■多角数の逆数和の問題(その12)

 八角数:n(6n−4)/2について

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 Σ2/n(6n−4)  (n=1〜)

=Σ2/(n+1)(6n+2)  (n=0〜)

=12Σ1/(n+1)(n+2/6)  (n=0〜)

=2Σ{1/(n+1/3)−1/(n+1)}  (n=0〜)

  Σ{1/(n+p/q)−1/(n+1)}

=π/2・cotpπ/q+log2q−2Σcos2pkπ/q・logsinkπ/q  (0<k<q/2)

  Σ{1/(n+1/3)−1/(n+1)}=π/2・cot2π/3+log6−2{cos2π/3・logsinπ/3}

=π/2・(−1/√3)+log6−2{−1/2・log1/2}

=−π/2√3+log6−log2

=−π/2√3+log3

したがって,この2倍が解となる.(誤り)

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