Re: -- Limit of ratio of consecutive primes = 1 ?

On May 21, 9:34 am, David W. Cantrell <DWCantr...@xxxxxxxxxxx> wrote:
Ray Johnstone <r...@xxxxxxxxxxxx> wrote:
On 07 May 2008 23:32:48 GMT, David W. Cantrell
<DWCantr...@xxxxxxxxxxx> wrote:

Let p(n) denote the nth prime.
It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = 1.
Is that correct? If so, how can it be proven?

Does it perhaps depend on some result (unfamiliar to me) concerning how
big, for given n, the gap between p(n) and p(n + 1) can be?

So is there a limit? There seems to be no answer in this long thread.
If there is one it seems to be between 1 and 2.

The limit is 1. That was proven early in the thread.

David

The first order limit is => 1. Since P(n+1) cannot be less that P(n) ;
the limit
must be => 1. . .
..
Twin primes quickly push the limit to 1 exactly.

P(n+1) is asymptotic to (n+1) * ln (n+1) and P(n) is asymptotic to n*
ln (n)
For significantly large 'n', these two curves are virtually
identical. This
might be useful information.

Bill J



















































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