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Question Number 114043 by Dwaipayan Shikari last updated on 16/Sep/20
Σ_(n=1) ^∞ (1/(n^3 +1))
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} +\mathrm{1}} \\ $$
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
$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}−{j}\right)\left({n}−\overset{−} {{j}}\right)} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\left\{\frac{\mathrm{1}}{\mathrm{3}\left({n}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{3}{j}^{\mathrm{2}} \left({n}−{j}\right)}+\frac{\mathrm{1}}{\mathrm{3}\overset{−^{\mathrm{2}} } {{j}}\left({n}−\overset{−} {{j}}\right)}\right\} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\left\{\frac{\mathrm{1}}{\mathrm{3}\left({n}+\mathrm{1}\right)}+\frac{{j}}{\mathrm{3}\left({n}−\overset{−} {{j}}\right)}+\frac{\overset{−} {{j}}}{\mathrm{3}\left({n}−\overset{−} {{j}}\right)}\right\} \\ $$$${we}\:{can}\:{Write}\:{it} \\ $$$${as}\:\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\right)} {\sum}\left(\frac{\mathrm{1}}{\mathrm{3}{w}^{\mathrm{2}} \left({n}−{w}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\right)} {\sum}\left(\frac{−{w}}{\left({n}−{w}\right)}+\frac{{w}}{{n}+\mathrm{1}}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{3}} +\mathrm{1}} \\ $$$${since}\:\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}\right)} {\sum}{w}=\mathrm{0}\: \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\geqslant\mathrm{0}} {\sum}\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}\right)} {\sum}\left(\frac{−{w}}{{n}−{w}}+\frac{{w}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\underset{{w}} {\sum}{w}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{−\mathrm{1}}{{n}−{w}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\underset{{w}} {\sum}{w}\left(\Psi\left(−{w}\right)+\gamma\right) \\ $$$$\boldsymbol{\Psi}\:{digamma}\:{function}\:\Psi\left({x}\right)=\frac{\Gamma'\left({x}\right)}{\Gamma\left({x}\right)} \\ $$$$=\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\right)} {\sum}\frac{{w}\Psi\left(−{w}\right)}{\mathrm{3}}+\frac{\gamma}{\mathrm{3}}\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\right)} {\sum}\underset{=\mathrm{0}} {{w}} \\ $$$${we}\:{get} \\ $$$$\underset{{w}:\left({w}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\right)} {\sum}\frac{{w}}{\mathrm{3}}\Psi\left(−\boldsymbol{{w}}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\Psi\left(\mathrm{1}\right)+\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{6}}\Psi\left(−\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$+\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{6}}\Psi\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\backsimeq\mathrm{0}.\mathrm{6865} \\ $$$$ \\ $$

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