# Is $\sum\limits_n\frac{1}{2^{n}-1}$ irrational?

Can we prove that $$\sum_{n=1}^{\infty}\frac{1}{2^{n}-1}$$ is irrational?

Wolfram Alpha says that it is equal to $$1-\frac{\Psi_{1/2}(1)}{\log 2}$$ where $\Psi_{q}(z)$ is a $q$-polygamma function.

Yes, the irrationality of $\sum\limits_n\frac1{t^n-1}$ for any integer $|t|>1$ is a famous result of Erdős. The elementary proof may be found here.
We have $$S=\sum_{n\geq 1}\frac{1}{2^n-1}=\sum_{n\geq 1}\frac{d(n)}{2^n}=\frac{5}{4}+\sum_{k\geq 2}\frac{8^k+1}{(2^k-1)\,2^{k^2+k}} \tag{1}$$ hence the aperiodicity of the binary representation of $S$ (from which the irrationality of $S$ follows) can be proved by exploiting the acceleration formula or the fact that $d(n)=\sigma_0(n)$ is odd iff $n$ is a square, and the set of squares does not belong to any arithmetic progression.
• Can we use the same argument for $\sum 1/(10^n-1)$? – Seewoo Lee Jul 19 '17 at 13:36
• I'm sorry but I have more questions. Is the series $\sum d(n)/2^n$ gives its binary expansion? I think we need some regrouping when we adding them since $d(n)$ is bigger than 1. Also, I can't understand you first argument.. what is the acceleration formula? – Seewoo Lee Jul 19 '17 at 13:54
• Everything is standing just in front of you. The binary representation is given by $\sum_{n\geq 0}\frac{c_n}{2^n}$ with $c_n\in\{0,1\}$, hence the middle series is clearly related with the binary representation. The acceleration formula is the last equality in $(1)$: we convert a series whose general term behaves like $\frac{\log}{2^n}$ into a series whose general term behaves like $\frac{1}{2^{n^2}}$, with a massive improvement about the speed of convergence of partial sums. – Jack D'Aurizio Jul 19 '17 at 13:58