mellin transforms. –  Marko Riedel Apr 8 at 22:46

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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I almost see in each step in the link a possible nice question to ask. :D (+1) –  Chris's sis Dec 28 '12 at 19:24
    
Is it known whether the Erdos-Borwein Constant is transcendental or not? –  Rustyn Dec 28 '12 at 19:27
    
I have no idea. I guess no. –  Amr Dec 28 '12 at 19:35
    
@Rustyn: Please use markdown formatting for non-mathematical things like italics and bold in normal text. I've edited your comment. –  Zev Chonoles Dec 28 '12 at 19:42
    
@ZevChonoles Ok, I will in the future. –  Rustyn Dec 28 '12 at 19:44

$$ \displaystyle \sum _{k=1}^n \frac{1}{\left(\frac{1}{q}\right)^k-\frac{1}{r}}=\frac{r}{\log (q)} \left(\psi _q^{(0)}\left(1-\frac{\log (r)}{\log (q)}\right)-\psi _q^{(0)}\left(n+1-\frac{\log (r)}{\log (q)}\right)\right) $$

In trying to get Mathematica to solve the series, I eventually found the preceding form which assumes $0<q<1$. 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, $\psi _q^{(n)}(z)$.

$$ \displaystyle \lim_{n\to \infty } \, \frac{1}{\log (1/2)}\left(\psi _{\frac{1}{2}}^{(0)}(1)-\psi _{\frac{1}{2}}^{(0)}(n+1)\right)=1+\frac{\psi _{\frac{1}{2}}^{(0)}(1)}{\log (1/2)}=E $$

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(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.

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