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Question Number 125458 by mnjuly1970 last updated on 11/Dec/20

$$...advancedcalculus...$$$$evaluate:::$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\left\{\frac{\zeta \left(2n\right)}{2^n}\right\}=??$$$$$$

Answered by Dwaipayan Shikari last updated on 11/Dec/20

$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n{k}^{2n}}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n{\left(k^2\right)}^n}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{\frac{1}{4k^4}}{1-\frac{1}{2k^2}}$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2k^2-1}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2k^2(2k^2-1)}$$$$=\sum ^{\mathrm{\infty }}\frac{1}{2k^2-1}-\frac{1}{2k^2}=\frac{1}{2}\sum ^{\mathrm{\infty }}\frac{1}{k^2-\frac{1}{2}}-\frac{{\pi }^2}{12}$$$$=\frac{1}{2\sqrt{2}}\sum ^{\mathrm{\infty }}\frac{1}{k-\frac{1}{\sqrt{2}}}-\frac{1}{k+\frac{1}{\sqrt{2}}}-\frac{{\pi }^2}{12}=\frac{1}{2\sqrt{2}}(\psi (1+\frac{1}{\sqrt{2}})-\psi (1-\frac{1}{\sqrt{2}}))-\frac{{\pi }^2}{12}$$$$\sim 0.52$$

Commented by Dwaipayan Shikari last updated on 11/Dec/20

$$\zeta \left(2n\right)=\frac{{(-1)}^{n+1}{\beta }_{2n}{\left(2\pi \right)}^{2n}}{2\left(2n\right)!}$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{\beta }_{2n}{\left(2\pi \right)}^{2n}}{2^{n+1}\left(2n\right)!}=\frac{1}{2}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathit{n}+1}{\mathit{\beta }}_{2\mathit{n}}{\left(2{\mathit{\pi }}^2\right)}^n}{\left(2n\right)!}$$$${\mathit{\beta }}_{\mathit{n}}-\mathit{B}\mathit{e}\mathit{r}\mathit{n}\mathit{o}\mathit{u}\mathit{l}\mathit{l}\mathit{i}\mathit{N}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}$$

Commented by mnjuly1970 last updated on 11/Dec/20

$$peacebeuponyou...$$

Commented by Dwaipayan Shikari last updated on 11/Dec/20

$$Isitright?\left(Myanswer\right)$$

Commented by mnjuly1970 last updated on 11/Dec/20

$$yesofcourse.$$$$$$$$butoriginalquestionwasasbelow.$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{\zeta \left(n\right)}{2^n}\stackrel{??}{=}log\left(2\right)...$$

Commented by Dwaipayan Shikari last updated on 11/Dec/20

$$\sum ^{\mathrm{\infty }}\frac{\zeta \left(n\right)}{2^n}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2k\right)}^n}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2k(2k-1)}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..=log\left(2\right)$$

Commented by mnjuly1970 last updated on 11/Dec/20

$$perfect..$$

Commented by Dwaipayan Shikari last updated on 11/Dec/20

$$Withpleasure$$

Answered by mathmax by abdo last updated on 11/Dec/20

$$\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{\xi \left(2\mathrm{n}\right)}{2^{\mathrm{n}}}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{k}}^{2\mathrm{n}}}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}\frac{1}{{\left({\mathrm{k}}^2\right)}^{\mathrm{n}}}$$$$=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}{\left(\frac{1}{{2\mathrm{k}}^2}\right)}^{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\sum _{\mathrm{p}=0}^{\mathrm{\infty }}{\left(\frac{1}{{2\mathrm{k}}^2}\right)}^{\mathrm{p}+2}$$$$=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{4\mathrm{k}}^4}\sum _{\mathrm{p}=0}^{\mathrm{\infty }}{\left(\frac{1}{{2\mathrm{k}}^2}\right)}^{\mathrm{p}}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{4\mathrm{k}}^4}\times \frac{1}{1-\frac{1}{{2\mathrm{k}}^2}}$$$$=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{{2\mathrm{k}}^2}{{4\mathrm{k}}^4({2\mathrm{k}}^2-1)}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{2\mathrm{k}}^2({2\mathrm{k}}^2-1)}$$$$=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}(\frac{1}{{2\mathrm{k}}^2-1}-\frac{1}{{2\mathrm{k}}^2})=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{2\mathrm{k}}^2-1}-\frac{1}{2}\times \frac{{\pi }^2}{6}$$$$=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{2\mathrm{k}}^2-1}-\frac{{\pi }^2}{12}\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{value}\mathrm{of}\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{1}{{2\mathrm{k}}^2-1}$$$$...\mathrm{be}\mathrm{continued}...$$$$$$

Commented by mnjuly1970 last updated on 11/Dec/20

$$thanksalotsir...$$

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

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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