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If I understand well, your question decomposes in two parts the second one being divided in two subparts

Q1) Is $f=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1}$ rational ?

Q2) Let $\sum_{n=0}^{\infty} a_nz^n$ be a function

Q2.1) How to know if it is rational ?

Q2.2) How to check that it has simple pôles ?

Here are my answers (in part not different from what has been written though, save the last part).

R1) $f$ has radius of convergence $2$ because $\frac{1}{2^n-1}\sim_{n\mapsto +\infty} \frac{1}{2^n}$ and $$\sum_{n=1}^{\infty} \frac{z^n}{2^n}=\frac{2z}{2-z}$$ Computing $$ f(2z)=\sum_{n=1}^{\infty} \frac{2^nz^n}{2^n-1}=\sum_{n=1}^{\infty} \Big(1+\frac{1}{2^n-1}\Big) z^n $$ one gets easily $f(2z)=f(z)+\frac{z}{1-z}$ hence, after having checked that there is no cancellation of the pôle $1$, you get pôles at $\{2^n\}_{n\geq 0}$ which is impossible for a rational fraction.

R2) You use the shift operator. Here there is only one and this question, for several noncommutative series, is closely related to automata theory (see here, unfortunately in french) and the theory of Sweedler's, duals, see here. The (one step) shift operator reads

$$\sigma(f)=\frac{f-f(0)}{z}=\sum_{n\geq 0}a_{n+1}z^n\ .$$

Then iterate $$f,\sigma(f),\sigma^2(f),\cdots \sigma^n(f)\cdots $$ then either the orbit $\{\sigma^n(f)\}$ is of infinite rank and $f$ is not rational or $\{\sigma^n(f)\}$ it is of finite rank and you find a (first) $p$ such that $$ \sigma^p(f)=\sum_{j=0}^{p-1}\alpha_j\sigma^j(f) $$ the coefficients $\alpha_j$ being complex. Then, you write the transfer matrix $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \vdots\\ \sigma^p(f) \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 & \cdots &\cdots\\ 0 & 0 & 1 & \cdots & \cdots\\ \vdots & \vdots & \ddots & \ddots &\vdots\\ \vdots & \ddots & \ddots & 0 &1\\ \alpha_0 & \cdots & \cdots & \alpha_{p-2} & \alpha_{p-1} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f)\\ \vdots\\ \sigma^{p-1}(f) \end{pmatrix} $$ then the (non-zero) eigenvalues of $T$ (and their multiplicities) give you the (inverses of the) pôles of $f$ and their multiplicities. Hope it helps.

If I understand well, your question decomposes in two parts the second one being divided in two subparts

Q1) Is $f=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1}$ rational ?

Q2) Let $\sum_{n=0}^{\infty} a_nz^n$ be a function

Q2.1) How to know if it is rational ?

Q2.2) How to check that it has simple pôles ?

Here are my answers (in part not different from what has been written though, save the last part).

R1) $f$ has radius of convergence $2$ because $\frac{1}{2^n-1}\sim_{n\mapsto +\infty} \frac{1}{2^n}$ and $$\sum_{n=1}^{\infty} \frac{z^n}{2^n}=\frac{2z}{2-z}$$ Computing $$ f(2z)=\sum_{n=1}^{\infty} \frac{2^nz^n}{2^n-1}=\sum_{n=1}^{\infty} \Big(1+\frac{1}{2^n-1}\Big) z^n $$ one gets easily $f(2z)=f(z)+\frac{z}{1-z}$ hence, after having checked that there is no cancellation of the pôle $1$, you get pôles at $\{2^n\}_{n\geq 0}$ which is impossible for a rational fraction.

