slightly OT: harmonic series revisited...

[Richard Guy]

Have come into this rather late, so apologize
if this is old hat.  The  nth  partial sum is
c1 \ln n  -  c2 \ln\ln n,   regardless of
ordering, which can only affect  c1  and  c2,
for which trivial bounds suffice.

R.

On Tue, 13 Dec 2005, Rob Arthan wrote:

>
> 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.
>>> 
>> 
>> 
>> 
>