■多角数の逆数和(その8)

 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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