slightly OT: harmonic series revisited...

[Rob Arthan]

On 13 Dec 2005, at 17:46, Rob Arthan wrote:

> 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).

And I was having a mental aberration at the time I wrote it. The 
reordering is not obviously harmless and it isn't obvious (at least to 
me) how to get a rigorous argument out of Hugo's estimate.

Apologies,

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.
>>
>
>
>