■多角数の逆数和(その33)
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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