Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
Why the expression appears so small? How can I enlarge that? : ( –  Ryan Dec 28 '12 at 18:54
use \displaystyle –  sxd Dec 28 '12 at 18:55
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
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
show 6 more comments

4 Answers 4

up vote 3 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 $x=\ln2$


share|improve this answer
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
add comment

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.

share|improve this answer
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
show 1 more comment

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

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, $\psi _q^{(n)}(z)$.

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

share|improve this answer
add comment

(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

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.