■素数と無限級数(その125)

 ウォリスの公式

 π/2=Π4n^2/(4n^2−1)=Π2n・2n/(2n−1)(2n+1)

=(2・2/1・3)(4・4/3・5)(6・6/5・7)・・・

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 πcotπz=1/x+2zΣ1/(z^2−n^2)

 Σ1/(n^2−z^2)=1/2z^2−πcotπz/2z

z=1/4とおくと

 Σ1/(n^2−1/16)=Σ16/(4n−1)(4n+1)

=16{1/3・5+1/7・9+・・・}=8−2π

 Σ1/(4n−1)(4n+1)

={1/3・5+1/7・9+・・・}=1/2−π/8

z=1/3とおくと

 Σ1/(3n−1)(3n+1)

={1/2・4+1/5・7+・・・}=1/2−π/6√3

z=1/2とおくと

 Σ1/(2n−1)(2n+1)

={1/1・3+1/3・5+・・・}=1/2

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