■ほぼ1の数の無限積(その46)

  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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