Talk:Nth Root Test

Why $\limsup$ and not $\lim$? --GFauxPas 08:49, 17 February 2012 (EST)


 * To cover common cases where the sequence is bounded but does not converge. For example when all the even terms are zero. Limsup then still makes sense, while lim doesn't. --Lord_Farin 09:25, 17 February 2012 (EST)


 * Got it, thanks --GFauxPas 09:38, 17 February 2012 (EST)

Generalization to $\C$
Anything else needed to generalize this result to $\C$? Abcxyz 17:21, 9 March 2012 (EST)


 * You see, there's still this problem. In order for this proof to be relevant for the complex numbers, all the proofs linked to must be established as to their relevance to the complex plane. We are still some way off this.


 * Also note that the citation in the Sources section at the bottom consists only of the one source which is "real numbers" only (it was the original source work from which the entire Real Analysis section was based). I did start putting some work together for the basics of complex analysis, but in order to rigorously demonstrate the truth of those fundamentals I got sidetracked into topology and never got back to it. So all these complex versions of practically all of the analysis results never really got started.


 * I reiterate, and will continue to do so, that "just" changing "$\R$" to "$\C$" throughout is not what ProofWiki is all about. a) We need to keep the real results as they are, and b) we need to ensure that the complex results are based on previously demonstrated proofs. --prime mover 17:55, 9 March 2012 (EST)


 * Over the past few days, I've checked all (I hope) of the proofs that the article uses and tried to make them relevant to the complex numbers. Could you point out which one is still not rigorously justified? Abcxyz 18:28, 9 March 2012 (EST)


 * For starters: Terms of Bounded Sequence Within Bounds, Comparison Test. --prime mover 18:32, 9 March 2012 (EST)


 * Those two proofs, which only make sense for real sequences, are applied to the real sequence $\langle\left\vert {a_n} \right\vert\rangle$. Maybe I need to make this clearer in the proof? Abcxyz 18:38, 9 March 2012 (EST)


 * Whoops. The comparison test makes sense for complex series as well. The complex case was just not used in the proof. Abcxyz 01:25, 10 March 2012 (EST)


 * Oh wot-EVV-ah. --prime mover 18:43, 9 March 2012 (EST)


 * I hope it's now clear how Terms of Bounded Sequence Within Bounds and Comparison Test are used in the proof. I guess I need to be a little more detailed. Abcxyz 18:58, 9 March 2012 (EST)

Actually, the entire assumption that $\langle {a_n} \rangle$ is a sequence in $\R$ (or $\C$) is not necessary. It is sufficient that the terms of the sequence are elements of a Banach space. Abcxyz 01:37, 10 March 2012 (EST)


 * It appears that this is indeed the case. But certainly the result for arbitrary Banach spaces will need to be separated from this particular case, because it is undesirable to bother people with interest in complex analysis with functional analysis. --Lord_Farin 09:17, 10 March 2012 (EST)