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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0Why the expression appears so small? How can I enlarge that? : ( – 2012-12-28
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1use \displaystyle – 2012-12-28
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1If I'm not wrong, this one appears in one volume of Ramanujan's notebook. (Chris) – 2012-12-28
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0$1.606695152415291$ – 2012-12-28
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0I saw this sum before – 2012-12-28
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0Is there a closed form? – 2012-12-28
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0No I remember that the constant was called E – 2012-12-28
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0How about if we change the infinity with k? – 2012-12-28
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0The numerators and denominators of the partial sums are OEIS sequences A087689 and A087690. – 2012-12-28
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2Not 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 $m$. – 2012-12-28
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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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2I almost see in each step in the link a possible nice question to ask. :D (+1) – 2012-12-28
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0Is it known whether the *Erdos-Borwein Constant* is transcendental or not? – 2012-12-28
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0I have no idea. I guess no. – 2012-12-28
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0@Rustyn: Please use [markdown formatting](http://stackoverflow.com/editing-help) for non-mathematical things like italics and bold in normal text. I've edited your comment. – 2012-12-28
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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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2I 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. – 2012-12-29
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0@Limitless: thank you for the details provided! (+1) – 2012-12-29
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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, $\psi _q^{(n)}(z)$.
$$ \displaystyle \lim_{n\to \infty } \, \frac{1}{\log (1/2)}\left(\psi _{\frac{1}{2}}^{(0)}(1)-\psi _{\frac{1}{2}}^{(0)}(n+1)\right)=1+\frac{\psi _{\frac{1}{2}}^{(0)}(1)}{\log (1/2)}=E $$