proof of radius of convergence

According to Cauchy’s root test a power series is absolutely convergent if

\limsup_{k\to\infty}\sqrt[k]{|a_{k}(x-x_{0})^{k}|}=|x-x_{0}|\limsup_{k\to% \infty}\sqrt[k]{|a_{k}|}<1.

This is obviously true if

|x-x_{0}|<\frac{1}{\limsup_{k\to\infty}\sqrt[k]{|a_{k}|}}=\liminf_{k\to\infty}% \frac{1}{\sqrt[k]{|a_{k}|}}=.

In the same way we see that the series is divergent if

|x-x_{0}|>\liminf_{k\to\infty}\frac{1}{\sqrt[k]{|a_{k}|}},

which means that the right hand side is the radius of convergence of the power series. Now from the ratio test we see that the power series is absolutely convergent if

\lim_{k\to\infty}\left|\frac{a_{k+1}(x-x_{0})^{k+1}}{a_{k}(x-x_{0})^{k}}\right% |=|x-x_{0}|\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_{k}}\right|<1.

Again this is true if

|x-x_{0}|<\lim_{k\to\infty}\left|\frac{a_{k}}{a_{k+1}}\right|.

The series is divergent if

|x-x_{0}|>\lim_{k\to\infty}\left|\frac{a_{k}}{a_{k+1}}\right|,

as follows from the ratio test in the same way. So we see that in this way too we can the radius of convergence.

Title	proof of radius of convergence
Canonical name	ProofOfRadiusOfConvergence
Date of creation	2013-03-22 13:21:50
Last modified on	2013-03-22 13:21:50
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	6
Author	mathwizard (128)
Entry type	Proof
Classification	msc 40A30
Classification	msc 30B10