--- In

primenumbers@yahoogroups.com, Mehmet "Ã‡oker"

<netmaster_64@y...> wrote:

> Hi

> i proved this i think this is not difficult but not easy...let f(n)=

(1+((-1)^((n-1/)2))/n) and p1,p2,p3,... denotes odd primes than

> f(p1)f(p2)f(p3).....=2/pi

>

>

You are evaluating this infinite product:

P = product {p prime}(1+ c(p)/p)

where c(p) is the following "character" [usually printed as a Greek chi]:-

c(p) = 1 if p = 1 mod 4

-1 if p = 3 mod 4

0 otherwise

P = {product p prime >= 3}(1 - [c(p)/p]^2)/(1 - c(p)/p)

= [{product p prime >= 3} (1 - 1/p^2)] / [{product p prime>=3}(1 - c(p)/p)]

= [(3/4) * zeta(2)]^(-1) / [1 - 1/3 + 1/5 -...]^(-1)

= (pi/4) / (3/4) * ((pi^2)/6),

using Leibnitz's formula;

and this simplifies to 2/pi, as you said.

[At last I got it right!]

Mike

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