I am always using the following result but I do not know why it is true. So: How to prove the following statement:
Suppose the complex power series $\sum_{n = 0}^\infty a_n(z-z_0)^n$ has radius of convergence $R > 0$. Then the function $f: B_R(z_0) \to \mathbb C$ defined by
\begin{align*}
f(z) := \sum_{n = 0}^\infty a_n (z-z_0)^n
\end{align*}
is differentiable in $B_R(z_0)$ and for any $z \in B_R(z_0)$ the derivative is given by the formula
\begin{align*}
f'(z) = \sum_{n = 1}^\infty na_n(z-z_0)^{n-1}.
\end{align*}
Thanks in advance for explanations.
Best Answer
It is a result of Abel that says that the power series converges in the ball $B_R(z_0)$. Moreover since there is uniform convergence in every concentric sub-ball, $B_r (z_0)$ with $r<R$ you can use term by term differentiation. This shows that the series is differentiable in $B_R(z_0)$, since any of its points lies in such a concentric sub-ball.