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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|. 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 for all sufficiently large $n$.2012-08-27

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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 $z$.

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    Thank you so much Sir! It is clear now..2017-07-18