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Given a complex function $f(z)=\sum f_n(z)$, I want to show that $f(z)$ is entire.

I usually did this by showing that on every $|z|<N$ the convergence is uniform, so that $f(z)$ converges uniformly on every compact subset of $\mathbb{C}$, hence $f(z)$ is entire.

But I found new arguments: For any $z$, find a neighborhood where the convergence is uniform. Then $f(z)$ is holomorphic on each neighborhood of $z$ so that $f(z)$ is entire.

Is the new argument right? Isn't it easier than the old one? (I think that the new arguments may be easier than the old one, but my textbook uses the old method usually)

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up vote 2 down vote accepted

Yes, that works.

In fact, if you can show every point has a neighborhood where the convergence is uniform, it follows that the sequence converges uniformly on compact sets. You might like to try proving this.

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