■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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