I am trying assignment problems in complex analysis and I couldn't deduce the reasoning behind a particular Statement.
Suppose f is holomorphic in an open neighborhood of $z_{0} $ $\epsilon $ $\mathbb{C} $. Given that the series $\sum_{n=0}^{\infty} f^{n} (z_{0}) $ converges absolutely, then how can we conclude that f can be extended to an entire function?
I know of a result that uniform convergence on compact sets implies analyticity. By M-test (Uniform convergence ) I can deduce that f is uniformly convergent locallyand on compact intervel it is entire.
But, can it be extended on whole of $\mathbb{C} $ ? I think it cannot be as convergence is not uniform everywhere. Am I right in my reasoning?
Convergence of $\sum f^{(n)}(z_0)$ implies that $f^{(n)}(z_0) \to 0$. In particular this implies boundedness of $(f^{(n)}(z_0))$ and this is enough to conclude that $\frac {\sum f^{(n)}(z_0)} {n!} (z-z_0)^{n}$ converges for every complex number $z$. This series has radius of convergence $\infty$ and its sum is an entire function.