Re: limit = 6?
- That's right. One could also say: q ~ p, for p/q ~ 1. Then 1/(q-p*ln
q) ~ 1/(p-p*ln p) and int(1/(x-x*ln x)) = -ln (ln x - 1) -> -inf.
Thanks. Can you nevertheless find out p when sum(1/(q-p*ln(q)) < 6
for the first time?
--- In email@example.com, "Adam" <a_math_guy@...> wrote:
> Does it even converge?
> If gap size at p is about ln(p) then (with all equalities
> being "approximately equal to") q=p+ln(p) so q-p*ln(q)=p+ln(p)-p*ln
> (p)) is on the order of p*ln(p) and the improper integral of 1/[x*ln
> (x)] diverges.
> It should it diverge to -oo about as fast as ln(ln(p)).
> > Let p < q be consecutive prime numbers.
> > Does sum(1/(q-p*ln q), p=2..inf, converge exactly to 6?