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The radius of convergence of any power series can be found by simply using the root test, ratio test etc.

I am confused as to how to find the radius of convergence for an analytic $f$ such as

$f(z)=\frac{4}{(z-1)(z+3)}$.

I can't imagine that I would have to find the power series representation of this, find the closed form, and then use one of the convergence tests. I am fairly certain that the radius of convergence would have to do with the singularities at $1$ and $-3$, however, I can't find a formula for the radius of convergence..

Masan
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    Radius of convergence of an analytic function doesn't really exist as a concept: an analytic function has a domain on which it is analytic, and its power series around a point will have a disk of some radius on which it converges, but for a function there's nothing to converge or diverge, hence no radius of convergence. I also want to point out that if you expand your function in power series, then compute the closed form of the power series, you get back the original expression. This also really isn't a computational science question, please consider moving it to mathematics stack exchange. – Kirill Feb 11 '15 at 11:44
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    I'm voting to close this question as off-topic because it is a pure mathematics question, hence more suitable to Math.SE. – Kirill Feb 11 '15 at 11:44
  • The short answer is, radius equals distance to nearest singularity. Of course you have to specify the point around which the power series is expanded (this need not be the origin). – hardmath Feb 11 '15 at 16:49
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    ...and don't forget to check the whole complex plane for singularities, even if you're only interested in real values. – Christian Clason Feb 11 '15 at 18:15

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