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Find the sum of the series:


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

Thanks in advance.


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

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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:


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:





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.


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