We say a power series in $\mathbb{C}[[z]]$ is analytic, if it has a positive radius of convergence.
Let $P(T) = \sum_{i=0}^n u_iT^i$ be a polynomial those coefficients $u_i \in \mathbb{C}[[z]]$ are analytic power series and let $f \in \mathbb{C}[[z]]$ be a root of $P$.
Question: Is $f$ analytic ?
Background: If the answer is "yes", any non-analytic power series would be transcendental over the ring of analytic power series. This would produce a counter-example to the second question in https://mathoverflow.net/questions/266198/how-to-pass-from-algebraic-power-series-to-the-analytic-ones/266216#266216.
The answer is yes.
Proof: Let $F$ be the field of Puiseux series over $\mathbb{C}$. Each $g \in F$ can be written as $g(z)=h(z^{1/r})$ where $h(z)=z^k\sum_{n=0}^\infty a_nz^n$ and $k$ an integer. $g$ is called an analytic Puiseux series, if $\sum_{n=0}^\infty a_nz^n$ is analytic. Denote the set of analytic Puiseux series by $F_a$. We will use the Newton-Puiseux Theorem from "Nowak: Some elementary proofs of Puiseux's Theorem" (in the paper "analytic" is called "convergent"):
$P$ (from the question) is a polynomial over $F_a\subseteq F$, so its roots $f_1,...,f_n$ are in $F_a$. Since $f \in \mathbb{C}[[z]] \subseteq F$ is also a root of $P$, there is an $i$ such that $f=f_i$. Hence $f \in F_a$ is an analytic Puiseux series and because $f$ is a power series, it's an analytic power series. Q.E.D.