■多角数の逆数和(その15)

 五角数について検算.

 Σ2/n(3n−1)  (n=1〜)

=Σ2/(n+1)(3n+2)  (n=0〜)

=2/3Σ1/(n+1)(n+2/3)  (n=0〜)

=2Σ{1/(n+2/3)−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+2/3)−1/(n+1)}=

=π/2・cot2π/3+log6−2{cos4π/3・logsinπ/3}

=−π/2√3+log6−2{−1/2・log√3/2}

=−π/2√3+log2+log3+1/2log3−log2

=−π/2√3+3/2・log3

 したがって,

 Σ2/n(3n−1)=3log3−π/√3  (OK)

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