(12C+) Bernoulli Number

[Albert Chan]

(07-27-2019 06:41 AM)Gamo Wrote:  Formula use to calculate Bernoulli Number

B(n) = [2(2n)! ÷ ((2^2n) - 1)(Pi^2n)] [1 + (1/3^2n) + (1/5^2n) + ...]

Hi, Gamo

I think there is a typo: B(n) should be abs(B(2n))

Can you provide source ?

From https://wstein.org/edu/fall05/168/projec...rnproj.pdf

zeta(x) = sum(k^-x, where k = 1 to ∞) = 1 / product(1 - p^-x, where p is prime)

B(2n) = (-1)^(n+1) * 2*(2n)!/(2 pi)^(2n) * zeta(2n)