1
$\begingroup$

This question already has an answer here:

Find the sum of the series:

$$\sum_{n=1}^{\infty}\frac{1}{2^n-1}.$$

The series converges (using either the comparison test, or the ratio test). But can the sum be somehow calculated?

Thanks in advance.

$\endgroup$

marked as duplicate by Arnaud D., Delta-u, José Carlos Santos calculus Oct 26 '18 at 12:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

9
$\begingroup$

Your series converges to a value known as the Erdős-Borwein Constant.

It's irrational and as of $2018$ has no known "closed form" in terms of elementary functions.

The only place I know it comes up is in the analysis of I believe several sorting algorithms.


However if one uses special functions related to the theory of $q$ series it can be re-written with what is known as the $q$-polygamma function $\psi_q(z)$ which is also analogously the logarithmic derivative of the $q$-gamma function $\Gamma_q(z)$. Namely we have that: $$\sum_{n=1}^\infty\frac{1}{2^n-1}=1-\frac{\psi_{\frac{1}{2}}(1)}{\log(2)}$$

Though here are some other representations if you're interested:

$$\sum_{n=1}^\infty\frac{1}{2^n-1}=\sum_{n=1}^{\infty}\frac{d(n)}{2^n}=\sum_{n=1}^{\infty}\frac{1}{2^{n^2}}\frac{(2^n+1)}{(2^n-1)}=-\frac{1}{2}+\frac{1}{2}\sum_{n=-\infty}^{\infty}\frac{1}{2^{n^2}}+2\sum_{n=1}^{\infty}\frac{1}{2^{n^2}(2^n-1)}\\=-\frac{1}{2}+\prod_{n=1}^{\infty}\left(1-\frac{1}{4^n}\right)\left(1+\frac{2}{4^n}\right)^2+2\sum_{n=1}^{\infty}\frac{1}{2^{n^2}(2^n-1)}$$

Where for any $m\in \mathbb{N}$ the function $d(m)=\sum_{d\mid m}1$ counts the positive divisors of $m$.


For additional information on its usage/properties:

http://mathworld.wolfram.com/Erdos-BorweinConstant.html

http://mathworld.wolfram.com/TreeSearching.html

https://books.google.com/books?id=cYULBAAAQBAJ&pg=PA155&lpg=PA155

$\endgroup$
2
$\begingroup$

Since $\frac{1}{2^n-1}=\frac{2^{-n}}{1-2^{-n}}=2^{-n}+2^{-2n}+\cdots$, your series is $\sum_{n\ge 1}2^{-n}\tau(n)$, where $\tau(n)$ is the number of positive factors of $n$. If we define $\tau(0):=0$, the generating function is $f_\tau(x):=\sum_{n\ge 1}x^n\tau(n)$, so your series is $f_\tau(\frac{1}{2})$. However, I doubt there's a closed form for this.

$\endgroup$

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