Question Number 114043 by Dwaipayan Shikari last updated on 16/Sep/20

Σ_(n=1) ^∞ (1/(n^3 +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