if we adopte the following definition for a analytic function :
$\textbf{Definition }$
$\bullet$ A function $f$ is called analytic at a point $z_{0} \in \mathbb{C}$ if there exist $r>0$ and a sequence $\{a_n\}_n$ such that $\forall z\in D(z_0,r) : f(z)=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n}$.
$\bullet$ A function is called analytic in an open set $U \subseteq \mathbb{C}$ if it is analytic at each point of $U$.
$\textbf{My question is : }$
if $f$ is analytic on $D(0,R)$
can we find a sequence $\{a_n\}_n$ such that $\forall z\in D(0,R) : f(z)=\sum_{n=0}^{\infty} a_{n}z^{n}$?
$\textbf{My try is : }$
Yes we can, because there is a $r>0$ and a sequence sequence $\{a_n\}_n$ such that $\forall z\in D(0,r) : f(z)=\sum_{n=0}^{\infty} a_{n}z^{n}$.
Then if $r<R$, we can have by Abel lemma another $r'>r$ such that $\forall z\in D(0,r') : f(z)=\sum_{n=0}^{\infty} a_{n}z^{n}$.
And so on.
What what I want is rigorous proof.
