TITLE: Is this a rational function?
QUESTION [19 upvotes]: Is $$\sum_{n=1}^{\infty} \frac{z^n}{2^n-1} \in \mathbb{C}(z)\ ?$$
In a slightly different vein, given a sequence of real numbers $\{a_n\}_{n=0}^\infty$, what are some necessary and sufficient conditions for $\sum a_nz^n$ to be in $\mathbb{C}(z)$ with all poles simple?

REPLY [14 votes]: I would not argue with the irrationality of the function following from the functional equation relating $f(2z)$ to $f(z)$. In fact this function is a particular case of the so-called $q$-logarithm which received a quite special attention. Note that being rational over $\mathbb C(z)$ or over $\mathbb Q(z)$ for a power series with rational coefficients is equivalent.
If $f(z)$ where rational then $$f(1)=\sum_{n=1}^\infty \frac1{2^n-1}=\sum_{m=1}^\infty \tau(m)2^{-m}$$ would be a rational number, where $\tau(m)$ denotes the number of divisors of $m$. Erdős proved already in 1948 that this is not the case by showing that the base $2$ expansion of the number is not periodic. (This is still a nice exercise in analytic number theory!)
The irrationality of the values of $f(z)$ for rational $z\ne0$ was established by P. Borwein in 1992 using the Padé approximations to the function.