Expand Messages
• ... 27/77, then 27/91, etc - also converges. But does the product converge to zero, or to a small but finite number? ... It converges to zero as well as the
Message 1 of 8 , Apr 2, 1938
> Clearly the terms converge to 1, and the product - 3/5, then 3/7, then
27/77, then 27/91, etc - also converges. But does the product converge to
zero, or to a small but finite number?
>
> I have been unable to answer him definitively. Can anyone else here do so?

It converges to zero as well as the product of (p-1)/p.

Best regards,

Andrey
• Thanks Andrey. Is there a simple proof of this? Tim
Message 2 of 8 , Apr 2, 1938
Thanks Andrey. Is there a simple proof of this?

Tim

--- In primenumbers@yahoogroups.com, Andrey Kulsha <Andrey_601@...> wrote:
>
> > Clearly the terms converge to 1, and the product - 3/5, then 3/7, then
> 27/77, then 27/91, etc - also converges. But does the product converge to
> zero, or to a small but finite number?
> >
> > I have been unable to answer him definitively. Can anyone else here do so?
>
> It converges to zero as well as the product of (p-1)/p.
>
> Best regards,
>
> Andrey
>
• ... See Hardy and Wright Sections 22.7 and 22.8 for the Mertens product prod(p
Message 3 of 8 , Apr 3, 1938
"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
• ... So let s work out the constant, say K, in prod(2
Message 4 of 8 , Apr 3, 1938

> 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
• ... Hmm.... And I thought I had proven that the product converged to zero. David, what is wrong with my Proof below which I had already sent to Tim? Kermit
Message 5 of 8 , Apr 4, 1938
>
>
>
> Messages in this topic (5)
> ________________________________________________________________________
> 1b. Re: product convergence
> Date: Tue May 4, 2010 1:17 pm ((PDT))
>
>
>
>
>
>> 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
>
>
>

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

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
>
• ... Now let x tend to infinity and I think that you will see that K/log(x)^2 vanishes, not so :-? David
Message 6 of 8 , Apr 4, 1938
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
> K = 0.832429065661945278030805943531465575045445318077417053240894...

Now let x tend to infinity and I think that you
will see that K/log(x)^2 vanishes, not so :-?

David
• ... 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
Message 7 of 8 , Apr 7, 1938
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
Your message has been successfully submitted and would be delivered to recipients shortly.