Let $\alpha,\beta\in\overline{\mathbb Q}$ be such that $0<|a|<1$ and $|\beta|=1$. One assumes that $\beta$ is not a root of unity. Is $A:=\sum_{n\ge0}\frac{\alpha^{2^n}}{2-\beta^{2^n}}$ transcendental?
If one introduces the function $f(X,Y)=\sum_{n\ge0}\frac{X^{2^n}}{2-Y^{2^n}}$, one sees that $f(X^2,Y^2)=f(X,Y)-\frac{X}{2-Y}$, $f$ is analytic on $\{(X,Y)\in\mathbb C^2\mid |X|<1\text{ and }|Y|<1\}$ and is definite on $\{(X,Y)\in\mathbb C^2\mid |X|<1\text{ and }|Y|=1\}$
But one can not use Mahler's theorem (see Nishioka's book "Mahler functions and transcendence", Chpater II) to obtain the transcendence of $A$ since $|\beta|\ge1$.
Does anyone know a way to prove $A$ is transcendental?
