# uniform convergence of sequences of analytic functions

Suppose you have a sequence $f_n$ of analytic functions on an open set $\Omega$, which converges uniformly on compact subsets of $\Omega$. Can you conclude that $f_n$ converges uniformly on the whole open $\Omega$?

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No: if $\Omega:=\{z,|z|<1\}$, and $f_n(z):=z^n$, this sequence converges uniformly to $0$ on compact sets (because such a set is contained in $B(0,r),r<1$) but not on $\Omega$ as $f_n(1-n^{—1)}\to e$.
and the simple convergence? if we have uniform convergence on compacts, do we have convergence on $\Omega$? –  Federica Maggioni Jan 5 '13 at 13:03
Yes, we have simple/pointwise convergence as $\{z\}$ is compact for all $z\in\Omega$. –  Davide Giraudo Jan 5 '13 at 13:11
Another example: the series $\displaystyle\sum \frac{z^k}{k!}$ converges uniformly on compact subsets of $\Omega$, but not on the whole $\mathbb{C}$, as $$e^{n} - \sum_{k=0}^{n-1}\frac{n^k}{k!} \geq \frac{n^n}{n!}$$