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Fungrim entry: 72ccda

ζ ⁣(2n)=(1)n+1B2n(2π)2n2(2n)!\zeta\!\left(2 n\right) = \frac{{\left(-1\right)}^{n + 1} B_{2 n} {\left(2 \pi\right)}^{2 n}}{2 \left(2 n\right)!}
Assumptions:nZ  and  n1n \in \mathbb{Z} \;\mathbin{\operatorname{and}}\; n \ge 1
\zeta\!\left(2 n\right) = \frac{{\left(-1\right)}^{n + 1} B_{2 n} {\left(2 \pi\right)}^{2 n}}{2 \left(2 n\right)!}

n \in \mathbb{Z} \;\mathbin{\operatorname{and}}\; n \ge 1
Fungrim symbol Notation Short description
RiemannZetaζ ⁣(s)\zeta\!\left(s\right) Riemann zeta function
Powab{a}^{b} Power
BernoulliBBnB_{n} Bernoulli number
Piπ\pi The constant pi (3.14...)
Factorialn!n ! Factorial
ZZZ\mathbb{Z} Integers
Source code for this entry:
    Formula(Equal(RiemannZeta(Mul(2, n)), Div(Mul(Mul(Pow(-1, Add(n, 1)), BernoulliB(Mul(2, n))), Pow(Mul(2, Pi), Mul(2, n))), Mul(2, Factorial(Mul(2, n)))))),
    Assumptions(And(Element(n, ZZ), GreaterEqual(n, 1))))

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2021-03-15 19:12:00.328586 UTC