I may be overlooking a small detail, but this question has me stumped. Suppose $f:\Bbb{C}\to \Bbb{C}$ is analytic. Then, by definition of analyticity, for every $z_0\in \Bbb{C}$, there exists a formal power series $S(X)=\sum\limits_{n\geq 0}a_nX^n$ with radius of convergence $\rho_{z_0}>0$ such that for $z\in \Bbb{C}$ with $|z-z_0|<\rho_{z_0}$, we have \begin{align} f(z)=\sum_{n\geq 0}a_n(z-z_0)^n. \end{align} I'm trying to show that since $f$ is analytic on $\Bbb{C}$, it is true that for every $z_0$, we have $\rho_{z_0}=\infty$.
I used to think this was a trivial consequence of the definitions, but now it doesn't seem to be so. I would appreciate it if someone could provide a proof (which hopefully doesn't use any complex integration theory, because I haven't gotten to that point yet).
Intuitively, this makes sense to me because if we fix $z_0$ as above and suppose that $\rho_{z_0}<\infty$, then we choose a point $z_1$ on the boundary of the disc $|z-z_0|\leq \rho_{z_0}$. Since $f$ is analytic on $\Bbb{C}$, we can find another formal series $\sum\limits_{n\geq 0}b_n X^n$ with radius of convergence $\rho_{z_1}>0$ such that for $|z-z_1|<\rho_{z_1}$, we have \begin{align} f(z)=\sum_{n\geq 0}b_n(z-z_1)^n \end{align} Now, there will be some points of overlap in the discs $\Delta_{\rho_{z_0}}(z_0)$ and $\Delta_{\rho_{z_1}}(z_1)$, so maybe this and the principle of uniqueness of analytic continuation somehow provide a contradiction to $\rho_{z_0}$ being finite? I'm not sure how to fill in the details or if this is even the right approach.
I think the easiest way to see this is Cauchy's Integral Formula. Write $\gamma_{a,r}$ for the circular contour of radius $r$ around $a$. Then, by Cauchy's Integral Formula, for $w \in B(a,r)$,
$$ f(w) = \frac{1}{2 \pi i} \int_{\gamma_{a,r}} \frac{f(z)}{z-w} dz$$
Now note that
$$ \frac{1}{z-w} = \frac{1}{z-a} \frac{1}{1-\frac{z-w}{z-a}} = \frac{1}{z-a} \sum_{n=0}^\infty \left( \frac{z-w}{z-a} \right)^n$$
And we get absolute convergence since $|z-w|/|z-a| < 1$ for $z \in \partial B(a,r)$ (and you can get a uniform bound $ < 1$ using compactness). Since we have absolute convergence, you can change the order of integration (note that $f$ is bounded on a compact set), so we get
$$ f(w) = \frac{1}{2 \pi i} \sum_{n=0}^\infty (z-w)^n \int_{\gamma_{a,r}} \left( \frac{f(z)}{(z-a)^{n+1}} \right) dz$$
Since this is true for any $r > 0$, the radius of convergence must be infinite (by identity theorem you can show this expansion is unique).