Talk:Convergence of P-Series

Should this result be merged with Sum of Infinite Geometric Progression? Also, is it worth specifying the domain of the series (i.e. presumably the complex numbers)? Also, what category would you put it in - Complex Analysis? --Matt Westwood 22:14, 23 April 2009 (UTC)

I would NOT merge it with sum of infinite geometric progression. That index of the summation in that case is found in the exponent, and in this is found in the base. As for the category, I would say Series and maybe Complex Analysis. We also might want to make Category:Convergence --Cynic (talk) 22:21, 23 April 2009 (UTC)

Agh, no you're right. Mind, I'm sure we've done this one before. Lemme go and think about it ... --Matt Westwood 22:26, 23 April 2009 (UTC)


 * This is particuarly related to the Zeta function, and so it's definitely complex analysis, especially since it pertains to the right half-plane $\Re(z)>1 \ $. Also, this is not a geometric progression, nor do we say anything about what the series actually converges TO, merely that it does.  Zelmerszoetrop 22:29, 23 April 2009 (UTC)

Just that I'm sure there's a result out there somewhere that does this in the real domain. Unless I dreamed it. ;-) No worries. --Matt Westwood 05:23, 24 April 2009 (UTC)


 * You just might have. I searched just looked at every page on ProofWiki with the word "series" somewhere in it, and the only thing that might be construed as this is the Basel Problem page, which is the p-series for p=2, aka $\zeta(2) \ $. Zelmerszoetrop 18:30, 24 April 2009 (UTC)

AHA! That was the baby: Sum of Reciprocals is Divergent (see the opening comment on this page). For some reason I thought there was a proof out there which had generalized it. --Matt Westwood 22:43, 24 April 2009 (UTC)

Lemma
I feel the lemma is important enough to be a theorem of its own right. And if it gets its own page, it might as well be extended to say that if $0 < x \le 1$ then the improper integral diverges. It seems unfair to label him a lemma. --GFauxPas 07:23, 5 February 2012 (EST)
 * I'm sorry, what lemma? --prime mover 09:38, 5 February 2012 (EST)
 * err sorry,
 * $\displaystyle \int_1^{\to \infty} \frac {\mathrm dt}{t^x}$ converges for $x > 1$. --GFauxPas 09:40, 5 February 2012 (EST)
 * Okay, can think about it ... --prime mover 09:47, 5 February 2012 (EST)

Would it be a good idea to somehow emphasize that this proof only shows absolute divergence of the complex series on 0<Re(p)<1? The title emphasizes that for convergence, but it may not be as clear on divergence. The absolute divergence is not nearly as interesting as whether the series converges conditionally over that interval. That question is not relevant for the real series, but is quite relevant for the complex series.

Why is it necessary to redo the integral test? Isn't the fact that the absolute value of the complex terms is the same as the terms of the real series sufficient to prove both conclusions with a simple comparison test?