proof of radius of convergence of a complex function

Without loss of generality, it may be assumed that $z_{0}=0$ .

Let $c_{n}$ denote the coefficient of the $n$ -th term in the Taylor series of $f$ about $0$ . Let $r$ be a real number such that $0<r<R$ . Then $c_{n}$ may be expressed as an integral using the Cauchy integral formula.

$c_{n}={1\over 2\pi i}\oint_{|z|=r}{f(z)\over z^{n+1}}\,dz={1\over 2\pi r^{n}}% \int_{-\pi}^{+\pi}e^{-n\theta}f(re^{i\theta})\,d\theta$

Since $f$ is analytic, it is also continuous. Since a continuous function on a compact set is bounded, $|f|<B$ for some constant $B>0$ on the circle $|z|=r$ . Hence, we have

$|c_{n}|={1\over 2\pi r^{n}}\left|\int_{-\pi}^{+\pi}e^{-n\theta}f(re^{i\theta})% \,d\theta\right|\leq{1\over 2\pi r^{n}}\int_{-\pi}^{+\pi}|e^{-n\theta}f(re^{i% \theta})|\,d\theta\leq{1\over 2\pi r^{n}}\int_{-\pi}^{+\pi}Bd\theta={B\over r^% {n}}$

Consequently, $\sqrt[n]{c_{n}}\leq\sqrt[n]{B}/r$ . Since $\lim_{n\to\infty}\sqrt[n]{B}=1$ , the radius of convergence must be greater than or equal to $r$ . Since this is true for all $r<R$ , it follows that the radius of convergence is greater than or equal to $R$ .

Title	proof of radius of convergence of a complex function
Canonical name	ProofOfRadiusOfConvergenceOfAComplexFunction
Date of creation	2013-03-22 14:40:35
Last modified on	2013-03-22 14:40:35
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	9
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 30B10