## Re: Check This!!

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• ... (1+((-1)^((n-1/)2))/n) and p1,p2,p3,... denotes odd primes than ... You are evaluating this infinite product: P = product {p prime}(1+ c(p)/p) where c(p)
Message 1 of 2 , Oct 26, 2003
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<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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