# An analytic function whose successive derivatives at a point converge in series

Qual season...

Let $\Omega$ be a region in which $g$ is analytic and suppose that for a $z_0\in\Omega$, $$\sum_{n=0}^\infty g^{(n)}(z_0)<\infty.$$

Prove that $g$ is entire and that the series converges uniformly on discs of the form $|z|<R$. This one has me stumped...

The function $g$ is going to have a convergent Taylor series expansion at all points in $\Omega$, but I'm guessing that the stronger convergence given in the problem statement imply that the Taylor series expanded at a point in our region converges across the plane. Does the M-test play a role here?

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Hint: given $R$, $R^n<n!$ for all sufficiently large $n$. – Chris Eagle Aug 27 '12 at 22:31

For each term in the Taylor expansion about $z_0$ of $g(z)$ we can write $$\left| \frac{g^{(n)}(z_0)}{n!}(z-z_0)^n\right|=\left|g^{(n)}(z_0)\right|\frac{|z-z_0|^n}{n!}$$ Since as Chris Eagle hints, $$\lim_{n\rightarrow\infty}\frac{R^n}{n!} = 0$$ for all $R>0$, we conclude that for any $z\in\mathbb{C}$, the coefficients above must eventually be bounded. Indeed, fix arbitrary $z\in\mathbb{C}$ such that $|z-z_0| < R$ for some $R$. Then there exists $N$ such that $$n>N \implies \frac{|z-z_0|^n}{n!}<1$$ Applying the Weirstrass M-Test, we can now conclude that the series for $g$ converges absolutely and uniformly for all $z$.