■ある無限級数(その114)
sinx/x=(1−x^2/π^2)(1−x^2/4π^2)(1−x^2/9π^2)・・・に対して,
sin2x/2x=sinx/x・cosx
より
cosx=(1−4x^2/π^2)(1−4x^2/9π^2)(1−4x^2/25π^2)
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(1−4/1)(1−4/9)(1−4/25)・・・(1−4/n^2)
=(−1・3/1・1)・(1・5/3・3)・(3・7)/(5・5)・・・(2n−3)(2n+1)/(2n−1)(2n−1)
=−(2n+1)/(2n−1)→−1
同様に
(1−16/1)(1−16/9)(1−16/25)(1−16/49)・・・=1
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