slightly OT: harmonic series revisited...

[franktaw at netscape.net]

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.
 
The earlier posted proof (looking at 1/p and 1/(p+1)) is correct; the sequence does diverge. 
 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645
 
 
-----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?"
>
> I'm not sure offhand whether P = sum{1/p} diverges, but it doesn't matter.
>...
>
> 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.
>

Nice argument! You only need the second bit because Euler proved that P 
diverges. There is some discussion and references at 
http://mathworld.wolfram.com/PrimeSums.html.

Regards,

Rob.
___________________________________________________
Try the New Netscape Mail Today!
Virtually Spam-Free | More Storage | Import Your Contact List
http://mail.netscape.com
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20051213/565e87a2/attachment-0001.htm>