R2) You use the shift operator. Here there is only one and this question, for several noncommutative series, is closely related to automata theory (see here, unfortunately in french) and the theory of Sweedler's, duals, see here. The (one step) shift operator reads

$$\sigma(f)=\frac{f-f(0)}{z}=\sum_{n\geq 0}a_{n+1}z^n\ .$$

Then iterate $$f,\sigma(f),\sigma^2(f),\cdots \sigma^n(f)\cdots $$ then either the orbit $\{\sigma^n(f)\}$ is of infinite rank and $f$ is not rational or $\{\sigma^n(f)\}$ it is of finite rank and you find a (first) $p$ such that $$ \sigma^p(f)=\sum_{j=0}^{p-1}\alpha_j\sigma^j(f) $$ the coefficients $\alpha_j$ being complex. Then, you write the transfer matrix $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \vdots\\ \sigma^p(f) \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 & \cdots &\cdots\\ 0 & 0 & 1 & \cdots & \cdots\\ \vdots & \vdots & \ddots & \ddots &\vdots\\ \vdots & \ddots & \ddots & 0 &1\\ \alpha_0 & \cdots & \cdots & \alpha_{p-2} & \alpha_{p-1} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f)\\ \vdots\\ \sigma^{p-1}(f) \end{pmatrix} $$ then the (non-zero) eigenvalues of $T$ (and their multiplicities) give you the (inverses of the) pôles of $f$ and their multiplicities.

A mini-example to illustrate the link between the non-zero spectrum of the transfer matrix and the poles.

Let $$ f=\sum_{n\geq 1} \frac{nz^n}{a^{n+1}},\ a\not=0 $$ then the shifts are $$ \sigma(f)=\sum_{n\geq 1} \frac{(n+1)z^n}{a^{n+2}},\ \sigma^2(f)=\sum_{n\geq 1} \frac{(n+2)z^n}{a^{n+3}} $$ which yields $$ \sigma^2(f)=\frac{2}{a}\sigma(f)-\frac{1}{a^2}f $$ the transfer matrix is then given by $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \end{pmatrix}= \begin{pmatrix} 0 & 1\\ -\frac{1}{a^2} & \frac{2}{a} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f) \end{pmatrix} $$ its characteristic polynomial is $(X-1/a)^2$.

In fact the fraction was $f=\frac{1}{(z-a)^2}$.

If I understand well, your question decomposes in two parts the second one being divided in two subparts

Q1) Is $f=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1}$ rational ?

Q2) Let $\sum_{n=0}^{\infty} a_nz^n$ be a function

Q2.1) How to know if it is rational ?

Q2.2) How to check that it has simple pôles ?

Here are my answers (in part not different from what has been written though, save the last part).

R1) $f$ has radius of convergence $2$ because $\frac{1}{2^n-1}\sim_{n\mapsto +\infty} \frac{1}{2^n}$ and $$\sum_{n=1}^{\infty} \frac{z^n}{2^n}=\frac{2z}{2-z}$$ Computing $$ f(2z)=\sum_{n=1}^{\infty} \frac{2^nz^n}{2^n-1}=\sum_{n=1}^{\infty} \Big(1+\frac{1}{2^n-1}\Big) z^n $$ one gets easily $f(2z)=f(z)+\frac{z}{1-z}$ hence, after having checked that there is no cancellation of the pôle $1$, you get pôles at $\{2^n\}_{n\geq 0}$ which is impossible for a rational fraction.

R2) You use the shift operator. Here there is only one and this question, for several noncommutative series, is closely related to automata theory (see here, unfortunately in french) and the theory of Sweedler's, duals, see here. The (one step) shift operator reads

$$\sigma(f)=\frac{f-f(0)}{z}=\sum_{n\geq 0}a_{n+1}z^n\ .$$

Then iterate $$f,\sigma(f),\sigma^2(f),\cdots \sigma^n(f)\cdots $$ then either the orbit $\{\sigma^n(f)\}$ is of infinite rank and $f$ is not rational or $\{\sigma^n(f)\}$ it is of finite rank and you find a (first) $p$ such that $$ \sigma^p(f)=\sum_{j=0}^{p-1}\alpha_j\sigma^j(f) $$ the coefficients $\alpha_j$ being complex. Then, you write the transfer matrix $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \vdots\\ \sigma^p(f) \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 & \cdots &\cdots\\ 0 & 0 & 1 & \cdots & \cdots\\ \vdots & \vdots & \vdots & \ddots &\vdots\\ \vdots & \ddots & \ddots & \ddots &1\\ \alpha_0 & \cdots & \cdots & \alpha_{p-2} & \alpha_{p-1} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f)\\ \vdots\\ \sigma^{p-1}(f) \end{pmatrix} $$ then the (non-zero) eigenvalues of $T$ (and their multiplicities) give you the (inverses of the) pôles of $f$ and their multiplicities. Hope it helps.

