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Aug 5, 2019 at 2:30 history edited Tom Copeland CC BY-SA 4.0
Missing eqn introduced
Aug 4, 2019 at 23:17 comment added Tom Copeland $\frac{2}{(2\pi)^{2n}}\:(2n-1)!\:\zeta(2n)=(-1)^{n+1}\frac{B_{2n}}{2n}=(-1)^{n}\zeta(1-2n)$
Aug 4, 2019 at 22:58 history answered Tom Copeland CC BY-SA 4.0