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Question Number 98118 by abdomathmax last updated on 11/Jun/20

calculate Σ_(n=1) ^∞ (ξ(2n)−1)x^(2n) ξ(x)=Σ_(n=1) ^∞ (1/n^x )

$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(\xi\left(\mathrm{2n}\right)−\mathrm{1}\right)\mathrm{x}^{\mathrm{2n}} \\ $$$$\xi\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{x}} } \\ $$

Answered by maths mind last updated on 11/Jun/20

ζ(2n)−1=Σ_(k≥2) (1/k^(2n) ) =Σ_(n≥1) .Σ_(k≥2) (x^(2n) /k^(2n) ) =Σ_(k≥2) .Σ_(n≥1) ((x^2 /k^2 ))^n =Σ_(k≥2) ((x^2 /k^2 )).(1/(1−(x^2 /k^2 ))) =Σ_(k≥2) (x^2 /(k^2 −x^2 ))=S(x) for x∉Z πxcot(πx)=1+2Σ_(k≥1) (x^2 /(x^2 −k^2 )) S(x)=((1−πxcot(πx))/2)+(x^2 /(x^2 −1))=Σ_(n≥1) (ζ(2n)−1)x^(2n)

$$\zeta\left(\mathrm{2}{n}\right)−\mathrm{1}=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{k}^{\mathrm{2}{n}} } \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}.\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{x}^{\mathrm{2}{n}} }{{k}^{\mathrm{2}{n}} } \\ $$$$=\underset{{k}\geqslant\mathrm{2}} {\sum}.\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right)^{{n}} \\ $$$$=\underset{{k}\geqslant\mathrm{2}} {\sum}\left(\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right).\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} }} \\ $$$$=\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} −{x}^{\mathrm{2}} }={S}\left({x}\right) \\ $$$${for}\:{x}\notin\mathbb{Z} \\ $$$$\pi{xcot}\left(\pi{x}\right)=\mathrm{1}+\mathrm{2}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{k}^{\mathrm{2}} } \\ $$$${S}\left({x}\right)=\frac{\mathrm{1}−\pi{xcot}\left(\pi{x}\right)}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}=\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\zeta\left(\mathrm{2}{n}\right)−\mathrm{1}\right){x}^{\mathrm{2}{n}} \\ $$$$ \\ $$$$ \\ $$

Commented by maths mind last updated on 11/Jun/20

x∈]1,2[

$$\left.{x}\in\right]\mathrm{1},\mathrm{2}\left[\right. \\ $$

Commented by mathmax by abdo last updated on 11/Jun/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\: \\ $$

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