Consider the meromorphic function $f$ on $\mathscr D=\{z\in\mathbb C\mid|z|<1\}$ definied by $\displaystyle f(z)=\sum_{n\ge1}\frac{z^{2^n}}{z^{2^n}-\frac12}$. Obviously $f$ admits infinitely many poles (at $z=\frac1{\sqrt[2^n]2}$). So $f$ is transcendental on $\mathbb C(z)$. Now consider the formal power series $\displaystyle F=\sum_{n\ge1}\frac{Z^{2^n}}{Z^{2^n}-\frac12}$. $F$ belongs to $\mathbb C((Z))$ since for every $n\in\mathbb N$, $\displaystyle \frac{Z^{2^n}}{Z^{2^n}-\frac12}$ belongs to $\mathbb C(Z)$ and $\displaystyle\lim_{n\to+\infty}\mathrm{ord}_0\left(\frac{Z^{2^n}}{Z^{2^n}-\frac12}\right)=+\infty$.
My question now: How can one prove RIGOROUSLY that $F$ is transcedental on $\mathbb C(Z)$? There is no sense to talk about poles for this formal power series.
Thanks in advance for any answer.