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• ... All factors 7/5, 23/17,...,p(n+3)/p(n+2),... are larger than 1 so how their products can converge to zero? 5/7*17/23* ... seems to converge to 0, so its
Message 1 of 11 , Dec 2, 2005
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On Fri, 2 Dec 2005, Bob Gilson wrote:

>
>
> "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

All factors 7/5, 23/17,...,p(n+3)/p(n+2),... are larger than 1 so how
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@...
• 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
Message 2 of 11 , Dec 2, 2005
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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

> 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).
>
• ... Hi, I ve calculated very rough estimation of products p1=2/3*11/13*... and p2=7/5*19/17*... up to LIM primes (it is very rough since I used Nextprime from
Message 3 of 11 , Dec 2, 2005
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On Fri, 2 Dec 2005, Bob Gilson wrote:

>
>
> "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
>
Hi,
I've calculated very rough estimation of products p1=2/3*11/13*...
and p2=7/5*19/17*... up to LIM primes (it is very rough since I used
Nextprime from GMP lib):
LIM= 1000 p1=0.131402 p2= 6.980669 p1p2=0.917275
LIM= 2000 p1=0.108394 p2= 8.584098 p1p2=0.930461
LIM= 4000 p1=0.088371 p2=10.297879 p1p2=0.910038
LIM= 8000 p1=0.073380 p2=12.474514 p1p2=0.915374
LIM= 16000 p1=0.060168 p2=15.118286 p1p2=0.909637
LIM= 32000 p1=0.049730 p2=18.303402 p1p2=0.910235
LIM= 64000 p1=0.041075 p2=22.084929 p1p2=0.907139
LIM= 128000 p1=0.033984 p2=26.647697 p1p2=0.905591
I have no time to analyze these series (e.g. quotients
18.303/15.118, etc), but I have another idea:

You investigate

Prod_{n=0} p(n)/p(n+1) * p(n+3)/p(n+2)
(I assume p(0)=2, for convenience). Another formulation is
Prod_{n=0} a_n/a_{n+1}, where
a_n=p(2n)/p(2n+1).
What about log of this product? I'm not good in logs for infinite products
but for finite one we have

log(Prod_{n=0} a_n/a_{n+1}) = Sum_{n=0} log a_n - log a_{n+1},
so it is an alternating series. Is it enough to apply Leibniz criterion?

log a_n= log [ p(2n)/p(2n+1)] = log p(2n) - log p(2n+1).

negative, but what more? |log (a_n)| > |log(a_{n+1}| ????
There are large gaps in primes, so sometimes a_n is quite small, but for
twin primes it tends to 1.

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@...
• Hello Bob, you are not allowed to handle an infinite product like that. Did you already calculate step by step? Your argumentation is as if you would say: 1 -
Message 4 of 11 , Dec 2, 2005
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Hello Bob,

you are not allowed to handle an infinite product like that. Did you
already calculate step by step? Your argumentation is as if you would
say:
1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 +- ... = inf - inf = indefinite,
actually it is pi/4.

Werner
• Hi Jacques and Wojtek, thanks for computing. No chance to reach limit=1? Good idea to do the log, I will try. Werner
Message 5 of 11 , Dec 2, 2005
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Hi Jacques and Wojtek,

thanks for computing. No chance to reach limit=1?
Good idea to do the log, I will try.

Werner
• ... 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
Message 6 of 11 , Dec 2, 2005
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<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
• I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge to 1, oscillating around 1.
Message 7 of 11 , Dec 4, 2005
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I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge
to 1, oscillating around 1.
• From: Werner D. Sand Date: 12/04/05 04:58:46 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] Re: product I suspect the product 2/3 * 7/5 * 11/13 *
Message 8 of 11 , Dec 4, 2005
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From: Werner D. Sand
Date: 12/04/05 04:58:46

I suspect the product 2/3 * 7/5 * 11/13 * 19/17 * ... to converge
to 1, oscillating around 1.

Kermit says:
With primes, anything might be possible.

[Non-text portions of this message have been removed]
• ... This product is not very difficult to compute if you know about the function gamma (the extension of factorials to the complex domain). You are trying to
Message 9 of 11 , Dec 5, 2005
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<mark.underwood@s...> wrote:
>
> --- 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
>

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