■マーダヴァの無限級数(その6)

四角数の逆数和

Σ1/n^2 =1/1^2+1/2^2+1/3^2+1/4^2+・・・

はπ^2/6に収束する.

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多角数の逆数和は

[1]三角数:n(n+1)/2→Σ2/n(n+1)=2

[2]四角数:n^2→Σ1/n^2=π^2/6

[3]五角数:(3n^2-n)/2→Σ2/n(3n-1)=3log3-π/√3

Σ2/n(3n-1)  (n=1~)

=Σ2/(n+1)(3n+2)  (n=0~)

=6Σ1/(n+1)(n+2/3)  (n=0~)

=2Σ{1/(n+2/3)-1/(n+1)}  (n=0~)

  Σ{1/(n+2/3)-1/(n+1)}=-π/2√3+log12-2{-1/2・log1/2-1/2・log√3/2}

=-π/2√3+2log2+log3-log2+(1/2・log3-log2)

=-π/2√3+3/2・log3

 したがって,

 Σ2/n(3n-1)=3log3-π/√3

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[4]六角数:2n^2-n→Σ1/2n(2n-1)=2log2

[5]七角数:n(5n-3)/2→Σ2/n(5n-3)=π/3・(1-2/√5)^1/2+5/6・log5-√5/3・log(φ)

[6]八角数:n(6n-4)/2→Σ2/n(6n-4)π/4√3+3/4・log3

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