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• Werner D. Sand wrote: Hi grostoon, my beginning of the product 2/3 * 7/5 * ... is quite correct. I thought I could avoid the ugly
Message 1 of 11 , Dec 2, 2005
"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

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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 2 of 11 , Dec 2, 2005
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 3 of 11 , Dec 2, 2005
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 4 of 11 , Dec 2, 2005
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 5 of 11 , Dec 2, 2005
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 6 of 11 , Dec 2, 2005
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 7 of 11 , Dec 2, 2005
<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 8 of 11 , Dec 4, 2005
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 9 of 11 , Dec 4, 2005
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 10 of 11 , Dec 5, 2005
<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)