Suppose we have a complex-analytic function $f$ given by \begin{equation} \tag{1} f(z) = \sum_{n=0}^{\infty}\, a_n \mspace{.5mu} (z-z_0)^n, \qquad a_n \in \mathbb{C}, \end{equation} where $z_0$ is a point in the complex plane. It is then straightforward to show that $f$ must be absolutely convergent in an open disk $D$ centred at $z_0$ and extending to at least one singularity $s_0$ at the circumference $\partial D$.
Since $f$ is analytic, we can write down its Taylor series expansion about any point $z_1\in D$. This amounts to introducing the function \begin{equation} \tag{2} f_1(z) = \sum_{n=0}^{\infty} \, b_n \mspace{.5mu} (z-z_1)^n , \qquad b_n=\frac{f^{(n)}(z_1)}{n!} , \end{equation} where $f^{(n)}(z_1)$ is the $n$-th derivative of $f$ at $z_1$. Again we see that this new function must be absolutely convergent on some open disk $D_1$, centred at $z_1$ rather than $z_0$. Using Cauchy's integral formula, it is possible to show that the radius of $D_1$ — which we will call $r_1$ — must be \begin{equation} \tag{3} r_1 \geq |s_0-z_0|-|z_1-z_0|, \end{equation} since $f_1(z)=f(z)$ within a disk $D_1'$ strictly contained in $D$ and centred on $z_1$.
Given this setup, my question is simply this: Assuming $s_0$ is the closest non-removable singularity of $f$ to $z_1$, what additional requirements can we put on $f$ in order to be able to show that $r_1\geq|s_0-z_1|$, and how would we go about doing so? Or in other words, when will the Taylor series $f_1$ analytically continue $f$ to $D_1 \nsubseteq D$?
Edit: The original wording was too restrictive; exchanged a "must" with a "can" to make my intent clearer. I would be happy with sufficient (non-trivial) requirements even if they are not, in fact, necessary. My main goal is to obtain some justification for the assertion, seen here and elsewhere, that "a Taylor series of $f$ about $z_1$ will converge on a disk extending to the nearest singularity of $f$".