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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", and Lemma 9 in Tao's blog. 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 (the proof was published by Heine 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.)

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", and Lemma 9 in Tao's blog. 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 (the proof was published by Heine 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.)

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", and Lemma 9 in Tao's blog. 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 (due to Eisenstein's early death the proof was published by Heine.) 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.)

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", and Lemma 9 in Tao's blog. 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 (the proof was published by Heine 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.)

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", and Lemma 9 in Tao's blog. 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 (due to Eisenstein's early death the proof was published by Heine.) 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 (cf. Fermat's little theorem).

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", and Lemma 9 in Tao's blog. 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 (due to Eisenstein's early death the proof was published by Heine.) 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.)