Given an analytic continuation along $\gamma$ such that $R(t)\equiv \infty$ for some $t$, then $R(s)\equiv \infty$ for each $s\in [0,1]$

Question

Definition: A function element is a pair $(f,U)$ where $U$ is a region and $f$ is an anaytic function on $U$. For a given function element $(f,U)$ define the germ of $f$ at $a$ to be the collection of all function elements $(g,v)$ such that $a\in V$ and $f(z)=g(z)$ for all $z$ in a neighborhood of $a$. Denote the germ by $\left[f\right]_a$.

This allows to define the following

Definition Let $\gamma:[0,1]\rightarrow \mathbb{C}$ be a path and suppose that for each $t\in [0,1]$ there is a function element $(f_t,U_t)$ such that:

1. $\gamma(t)\in U_t$.
2. for each $t\in [0,1]$ there is $\delta >0$ such that $\left|s-t\right|<\delta$ implies $\gamma(s)\in U_t$ and $$\left[f_s\right]_{\gamma(s)}=\left[f_t\right]_{\gamma(s)}.$$

Then $(f_1,U_1)$ is the analytic continuation of $(f_0,U_0)$ along the path $\gamma$.

The above definitions were introduced to contextualize my question:

My Question: Let $\gamma:[0,1]\rightarrow \mathbb{C}$ be a path and let $\left\{(f_t,U_t)\: : 0\leq t\leq 1\right\}$ be an analytic continuation along $\gamma$. For $0\leq t\leq 1$ let $R(t)$ be the radius of convergence of the power series expansion of $f_t$ about $z=\gamma(t)$. Suppose that $R(t)\equiv \infty$ for some value value of $t$. Can I say $R(s)\equiv \infty$ for each $s\in [0,1]$?