If I understand well, your question decomposes in two parts the second one being divided in two subparts

Q1) Is $f=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1}$ rational ?

Q2) Let $\sum_{n=0}^{\infty} a_nz^n$ be a function

Q2.1) How to know if it is rational ?

Q2.2) How to check that it has simple pôles ?

Here are my answers (in part not different from what has been written though, save the last part).

R1) $f$ has radius of convergence $2$ because $\frac{1}{2^n-1}\sim_{n\mapsto +\infty} \frac{1}{2^n}$ and $$\sum_{n=1}^{\infty} \frac{z^n}{2^n}=\frac{2z}{2-z}$$ Computing $$ f(2z)=\sum_{n=1}^{\infty} \frac{2^nz^n}{2^n-1}=\sum_{n=1}^{\infty} \Big(1+\frac{1}{2^n-1}\Big) z^n $$ one gets easily $f(2z)=f(z)+\frac{z}{1-z}$ hence, after having checked that there is no cancellation of the pôle $1$, you get pôles at $\{2^n\}_{n\geq 0}$ which is impossible for a rational fraction.

R2) You use the shift operator. Here there is only one and this question, for several noncommutative series, is closely related to automata theory (see here, unfortunately in french) and the theory of Sweedler's, duals, see here. The (one step) shift operator reads

$$\sigma(f)=\frac{f-f(0)}{z}=\sum_{n\geq 0}a_{n+1}z^n\ .$$

Then iterate $$f,\sigma(f),\sigma^2(f),\cdots \sigma^n(f)\cdots $$ then either the orbit $\{\sigma^n(f)\}$ is of infinite rank and $f$ is not rational or $\{\sigma^n(f)\}$ it is of finite rank and you find a (first) $p$ such that $$ \sigma^p(f)=\sum_{j=0}^{p-1}\alpha_j\sigma^j(f) $$ the coefficients $\alpha_j$ being complex. Then, you write the transfer matrix $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \vdots\\ \sigma^p(f) \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 & \cdots &\cdots\\ 0 & 0 & 1 & \cdots & \cdots\\ \vdots & \vdots & \ddots & \ddots &\vdots\\ \vdots & \ddots & \ddots & 0 &1\\ \alpha_0 & \cdots & \cdots & \alpha_{p-2} & \alpha_{p-1} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f)\\ \vdots\\ \sigma^{p-1}(f) \end{pmatrix} $$ then the (non-zero) eigenvalues of $T$ (and their multiplicities) give you the (inverses of the) pôles of $f$ and their multiplicities. Hope it helps.

If I understand well, your question decomposes in two parts the second one being divided in two subparts

Q1) Is $f=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1}$ rational ?

Q2) Let $\sum_{n=0}^{\infty} a_nz^n$ be a function

Q2.1) How to know if it is rational ?

Q2.2) How to check that it has simple pôles ?

Here are my answers (in part not different from what has been written though, save the last part).

R1) $f$ has radius of convergence $2$ because $\frac{1}{2^n-1}\sim_{n\mapsto +\infty} \frac{1}{2^n}$ and $$\sum_{n=1}^{\infty} \frac{z^n}{2^n}=\frac{2z}{2-z}$$ Computing $$ f(2z)=\sum_{n=1}^{\infty} \frac{2^nz^n}{2^n-1}=\sum_{n=1}^{\infty} \Big(1+\frac{1}{2^n-1}\Big) z^n $$ one gets easily $f(2z)=f(z)+\frac{z}{1-z}$ hence, after having checked that there is no cancellation of the pôle $1$, you get pôles at $\{2^n\}_{n\geq 0}$ which is impossible for a rational fraction.

