バーゼル問題
{\displaystyle {\frac {\sin x}{x}}=\left(1-{\frac {x}{1\pi }}\right)\left(1+{\frac {x}{1\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots .}
隣接する2項を掛け合わせると
( 2 ) sin  x x = ( 1 − x 2 1 2 π 2 ) ( 1 − x 2 2 2 π 2 ) ( 1 − x 2 3 2 π 2 ) ⋯ . {\displaystyle (2)\qquad {\frac {\sin x}{x}}=\left(1-{\frac {x^{2}}{1^{2}\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{2^{2}\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{3^{2}\pi ^{2}}}\right)\cdots .}
(1) と (2) の右辺の x の係数は
( 1 ) : − 1 3 ! = − 1 6 , {\displaystyle (1)\colon -{\frac {1}{3!}}=-{\frac {1}{6}},}