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

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3+1}$$

Answered by maths mind last updated on 21/Sep/20

$$\sum _{n⩾1}\frac{1}{(n+1)(n-j)(n-\stackrel{-}{j})}$$$$\sum _{n⩾1}\{\frac{1}{3(n+1)}+\frac{1}{3j^2(n-j)}+\frac{1}{3\stackrel{{-}^2}{j}(n-\stackrel{-}{j})}\}$$$$=\sum _{n⩾1}\{\frac{1}{3(n+1)}+\frac{j}{3(n-\stackrel{-}{j})}+\frac{\stackrel{-}{j}}{3(n-\stackrel{-}{j})}\}$$$$wecanWriteit$$$$as\sum _{n⩾1}\sum _{w:(w^3+1=0)}\left(\frac{1}{3w^2(n-w)}\right)$$$$=\frac{1}{3}\sum _{n⩾1}\sum _{w:(w^3+1=0)}(\frac{-w}{(n-w)}+\frac{w}{n+1})=\sum _{n⩾0}\frac{1}{n^3+1}$$$$since\sum _{w:(w^3+1)}w=0$$$$\sum _{n⩾0}\frac{1}{n^3+1}=\frac{1}{3}\sum _{n⩾0}\sum _{w:(w^3+1)}(\frac{-w}{n-w}+\frac{w}{n+1})$$$$=\frac{1}{3}\sum _{w}w\sum _{n⩾0}(\frac{-1}{n-w}+\frac{1}{n+1})$$$$=\frac{1}{3}\sum _{w}w(\mathrm{\Psi }(-w)+\gamma )$$$$\mathbf{\Psi }digammafunction\mathrm{\Psi }\left(x\right)=\frac{{\mathrm{\Gamma }}^{\prime }\left(x\right)}{\mathrm{\Gamma }\left(x\right)}$$$$=\sum _{w:(w^3+1=0)}\frac{w\mathrm{\Psi }(-w)}{3}+\frac{\gamma }{3}\sum _{w:(w^3+1=0)}\underset{=0}{w}$$$$weget$$$$\sum _{w:(w^3+1=0)}\frac{w}{3}\mathrm{\Psi }(-\mathit{w})=-\frac{1}{3}\mathrm{\Psi }\left(1\right)+\frac{1+i\sqrt{3}}{6}\mathrm{\Psi }(-\frac{1+i\sqrt{3}}{2})$$$$+\frac{1-i\sqrt{3}}{6}\mathrm{\Psi }\left(\frac{-1+i\sqrt{3}}{2}\right)\backsimeq 0.6865$$$$$$

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