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

On Mon, 26 May 2008 21:55:40 -0700 (PDT), bill <b92057@xxxxxxxxx>
wrote:

On May 26, 3:48 am, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Sun, 25 May 2008 17:44:10 -0700 (PDT), bill <b92...@xxxxxxxxx>
wrote:



On May 24, 5:31 am, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Fri, 23 May 2008 13:26:39 -0700 (PDT), bill <b92...@xxxxxxxxx>
wrote:

On May 22, 5:12 am, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Wed, 21 May 2008 11:48:23 -0700 (PDT), bill <b92...@xxxxxxxxx>
wrote:

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 nthprime.
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.

What in the world does that mean?

_The_ _limit_ of P(n+1)/P(n) _is_ 1.
Another way to say that would be to say that
P(n+1)/P(n) -> 1. But the _limit_ _is_ 1,
it does not "tend to 1". And nobody but
you knows what a "first order limit" is.

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

Huh? That shows that limit is 1 _assuming_ two
things: (i) there are infinitely many twin primes,
(ii) the limit _exists_. But nobody knows whether
(i) is true, and (ii) is not something you're allowed
to assume here, it needs to be proved.

I understand now why you did not bother to explain this to me
earlier..
I would have to be able think like a mathematician to make any sense
out of the above

The reason I didn't say anything about those points earlier
is that they were not relevant to anything you'd said. The
same applies to almost all of the math I know.

All of this has been based on a certain theorem known
as thePrimeNumber Theorem, which says that a certain
other limit is 1. That is a very deep theorem. But if you
simply assume that the limit in question exists the PNT
becomes quite simple.

Way over my head!

Not every sequence _has_ a limit. An example of a seuqence
that does not have a limit is given by one of your other
errorneous comments about the limit of sin(n); that sequence
does not have a limit. A simpler example is this:

1, -1, 1, -1, 1, -1, ...

That sequence does not have a limit (very informally, the
limit can't be 1 because there are infinitely many -1's in
the sequence, the limit can't be -1 for similar reasons,
and the limit certainly can't be anything other than 1 or
-1.)

If pi(n) is the number of primes less than or equal to n
thePrimeNumber Theorem says that the sequence
defined by

pi(n)*ln(n)/n

has limit 1. That theorem is very difficult. But if we
assume that that sequence _has_ a limit it's "easy"
to prove that the limit must be 1; the very hard part
of the proof is showing that the sequence _has_ a limit.

I have forgotten what this has to do with the
limit [P(n+1) / P(n)].

The connection is visible directly above, in the stuff
you quoted.

To spare you the trouble of reading what you posted:

You said that twin primes showed that the limit of
P(n+1)/P(n) was 1. This was wrong for two reasons:
First, it only works if there are infinitely many twin
primes, which is not know. Second, even if we did
inow that there were infinitely many twin primes
this would only prove that the limit of P(n+1)/P(n)
was 1 _if_ we _also_ assume that the limit _exists_.
Examples of sequences that don't have limits
seemed relevant in explaining why we can't just
assume that.



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

David C. Ullrich

David C. Ullrich

David C. Ullrich

What are the criteria for the existence or non existence of a limit?

A sequence a_n has a limit if and only if there exists L such
that for every epsilon > 0 there exists N such that |a_n - L| <
epsilon for every n > N.

The ratio P(n+1)/P(n) must have a minimum value.

Why? I don't see any reason that should be true, and in
fact it seems unlikely.

Is there a unique
minimum value?

What does "unique miniimum value" mean?
If a sequence has a minimum it has only one
minimum.

Oh. Looking below it seems that you meant that a_n
has a unique minimum if there is a unique n such that
a_n is the mimumum of the sequence. Fine.

P(n+1)/P(n) = 1 + (gap_n) /P(n). Since the denominator of the
fraction .
(gap_n) /P(n) is a prime number; it is impossible? for two exactly
equal
fractions to exist. Therefore, it is possible that there is a unique
minimum.

Is the existence of a unique minimum value sufficient for the
existence of
a limit,

Of course not. Do you think the sequence

-2,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,...

has a limit?

or there other considerations?

David C. Ullrich
.

Relevant Pages

  • Re: -- Limit of ratio of consecutive primes = 1 ?
    ... Twin primes quickly push the limit to 1 exactly. ... as thePrimeNumber Theorem, ... That sequence does not have a limit (very informally, ... What are the criteria for the existence or non existence of a limit? ...
    (sci.math)
  • Re: -- Limit of ratio of consecutive primes = 1 ?
    ... Twin primes quickly push the limit to 1 exactly. ... as thePrimeNumber Theorem, ... errorneous comments about the limit of sin; that sequence ... What are the criteria for the existence or non existence of a limit? ...
    (sci.math)
  • Re: -- Limit of ratio of consecutive primes = 1 ?
    ... Twin primes quickly push the limit to 1 exactly. ... as thePrimeNumber Theorem, ... Not every sequence _has_ a limit. ... This was wrong for two reasons: ...
    (sci.math)
  • Re: -- Limit of ratio of consecutive primes = 1 ?
    ... Twin primes quickly push the limit to 1 exactly. ... as thePrimeNumber Theorem, ... Not every sequence _has_ a limit. ... might be useful information. ...
    (sci.math)
  • Re: The renascence of mathematics
    ... only if there is an algoritm to obtain it from the sequence of natural ... That is your definition of existence of sequences of integers. ... the existence of infinitely many twin primes results ... because there is an algorithm to obtain the twin primes. ...
    (sci.math)
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