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Question Number 130109 by Adel last updated on 22/Jan/21

$$2+\frac{3}{2^3}+\frac{4}{3^3}+\frac{5}{4^3}+\frac{6}{5^3}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot =?$$

Answered by Olaf last updated on 22/Jan/21

$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n+1}{n^3}$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}$$$$\mathrm{S}=\zeta \left(2\right)+\zeta \left(3\right)=\frac{{\pi }^2}{6}+\zeta \left(3\right)$$$$\xi \left(3\right):\mathrm{Apery}\mathrm{constant}\approx 1,202$$

Commented by Adel last updated on 22/Jan/21

$$\mathrm{what}\mathrm{thes}\varsigma \left(2\right)\mathrm{and}\varsigma \left(3\right)?$$

Commented by Ar Brandon last updated on 22/Jan/21

$$\zeta \left(\mathrm{n}\right)=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{k}}^{\mathrm{n}}}$$

Commented by Olaf last updated on 22/Jan/21

$$\zeta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}$$$$\zeta \left(s\right)=\mathrm{Riemann}\mathrm{zeta}\mathrm{function}.$$$$\mathrm{For}\mathrm{some}\mathrm{values}\mathrm{of}s{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{known}.$$$$\mathrm{For}\mathrm{example}\zeta \left(2\right)=\frac{{\pi }^2}{6},\zeta \left(4\right)=\frac{{\pi }^4}{90}$$$$\mathrm{But}\mathrm{for}\mathrm{some}\mathrm{values}\mathrm{of}s{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{approximative}.$$$$\mathrm{For}\mathrm{example}\zeta \left(3\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}\approx 1,202056903.$$

Commented by talminator2856791 last updated on 22/Jan/21

$$\mathrm{is}\mathrm{there}\mathrm{proof}\mathrm{that}\mathrm{not}\mathrm{all}\mathrm{riemann}\mathrm{zeta}\mathrm{functions}$$$$\mathrm{have}\mathrm{closed}\mathrm{forms}?$$

Answered by Ar Brandon last updated on 22/Jan/21

$$\mathrm{S}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{k}+1}{{\mathrm{k}}^3}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\{\frac{1}{{\mathrm{k}}^2}+\frac{1}{{\mathrm{k}}^3}\}$$$$=\zeta \left(2\right)+\zeta \left(3\right)$$

Answered by Dwaipayan Shikari last updated on 22/Jan/21

$$Approximationof\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}$$$$f\left(x\right)=\frac{1}{x^3}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}=\underset{z\to \mathrm{\infty }}{\lim }{\int }_1^{\mathrm{\infty }}\frac{1}{x^3}+\frac{f\left(z\right)+f\left(1\right)}{2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{B_{2n}}{\left(2n\right)!}(f^{2k-1}\left(z\right)-f^{2k-1}\left(1\right))$$$$=\frac{1}{2}+\frac{1}{2}+(\frac{3B_2}{2!}+\frac{3.4.5}{4!}{B}_4+\frac{3.4.5.6.7}{6!}{B}_6+\frac{3.4.5.6.7.8.9}{8!}{B}_8+...)$$$$=1.2020....(B_n=Bernoullinumber)$$$$Evenzetafunction\zeta \left(2n\right)(n\in \mathbb{Z})$$$$\zeta \left(2\right)=\frac{{\pi }^2}{6},\zeta \left(4\right)=\frac{{\pi }^4}{90},\zeta \left(6\right)=\frac{{\pi }^6}{945},\zeta \left(8\right)=\frac{{\pi }^8}{9450},\zeta \left(10\right)=\frac{{\pi }^{10}}{93555}....$$

Commented by Adel last updated on 23/Jan/21

$$\mathrm{thanks}\mathrm{bro}$$

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