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Question Number 97956 by  M±th+et+s last updated on 10/Jun/20

prove Σ_(n=1) ^∞ ((9n+4)/(3n(3n+1)(3n+2)))=(3/2)−ln(3)

prove∑∞n=19n+43n(3n+1)(3n+2)=32−ln(3)

Answered by maths mind last updated on 13/Jun/20

=Σ_(n≥1) ((4(3n+1)−3n)/(3n(3n+1)(3n+2))) =4Σ(1/(3n(3n+2)))−Σ(1/((3n+1)(3n+2))) =(4/9)Σ_(n≥1) (1/(n(n+(2/3))))−(1/9)Σ_(n≥1) (1/((n+(1/3))(n+(2/3)))) =(4/9)Σ_(n≥0) (1/((n+1)(n+(5/3))))−(1/9)Σ_(n≥0) (1/((n+(4/3))(n+(5/3)))) =(4/9).((Ψ((5/3))−Ψ(1))/((5/3)−1))−(1/9).((Ψ((5/3))−Ψ((4/3)))/((5/3)−(4/3))) S=(2/3)(Ψ((5/3))−Ψ(1))−(1/3)(Ψ((5/3))−Ψ((4/3))) (1/3)(Ψ((5/3))+Ψ((4/3)))−(2/3)Ψ(1) Ψ((p/q))=−γ−ln(2q)−(π/2)cot(((pπ)/q))+2Σ_(k=0) ^([((q−1)/2)]) cos(((2pπk)/q))ln(sin(((pkπ)/q))) Ψ((5/3))=(3/2)+Ψ((2/3)),Ψ((4/3))=3+Ψ((1/3)) Ψ((2/3))=−γ−ln(6)−(π/2)cot(((2π)/3))+2cos(((4π)/3))ln(sin(((2π)/3))) =−γ−ln(6)+(π/(2(√3)))−ln(((√3)/2)) Ψ((1/3))=−γ−ln(6)−(π/(2(√3)))−ln(((√3)/2)) S=(1/3)(−2γ−ln(27)+(3/2)+3)−(2/3)(−γ) S=ln((1/3))+(3/2)=(3/2)−ln(3)

=∑n⩾14(3n+1)−3n3n(3n+1)(3n+2)=4Σ13n(3n+2)−Σ1(3n+1)(3n+2)=49∑n⩾11n(n+23)−19∑n⩾11(n+13)(n+23)=49∑n⩾01(n+1)(n+53)−19∑n⩾01(n+43)(n+53)=49.Ψ(53)−Ψ(1)53−1−19.Ψ(53)−Ψ(43)53−43S=23(Ψ(53)−Ψ(1))−13(Ψ(53)−Ψ(43))13(Ψ(53)+Ψ(43))−23Ψ(1)Ψ(pq)=−γ−ln(2q)−π2cot(pπq)+2∑[q−12]k=0cos(2pπkq)ln(sin(pkπq))Ψ(53)=32+Ψ(23),Ψ(43)=3+Ψ(13)Ψ(23)=−γ−ln(6)−π2cot(2π3)+2cos(4π3)ln(sin(2π3))=−γ−ln(6)+π23−ln(32)Ψ(13)=−γ−ln(6)−π23−ln(32)S=13(−2γ−ln(27)+32+3)−23(−γ)S=ln(13)+32=32−ln(3)

Commented by  M±th+et+s last updated on 13/Jun/20

well done prof.math mind you are intellignt

welldoneprof.mathmindyouareintellignt

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