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.
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.
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.
It is relevant to mention that Tachiya has proved an interesting generalization of the above claim, by producing acceleration formulas through umbral calculus techniques.
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.
See also the Wikipedia page on the Erdős–Borwein constant.