- On Fri, 2 Dec 2005, Bob Gilson wrote:

>

All factors 7/5, 23/17,...,p(n+3)/p(n+2),... are larger than 1 so how

>

> "Werner D. Sand" <Theo.3.1415@...> wrote: Hi grostoon,

>

> my beginning of the product 2/3 * 7/5 * ... is quite correct.

> I thought I could avoid the ugly formula:

> product(p(n)/p(n+1)) * (p(n+3)/p(n+2)).

> It's an `alternating' product. Otherwise the limit would be simply = 0

> (I suspect).

>

> Werner

>

> Since 2/3*11/13*29/31 converges to zero (I suspect),

> surely your 7/5*23/17*41/37*zero, will also converge to zero

> Cheers

>

> Bob

their products can converge to zero?

5/7*17/23* ... seems to converge to 0, so its inverse to tends to

infinity, and Werner's product is of type 0*infinty, I suspect.

But there are also other possiblilites, e.g.

2; (1.5)*(1,5); (1.33)*(1.33)*(1.33); ...

All numbers are larger than 1 (and 1+1/n > 1 for n<infinty) but their

product is..., you know, I'm sure.

Wojtek

===============================================

Wojciech Florek (WsF)

Adam Mickiewicz University, Faculty of Physics

ul. Umultowska 85, 61-614 Poznan, Poland

Phone: (++48-61) 8295033 fax: (++48-61) 8257758

email: Wojciech.Florek@... - --- In primenumbers@yahoogroups.com, "Mark Underwood"

<mark.underwood@s...> wrote:>

This product is not very difficult to compute if you know about the

> --- In primenumbers@yahoogroups.com, "Jacques Tramu"

> <jacques.tramu@e...> wrote:

> >

> > I have computed the product and found

> > n= 200 000 000 : 0.9048530178

> > n= 1 000 000 0000 : 0.9048042986

> > n= 1 500 000 000 : 0.9047987573

> >

>

> I was just averaging some figures around the n = 40,000 area and it

> worked out to be around .9105. Looking a Jacque's findings the

> product seems to be ever so slowly shrinking.

>

> Contrast this the product 1/2 * 4/3 * 5/6 * 8/7 * ....

>

> and this appears to be converging to .599070...

>

> Mark

>

function gamma (the extension of factorials to the complex domain).

You are trying to find:

k=inf (4k+1)(4k+4) k=inf 2

Prod ------------ = Prod 1 - ------------

k=0 (4k+2)(4k+3) k=0 (4k+2)(4k+3)

This is clearly convergent. But what is the limit?

k=n (4k+1)(4k+4) k=n (k+1/4)(k+1)

Prod ------------ = Prod -------------- =

k=0 (4k+2)(4k+3) k=0 (k+2/4)(k+3/4)

gamma(k+5/4) gamma(k+2) gamma(2/4) gamma(3/4)

= ------------------------- ---------------------

gamma(k+6/4) gamma(k+7/4) gamma(1/4) gamma(1)

= A(k) x B

where A(k) is the first fraction and B the second.

We are interested in the value of A(k) as k->inf.

Fortunately it turns out that the limit is 1. This can be seen by

using Stirling approximation.

Since gamma(1) = 1 and gamma(1/2) = sqrt(pi) we finally get:

k=inf (4k+1)(4k+4) sqrt(pi) * gamma(3/4)

Prod ------------ = ----------------------

k=0 (4k+2)(4k+3) gamma(1/4)

This is about 0.5990701173677961037199612

Best regards,

Dario Alejandro Alpern

Buenos Aires - Argentina

http://www.alpertron.com.ar/ENGLISH.HTM