Editing Root test (section)

== Proof ==
The proof of the convergence of a series Σ''a''<sub>''n''</sub> is an application of the [[Direct comparison test|comparison test]].

If for all ''n'' ≥ ''N'' (''N'' some fixed [[natural number]]) we have <math>\sqrt[n]{|a_n|} \le k < 1</math>, then <math>|a_n| \le k^n < 1</math>. Since the [[geometric series]] <math>\sum_{n=N}^\infty k^n</math> converges so does <math>\sum_{n=N}^\infty |a_n|</math> by the comparison test. Hence Σ''a''<sub>''n''</sub> converges absolutely. 

If <math>\sqrt[n]{|a_n|} > 1</math> for infinitely many ''n'', then ''a''<sub>''n''</sub> fails to converge to 0, hence the series is divergent.

'''Proof of corollary''': 
For a power series Σ''a''<sub>''n''</sub> = Σ''c''<sub>''n''</sub>(''z''&nbsp;&minus;&nbsp;''p'')<sup>''n''</sup>, we see by the above that the series converges if there exists an ''N'' such that for all ''n'' ≥ ''N'' we have

:<math>\sqrt[n]{|a_n|} = \sqrt[n]{|c_n(z - p)^n|} < 1,</math>

equivalent to

:<math>\sqrt[n]{|c_n|}\cdot|z - p| < 1</math>

for all ''n'' ≥ ''N'', which implies that in order for the series to converge we must have <math>|z - p| < 1/\sqrt[n]{|c_n|}</math> for all sufficiently large ''n''. This is equivalent to saying

:<math>|z - p| < 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}},</math>

so <math>R \le 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}.</math> Now the only other place where convergence is possible is when

:<math>\sqrt[n]{|a_n|} = \sqrt[n]{|c_n(z - p)^n|} = 1,</math>

(since points &gt; 1 will diverge) and this will not change the radius of convergence since these are just the points lying on the boundary of the interval or disc, so

:<math>R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}.</math>