円周率を含む数式
ζ ( 2 ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 {\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}} (バーゼルの問題、リーマンゼータ関数)
ζ ( 4 ) = 1 1 4 + 1 2 4 + 1 3 4 + 1 4 4 + ⋯ = π 4 90 {\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}}
ζ ( 2 n ) = ∑ k = 1 ∞ 1 k 2 n = 1 1 2 n + 1 2 2 n + 1 3 2 n + 1 4 2 n + ⋯ = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!