Hi, wouter meeussen wrote:

or possibly a true identity (?) negaodd posieven Catalan == {0.9992596423851822204, 0.9999976893644401925, 0.9999999241377495063, 0.9999999985217359313, 0.9999999999896304167}

with "negaodd" defined as sum(p prime, p mod 4=1, 1-1/p^2) and "posieven" defined as sum(p prime, p mod 4=3, 1+1/p^2)

negaodd= Table[N[ Product[p=Prime[q];w=If[Mod[p+1,4]===2,p^2, Infinity];1-1/w,{q,1,10^w}] ,20], {w,5}] posieven= Table[N[ Product[p=Prime[q];w=If[Mod[p+1,4]===0,p^2, Infinity];1+1/w,{q,1,10^w}] ,20], {w,5}]

nothing under http://mathworld.wolfram.com/CatalansConstant.html

I have found a simple proof for this identity (!) by myself, and some related identities, but I don't know whether this is all new or not. Definition: Catalan = sum_{k=0}^{\infty} (-1)^k/(2k+1)^2, Theorem: 1/Catalan=prod_{odd primes p} (1 - (-1)^((p-1)/2)/p^2), where the last product is negaodd posieven of above definitions. Note that (-1)^((p-1)/2) is the Legendre symbol (-1/p). Proof (sketch): Remember Eulers product identity for the zeta function zeta(s):=sum_n 1/n^s: zeta(s) = 1 / prod_p (1-1/p^s), where the product goes over all primes p. This product identity can be proven by multiplying the series for zeta by (1-1/2^s), then by (1-1/3^s), (1-1/5^s) ... (1-1/p^s) , whereby each term cancels out all remaining inverses of (multiples of p)^s, see e.g. Eq. (53)-(57) on http://mathworld.wolfram.com/RiemannZetaFunction.html. Exactly the same method works for the Catalan identity: Multiplying the series for Catalan by (1+1/3^2), cancels -1/3^2, 1/9^2, -1/15^2, 1/21^2, ... Multiplying the series for Catalan by (1-1/5^2), cancels 1/5^2, 1/25^2, -1/35^2, ... This continues with (1+1/7^2), (1+1/11^2), (1-1/13^2) and so on which finally cancel every term (except the leading 1). This proofs Catalan prod_{odd primes p} (1 - (-1)^((p-1)/2)/p^2) = 1. Note that the sign for 1/p^2 is simply derived from Catalan = sum_{k=0}^{\infty} (-1)^k/(2k+1)^2: For p=2k+1, the sign (-1)^k is given by (-1)^((p-1)/2). The same method can be applied to the following sums: (-1)^k/(2k+1)^s, only p^2 has to be replaced by p^s in the product equation. for s=1, one obtains an identity for pi/4, which is similar to the one given by Euler (see Eq. (5) on http://mathworld.wolfram.com/InfiniteProduct.html). Only the sign of 1/p is opposite (which can be understood since the product of both expressions is a multiple of Zeta(2)). While this is sure math-fun, I cannot believe that it is not already known (it must have been obvious to Euler already). Any pointers to the literature are welcome! Christoph