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?



Is this a rational function?
The function
$$ f(z)=\sum\_{n=1}^{\infty} \frac{z^n}{2^n-1} $$
defines a holomorphic function for $|z|<2$, and it satisfies
$$ f(2z) = f(z)+\frac{z}{1-z} $$
for $|z|<1$. Based on this identity, it is easy to prove that $f(z)$ extends to a meromorphic function on $\mathbb{C}$, and the set of poles is $\{2^n:\ n=1,2,\dots\}$. In particular, $f(z)$ does not define a rational function, because its meromorphic extension to $\mathbb{C}$ has infinitely many poles.


Regarding your second question, I recommend the work of Dwork (with which I am not familiar), e.g. (8) in [Alain Robert's article "Des adèles: pourquoi"](http://retro.seals.ch/cntmng?type=pdf&rid=ensmat-001:1974:20::62&subp=hires), and Lemma 9 in [Tao's blog](https://terrytao.wordpress.com/2014/05/13/dworks-proof-of-rationality-of-the-zeta-function-over-finite-fields/). See also Remark 2 below.


**Remark 1.** A more direct proof of the above claims follows from the identity
$$ \sum\_{m=1}^\infty\frac{z}{2^m-z} = \sum\_{n=1}^\infty \frac{z^n}{2^n-1},\qquad |z|<2. $$
Indeed, left hand side defines a meromorphic function on $\mathbb{C}$ with pole set $\{2^m:\ m=1,2,\dots\}$.


**Remark 2.** One can give a different, number theoretic proof using [Eisenstein's theorem on algebraic functions](http://en.wikipedia.org/wiki/Eisenstein%27s_theorem) (the proof was [published by Heine](http://www.digizeitschriften.de/download/PPN243919689_0045/log42.pdf) because of Eisenstein's early death). Indeed, the Taylor coefficients of $f(z)$ around the origin are rational, but their denominators $2^n-1$ are not supported on finitely many primes by Fermat's little theorem. (As Gerald Edgar remarked below, this argument proves that $f(z)$ is not even algebraic.)


