■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)・・・
===================================
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
===================================