Re: what would euler do with
- From: Matt <matt271829-news@xxxxxxxxxxx>
- Date: Wed, 14 Jan 2009 19:32:55 -0800 (PST)
On Jan 14, 8:26 pm, Matt <matt271829-n...@xxxxxxxxxxx> wrote:
On Jan 14, 7:33 pm, Denis Feldmann <denis.feldmann.sanss...@xxxxxxx>
wrote:
Matt a écrit :> On Jan 13, 11:49 pm, amy666 <tommy1...@xxxxxxxxxxx> wrote:
what would euler do with
log(log(log(oo)))
seriously
log(oo) = 1 + 1/2 + 1/3 + ...
log(log(oo)) = 1/2 + 1/3 + 1/5 + ... ( 1/ primes )
but log(log(log(oo))) = ?
Even though this formulation is imprecise and somewhat peculiar, it
seems to me that there's the bones of an interesting question here.
To stay in the spirit, you would need to find a sequence q_n of
*integers* such that q_n ~ n log n (log log(n)), ie almost as dense as
primes. I dont know of any good one ; do you ?
It is the sum of the reciprocals of the primes <= n that tends to log
(log(n)), right? I'm slightly confused because some references give
results such as Sum k=1^n 1/p_k = log(log(n)) + ..., which to me means
the sum of the reciprocals of the first n primes...
Well, it must be the former, I guess...
Maybe a sequence something like Sum i=2^oo 1/f(i) where f(i) = floor
(sum_j=2^i log(j)*log(log(j))) would work (I'm not 100% confident),
but even if true that's not in this context a "good" sequence by my
reckoning. We need something like "super-primes" that have a pleasing
and relevant construction and the correct asymptotic density.
.
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