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[parent] proof of radius of convergence (Proof)

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

$\displaystyle \limsup_{k\to\infty}\sqrt[k]{\vert a_k(x-x_0)^k\vert}=\vert x-x_0\vert\limsup_{k\to\infty}\sqrt[k]{\vert a_k\vert}<1.$
This is obviously true if
$\displaystyle \vert x-x_0\vert<\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\vert a_k\vert}}=\liminf_{k\to\infty}\frac{1}{\sqrt[k]{\vert a_k\vert}}=.$
In the same way we see that the series is divergent if
$\displaystyle \vert x-x_0\vert>\liminf_{k\to\infty}\frac{1}{\sqrt[k]{\vert a_k\vert}},$
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
$\displaystyle \lim_{k\to\infty}\left\vert\frac{a_{k+1}(x-x_0)^{k+1}}{a_k(x-x_0)... ...t=\vert x-x_0\vert \lim_{k\to\infty}\left\vert\frac{a_{k+1}}{a_k}\right\vert<1.$
Again this is true if
$\displaystyle \vert x-x_0\vert<\lim_{k\to\infty}\left\vert\frac{a_k}{a_{k+1}}\right\vert.$
The series is divergent if
$\displaystyle \vert x-x_0\vert>\lim_{k\to\infty}\left\vert\frac{a_k}{a_{k+1}}\right\vert,$
as follows from the ratio test in the same way. So we see that in this way too we can calculate the radius of convergence.



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Cross-references: ratio test, radius of convergence, right hand side, divergent, series, absolutely convergent, power series, Cauchy's root test

This is version 3 of proof of radius of convergence, born on 2003-01-10, modified 2005-07-04.
Object id is 3890, canonical name is ProofOfRadiusOfConvergence.
Accessed 4075 times total.

Classification:
AMS MSC30B10 (Functions of a complex variable :: Series expansions :: Power series )
 40A30 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences of functions)

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