# Showing a function analytic on open unit disk can be analytically extended

Suppose that $f$ is analytic on the open unit disk such that there exists a constant $M$ with $|f^k(0)| \leq k^4M^k$ for all $k \geq 0$. Show that $f$ can be extended to be analytic on $\mathbb{C}$.

I'm not sure how to apply the basic facts about analytic continuation that I know to this problem. Are there any results that might be helpful here? A hint would be appreciated.

Context: I'm studying for a qual, so just a hint at this point would be most helpful.

Hint: Look at the Maclaurin series for $f$. What can you say about its radius of convergence?
• So with the given bound we get $|f(z)| \leq \Sigma |f^n(0)z^n|/n! \leq \Sigma |n^4M^nz^n|/n! \leq \Sigma |(zm)^nn^4|/n!$, but I don't see why this last series converges. And thank you for the hint, that's exactly what I was looking for. – user19817 Aug 20 '15 at 23:32
• Surely you know some suitable tests for convergence? Use for example the ratio test or the root test to find the values of $z$ for which the series on the right hand side converges. – mrf Aug 21 '15 at 7:09