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Given a meromorphic function on $\mathbb{C}$, is the radius of convergence in a regular point exactly the distance to the closest pole?

As Robert Israel points out in his answer, that this is of course an upper bound by the Cauchy-Hadamard principle.

Theo Buehler in the comments gives a refernce for the non obvious direction: Remmert, Theory of complex functions, Chapter 7, §3, p.210ff (p. 164ff of my old German edition). Look for Cauchy-Taylor.

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    I can not even see why the radius $R = \infty$, if $f$ is entire.2011-06-22
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    There is a chapter in Wilf's book *generatingfunctionology* about the relation between the asymptotics of the coefficients of a power series and the nature of the singularities on the boundary of the circle of convergence of the function.2011-06-22
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    Dear GEdgar, you certainly mean e.g. Fatou's theorem or other summability conditions on the $a_n$'s. This comment might better fit to this question of mine: http://math.stackexchange.com/questions/46898/power-series-without-analytic-continuation, but I do not see how this should relate.2011-06-22
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    I suggest that you take a look at Remmert's book I recommended in another question of yours (or any other book on complex analysis that deserves its title). This is really basic material that is covered everywhere and I find it a bit too much to ask for a proof here.2011-06-22
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    By the way, the things you need to understand are the lemmas on exchange of summation and integration and the development $$\frac{1}{(1-q)^{k+1}} = \sum_{n=k}^{\infty} {n \choose k} q^{n-k}$$ valid inside the unit disk plus basics on Möbius transformations.2011-06-22
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    I have to ask: How can one know about the existence of such concepts as the spectrum of the Laplace operator on Riemann surfaces, Phragmén-Lindelöf, and Hardy spaces and not know about this?2011-06-22
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    @Theo Buehler: Sorry, if my question sounded a little bit rough and demanding in the beginning, I edited the question acordingly and now ask explicitely for a reference. I read the whole chapter about power series in Remmert no answer to my question, though.2011-06-24
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    As I said, it's in any book on complex analysis. E.g. Remmert, *Theory of complex functions*, Chapter 7, §3, p.210ff (p. 164ff of my old German edition). Look for Cauchy-Taylor.2011-06-24
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    Perfect answer, thank you very much. Why did you not provide it directly, and were teasing me so much?;) If you care to pos the answer, I will accept it immediately.2011-06-24

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Yes, it is (that should be "pole", not "pol"). If $r$ is the distance from $z_0$ to the closest pole, the function is analytic in $\{z: |z - z_0| < r\}$, so the radius of convergence is at least $r$, but it can't be more than $r$ because $|f(z)| \to \infty$ as $z$ approaches that pole.

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    This is, of course, the correct answer. But one might want to give some reminder or reference to the proof, as this is certainly not obvious (e.g. it fails over $\mathbb{R}$ and even more dramatically over $\mathbb{C}_p$, where the principle of analytic continuation is violated). I seem to recall that this has something to do with the Cauchy estimates? (Well, it's a safe guess....)2011-06-22
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    Cauchy's Estimate is one way but more typically directly using Cauchy's formula and then expanding the $1/(z-w)$ term in $$f(w)=1/2\pi i\int_C f(z)/(z-w) dz$$.2011-06-22
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    The basic theorem, that any function analytic in an open disk has a Taylor series (about the centre of the disk) that converges in that disk, should appear in any introductory Complex Variables text. Caution: the result is often quoted as something like "the radius of convergence is the distance from the centre to the nearest singularity". The problem with this is that if you start with a given function $f$, the sum of the Taylor series agrees with $f$ in a neighbourhood of the centre, but might be different at some points within the circle of convergence (see next comment)2011-06-24
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    For example, consider $f(z) = \sqrt{z}$ (using the principal branch, which has a branch cut on the negative real axis) and $a = -1+i$. The largest disk around $a$ on which $f$ is analytic has radius 1, but the radius of convergence of the Taylor series is $\sqrt{2}$. The sum of the Taylor series is a branch of $\sqrt{z}$, but not the principal branch.2011-06-24
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    Interesting counterexample2011-06-25
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    @Robert Israel: But when you say "at least", doesn't that mean greater than or equal to?2014-07-20
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    @Libertron That is indeed what it means. The radius of convergence $\ge r$ because the function is analytic in $\{z: |z - z_0| < r\}$, and $\le r$ because $|f(z)| \to \infty$ as $z$ approaches the pole, so you conclude it is exactly $r$.2014-07-21