八角数について検算.
Σ2/n(6n-4) (n=1~)
=Σ2/(n+1)(6n+2) (n=0~)
=Σ1/(n+1)(3n+1) (n=0~)
=1/3Σ1/(n+1)(n+1/3) (n=0~)
=1/2・Σ{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・cotπ/3+log6-2{cos2π/3・logsinπ/3}
=π/2・(1/√3)+log6-2{-1/2・log√3/2}
=π/2√3+log3+log2+1/2log3-log2
=π/2√3+3/2・log3
したがって,この1/2倍が解となる.
π/4√3+3/4・log3=1.27741・・・
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