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

$$\mathrm{calculate}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}(\xi \left(2\mathrm{n}\right)-1){\mathrm{x}}^{2\mathrm{n}}$$$$\xi \left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^{\mathrm{x}}}$$

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

$$\zeta \left(2n\right)-1=\sum _{k⩾2}\frac{1}{k^{2n}}$$$$=\sum _{n⩾1}.\sum _{k⩾2}\frac{x^{2n}}{k^{2n}}$$$$=\sum _{k⩾2}.\sum _{n⩾1}{\left(\frac{x^2}{k^2}\right)}^n$$$$=\sum _{k⩾2}\left(\frac{x^2}{k^2}\right).\frac{1}{1-\frac{x^2}{k^2}}$$$$=\sum _{k⩾2}\frac{x^2}{k^2-x^2}=S\left(x\right)$$$$forx\notin \mathbb{Z}$$$$\pi xcot\left(\pi x\right)=1+2\sum _{k⩾1}\frac{x^2}{x^2-k^2}$$$$S\left(x\right)=\frac{1-\pi xcot\left(\pi x\right)}{2}+\frac{x^2}{x^2-1}=\sum _{n⩾1}(\zeta \left(2n\right)-1)x^{2n}$$$$$$$$$$

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

$$x\in ]1,2[$$

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

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

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