- --- In primenumbers@yahoogroups.com,

"tsr21" <timro21@...> wrote:

> Is there a simple proof of this?

See Hardy and Wright Sections 22.7 and 22.8

for the Mertens product

prod(p<x, (p-1)/p) ~ exp(-Euler)/log(x)

where the product runs over primes.

As Andrey remarked, your product tends to zero faster:

prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

David - --- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

So let's work out the constant, say K, in

prod(2<p<x, (p-2)/p) ~ K/log(x)^2

We should use the twin-prime constant

C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...

and then use the square of Mertens' formula,

remembering that the latter includes p = 2.

K = C2*(exp(-Euler)*2)^2 =

0.832429065661945278030805943531465575045445318077417053240894...

Sanity check:

default(primelimit,10^8);

\p5

P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);

0.83242

Looks OK to me...

David - primenumbers@yahoogroups.com wrote:
>

Hmm.... And I thought I had proven that the product converged to zero.

>

>

> Messages in this topic (5)

> ________________________________________________________________________

> 1b. Re: product convergence

> Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst

> Date: Tue May 4, 2010 1:17 pm ((PDT))

>

>

>

> --- In primenumbers@yahoogroups.com,

> "djbroadhurst" <d.broadhurst@...> wrote:

>

>

>> prod(2<p<x, (p-2)/p) = O(1/log(x)^2)

>>

>

> So let's work out the constant, say K, in

>

> prod(2<p<x, (p-2)/p) ~ K/log(x)^2

>

> We should use the twin-prime constant

>

> C2 = prod (2<p, p*(p-2)/(p-1)^2) = 0.6601618158...

>

> and then use the square of Mertens' formula,

> remembering that the latter includes p = 2.

>

> K = C2*(exp(-Euler)*2)^2 =

> 0.832429065661945278030805943531465575045445318077417053240894...

>

> Sanity check:

>

> default(primelimit,10^8);

> \p5

> P=1.;x=10^8;forprime(p=3,x,P*=1-2/p);print(P*log(x)^2);

>

> 0.83242

>

> Looks OK to me...

>

> David

>

>

>

David, what is wrong with my "Proof" below which I had already sent to Tim?

Kermit

> *****************************************************

>

>

>

> Hello Tim.

>

>

>

> http://en.wikipedia.org/wiki/Infinite_product

>

>

>

> The infinite product 3/5 5/7 9/11 ...

>

> converges to zero.

>

>

>

> would converge to a positive number between 0 and 1 only if

>

> the infinite product

>

> 5/3 7/5 11/9 ..... converged to a positive number > 1.

>

>

>

> 5/3 7/5 11/9 ..... = (1 + 2/3) (1 + 2/5) (1 + 2/9) (1 + 2/11) (1 +

> 2/15) (1 + 2/17) ...

>

>

> 1 + 2/3 + 2/5 + 2/9 + 2 / 11 + .... =< (1 + 2/3) (1 + 2/5) (1 + 2/9)

> (1 + 2/11) (1 + 2/15) (1 + 2/17) ... =< exp( 2/3 + 2/5 + 2/9 + 2/11

> - --- In primenumbers@yahoogroups.com,

Kermit Rose <kermit@...> wrote:

> I thought I had proven that the product converged to zero.

It does. I wrote:

> prod(2<p<x, (p-2)/p) ~ K/log(x)^2

Now let x tend to infinity and I think that you

> K = 0.832429065661945278030805943531465575045445318077417053240894...

will see that K/log(x)^2 vanishes, not so :-?

David - --- In primenumbers@yahoogroups.com,

Kermit Rose <kermit@...> wrote:

> what is wrong with my "Proof"

It began OK and then fizzled out.

Here is an elementary proof.

Proposition: The product prod(p>2, (p-2)/p) over primes p>2 vanishes.

Proof: It suffices to show that the contrary proposition is absurd.

Suppose that prod(p>2, (p-2)/p) did not vanish.

Then, by taking logs, we would conclude that

sum(p>2, log(p) - log(p-2)) is finite. But

log(p) - log(p-2) > 2/p. Thus

sum(p>2, 1/p) would also be finite.

Yet that is easily proven to be absurd:

http://primes.utm.edu/infinity.shtml#punchline

Hence the product prod(p>2, (p-2)/p) does indeed vanish.

Comment: By the same argument, it follows that

the product prod(p>x, (p-x)/p) over primes p > x

vanishes for all real x > 0.

David