■連分数の測度論(その9)

  N=Πn^2/(n^2−1)=Πn/(n−1)・n/(n+1)

=2/1・2/3・3/2・3/4・・・n/(n−1)・n/(n+1)

はうまくキャンセルアウトして

  N=2/1・n/(n+1)→2

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[1]N=Πn^k/(n^k−1)  n=2〜∞

k=2:N=2

k=3:N=3πsech(π√3/2)

k=4:N=4πcosech(π√3/2)

k=6:N=6π^2(sech(π√3/2))^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

 なお,

k=4:N=sin((−1)^1/4π)sin((−1)^3/4π)/π^2

k=6:N=sin((−1)^1/6π)sin((−1)^5/6π)sinh(π)/π^3

と計算される.

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

[3]

  Π((n^3−1)/(n^3+1)=2/3   n=2〜∞

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