## limit = 6?

Expand Messages
• Let p
Message 1 of 3 , Aug 3, 2007
Let p < q be consecutive prime numbers.
Does sum(1/(q-p*ln q), p=2..inf, converge exactly to 6?
• 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+ln (p))
Message 2 of 3 , Aug 8, 2007
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+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?
>
• 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
Message 3 of 3 , Aug 9, 2007
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?

Werner

>
> 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+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)).
>