slightly OT: harmonic series revisited...

[Rob Arthan]

On Tuesday 13 Dec 2005 2:45 pm, franktaw at netscape.net wrote:
> Actually, that argument doesn't quite work.  The convergence of the
> original sequence is conditional, not absolute, so you can't arbitrarily
> reorder the terms and draw any conclusions about convergence.

But the reordering is harmless here. Hugo's argument gives diverging lower 
bound for the sum of the a_n from 1 to 6p (since the reordered subsequence 
contains all the negative terms).

Regards,

Rob.

> -----Original Message-----
> From: Rob Arthan <rda at lemma-one.com>
> To: hv at crypt.org; santi_spadaro at virgilio.it <santi_spadaro at virgilio.it>
> Cc: seqfan at ext.jussieu.fr
> Sent: Tue, 13 Dec 2005 13:19:12 +0000
> Subject: Re: slightly OT: harmonic series revisited...
>
> On Tuesday 13 Dec 2005 12:26 pm, hv at crypt.org wrote:
> > "santi_spadaro at virgilio.it" <santi_spadaro at virgilio.it> wrote:
> > :Anybody knows an answer (and a neat way to show that the answer is
> > :true)?
> > :
> > :"Define a_n = 1/n if n is composite and a_n = -(1/n) if n is
> > :prime. Does the series of a_n (sum from n to infinity of a_n) diverges?"
> >
> > If P diverges, consider the set {p, 2p, 3p, 4p, 6p}; this avoids
> > collisions for all odd primes p, and the contribution to A for these 5
> > numbers is (-1 + 1/2 + 1/3 + 1/4 + 1/6)/p = 1/4p, so A > P/4, and so A
> > again diverges.
>