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Let $f(z)=\sum_{n\geq 0}a_n z^n$ for $z\in D(0,1)$ and $|n!a_n|\leq M$ for all $n\geq 0$ and for some $M\geq 0$. Prove that $f$ is entire.

I feel that this question is missing some hypothesis. Perhaps $f$ is continuous on $\mathbb{C}$?

Edit: I was trying to ask a modified version of another question, but I made a simple mistake. The hypothesis is supposed to read $|n!a_n|\leq M$. In this case, I feel like this is an easy problem, but I'll ask again if I get stuck.

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There's definitely a hypothesis missing. What about $a_n = 1$ for all $n$, for example? – t.b. May 2 '11 at 6:21
There's actually a stronger bound on the coefficients than I realized. I edited it to reflect this. – John C May 2 '11 at 8:23
up vote 3 down vote accepted

Indeed something is missing. You should ask something stronger on the coefficients of the power series.

In particular the following must hold true: \begin{equation} \lim_{n\to\infty}\left( a_n^{1/n}\right) = 0 \end{equation} In this case, the radius of convergence of the power series is infinite and the power series defines an entire function.

Edit: if $|n!a_n| \leq M$, then the equation above holds true.

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This answers to your edited post. If the stronger hypotesis on the coefficients holds, then the limit above exists and is equal to zero. – Raziel May 2 '11 at 8:32
Yup, I realized this. My troubles with this question came from an improper deduction of facts. Thanks for the answer. – John C May 2 '11 at 8:49
Accept it and vote it up then :) – Raziel May 2 '11 at 9:00

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