■ウォリス積と・・・(その59)
Π(p^2+1)/(p^2−1)=5/2
Π((n^3−1)/(n^3+1)=2/3 n=2〜∞
それでは
Π((n^2−1)/(n^2+1)=?
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[1]N=Πn^k/(n^k−1) n=2〜∞
k=2:N=2
k=3:N=3πsech(π√3/2)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
[2]N=Π(n^k+1)/n^k n=1〜∞
k=2:N=sinh(π)/π
k=3:N=cosh(π√3/2)/π
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
[3]N=Π(n^k+1)/n^k n=2〜∞
k=2:N=sinh(π)/2π
k=3:N=cosh(π√3/2)/2π
したがって,n=2〜∞
Π((n^2+1)/(n^2−1)=sinh(π)/π
Π((n^2−1)/(n^2+1)=π/(sinh(π))
なお,
Π((n^3+1)/(n^3−1)=cosh(π√3/2)/2π・3πsech(π√3/2)=3/2
Π((n^3−1)/(n^3+1)=2/3
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