For a real sequence $(a_n)_{n\in\mathbb{N}}$ I am considering the generating function $s(z) = \sum_{n=0}^{\infty} a_n z^n$ and by some calculations I find that it has a representation $f:\mathbb{C}\to \mathbb{C}$ where $f$ has poles ${z_1,z_2,...}$ on the real line with $0< z_1< z_2<...$ and $f$ is analytic for $|z|<|z_1|$. Now I know that since $\sum_{n=0}^{\infty} a_n z^n$ is analytic within its radius of convergence and $f$ is analytic for $|z|<|z_1|$ they both coincide within the radius of convergence of the series $s$ by uniqueness of analytic continuations.
But can I also conclude that the radius of convergence of $s$ is just $|z_1|$ since if it were smaller there would be a pole somewhere in the disk $|z|<|z_1|$ which would contradict the equality with $f$ and the fact that $f$ is analytic?