R2) You use the shift operator. Here there is only one and this question, for several noncommutative series is closely related to automata theory (see here, unfortunately in french) and the theory of Sweedler's, duals, see here. The (one step) shift operator reads

$$\sigma(f)=\frac{f-f(0)}{z}=\sum_{n\geq 0}a_{n+1}z^n\ .$$

Then iterate $$f,\sigma(f),\sigma^2(f),\cdots \sigma^n(f)\cdots $$ then either the orbit $\{\sigma^n(f)\}$ is of infinite rank and $f$ is not rational or $\{\sigma^n(f)\}$ it is of finite rank and you find a (first) $p$ such that $$ \sigma^p(f)=\sum_{j=0}^{p-1}\alpha_j\sigma^j(f) $$ the coefficients $\alpha_j$ being complex. Then, you write the transfer matrix $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \vdots\\ \sigma^p(f) \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 & \cdots &\cdots\\ 0 & 0 & 1 & \cdots & \cdots\\ \vdots & \vdots & \vdots & \ddots &\vdots\\ \vdots & \ddots & \ddots & \ddots &1\\ \alpha_0 & \cdots & \cdots & \alpha_{p-2} & \alpha_{p-1} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f)\\ \vdots\\ \sigma^{p-1}(f) \end{pmatrix} $$ then the (non-zero) eigenvalues of $T$ (and their multiplicities) give you the (inverses of the) pôles of $f$ and their multiplicities. Hope it helps.

If I understand well, your question decomposes in two parts the second one being divided in two subparts

Q1) Is $f=\sum_{n=1}^{\infty} \frac{z^n}{2^n-1}$ rational ?

Q2) Let $\sum_{n=0}^{\infty} a_nz^n$ be a function

Q2.1) How to know if it is rational ?

Q2.2) How to check that it has simple pôles ?

Here are my answers (in part not different from what has been written though, save the last part).

R1) $f$ has radius of convergence $2$ because $\frac{1}{2^n-1}\sim_{n\mapsto +\infty} \frac{1}{2^n}$ and $$\sum_{n=1}^{\infty} \frac{z^n}{2^n}=\frac{2z}{2-z}$$ Computing $$ f(2z)=\sum_{n=1}^{\infty} \frac{2^nz^n}{2^n-1}=\sum_{n=1}^{\infty} \Big(1+\frac{1}{2^n-1}\Big) z^n $$ one gets easily $f(2z)=f(z)+\frac{z}{1-z}$ hence, after having checked that there is no cancellation of the pôle $1$, you get pôles at $\{2^n\}_{n\geq 0}$ which is impossible for a rational fraction.

R2) You use the shift operator. Here there is only one and this question, for several noncommutative series, is closely related to automata theory (see here, unfortunately in french) and the theory of Sweedler's, duals, see here. The (one step) shift operator reads

$$\sigma(f)=\frac{f-f(0)}{z}=\sum_{n\geq 0}a_{n+1}z^n\ .$$

Then iterate $$f,\sigma(f),\sigma^2(f),\cdots \sigma^n(f)\cdots $$ then either the orbit $\{\sigma^n(f)\}$ is of infinite rank and $f$ is not rational or $\{\sigma^n(f)\}$ it is of finite rank and you find a (first) $p$ such that $$ \sigma^p(f)=\sum_{j=0}^{p-1}\alpha_j\sigma^j(f) $$ the coefficients $\alpha_j$ being complex. Then, you write the transfer matrix $$ \begin{pmatrix} \sigma(f)\\ \sigma^2(f)\\ \vdots\\ \sigma^p(f) \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 & \cdots &\cdots\\ 0 & 0 & 1 & \cdots & \cdots\\ \vdots & \vdots & \vdots & \ddots &\vdots\\ \vdots & \ddots & \ddots & \ddots &1\\ \alpha_0 & \cdots & \cdots & \alpha_{p-2} & \alpha_{p-1} \end{pmatrix} \begin{pmatrix} f\\ \sigma(f)\\ \vdots\\ \sigma^{p-1}(f) \end{pmatrix} $$ then the (non-zero) eigenvalues of $T$ (and their multiplicities) give you the (inverses of the) pôles of $f$ and their multiplicities. Hope it helps.