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Question Number 114043 by Dwaipayan Shikari last updated on 16/Sep/20
Σ_(n=1) ^∞ (1/(n^3 +1))
∑∞n=11n3+1
Answered by maths mind last updated on 21/Sep/20
Σ_(n≥1) (1/((n+1)(n−j)(n−j^− ))) Σ_(n≥1) {(1/(3(n+1)))+(1/(3j^2 (n−j)))+(1/(3j^−^2 (n−j^− )))} =Σ_(n≥1) {(1/(3(n+1)))+(j/(3(n−j^− )))+(j^− /(3(n−j^− )))} we can Write it as Σ_(n≥1) Σ_(w:(w^3 +1=0)) ((1/(3w^2 (n−w)))) =(1/3)Σ_(n≥1) Σ_(w:(w^3 +1=0)) (((−w)/((n−w)))+(w/(n+1)))=Σ_(n≥0) (1/(n^3 +1)) since Σ_(w:(w^3 +1)) w=0 Σ_(n≥0) (1/(n^3 +1))=(1/3)Σ_(n≥0) Σ_(w:(w^3 +1)) (((−w)/(n−w))+(w/(n+1))) =(1/3)Σ_w wΣ_(n≥0) (((−1)/(n−w))+(1/(n+1))) =(1/3)Σ_w w(Ψ(−w)+γ) 𝚿 digamma function Ψ(x)=((Γ′(x))/(Γ(x))) =Σ_(w:(w^3 +1=0)) ((wΨ(−w))/3)+(γ/3)Σ_(w:(w^3 +1=0)) w_(=0) we get Σ_(w:(w^3 +1=0)) (w/3)Ψ(−w)=−(1/3)Ψ(1)+((1+i(√3))/6)Ψ(−((1+i(√3))/2)) +((1−i(√3))/6)Ψ(((−1+i(√3))/2))⋍0.6865
∑n⩾11(n+1)(n−j)(n−j−)∑n⩾1{13(n+1)+13j2(n−j)+13j−2(n−j−)}=∑n⩾1{13(n+1)+j3(n−j−)+j−3(n−j−)}wecanWriteitas∑n⩾1∑w:(w3+1=0)(13w2(n−w))=13∑n⩾1∑w:(w3+1=0)(−w(n−w)+wn+1)=∑n⩾01n3+1since∑w:(w3+1)w=0∑n⩾01n3+1=13∑n⩾0∑w:(w3+1)(−wn−w+wn+1)=13∑ww∑n⩾0(−1n−w+1n+1)=13∑ww(Ψ(−w)+γ)ΨdigammafunctionΨ(x)=Γ′(x)Γ(x)=∑w:(w3+1=0)wΨ(−w)3+γ3∑w:(w3+1=0)w=0weget∑w:(w3+1=0)w3Ψ(−w)=−13Ψ(1)+1+i36Ψ(−1+i32)+1−i36Ψ(−1+i32)⋍0.6865

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