Suppose the one has a sequence of rational functions $Q_n(z)\in\mathbb Q(z)$. Let $p$ be a prime number. Suppose that that there exists an infinite subset $X$ of $\mathbb Q_p$ such that:
- the functions $Q_n$ are defined on $X$
- for all $z\in X$, $\lim_{n\to+\infty}Q_n(z)=0$ in $\mathbb Q_p$
- the function $F(z)=\sum_{n\ge0}Q_n(z)$ is transcendental on $\mathbb Q(z)$; that is: for any polynomials $P_0,\cdots,P_n$ of $\mathbb Q[z]$ not all zero, there exists $z\in X$ such that $\sum_{i=0}^nP_i(z)F^i(z)\ne0$
My question now: considers an another prime $p'$ and $Y$ be an infinite subset of $\mathbb Q_{p'}$. Suppose that for all $n\in\mathbb N$, the functions $Q_n$ are definied on $Y$ and for every $z\in Y$, one has $\lim_{n\to+\infty}Q_n(z)=0$ in $\mathbb Q_{p'}$.
Is the function $G(z)=\sum_{n\ge0}Q_n(z)$ (considered as a function defined on $Y\subset\mathbb Q_{p'}$) transcendental on $\mathbb Q[z]$?
Thanks in advance for any answer.