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Question Number 127020 by bramlexs22 last updated on 26/Dec/20

$$supernice!$$$$showthat$$$$\zeta \left(6\right)=\frac{{\pi }^6}{945}$$

Commented by liberty last updated on 26/Dec/20

$$hahahaverynice$$

Answered by Olaf last updated on 26/Dec/20

$$\zeta \left(2k\right)=\frac{{(-1)}^{k-1}{B}_{2k}{\left(2\pi \right)}^{2k}}{2\left(2k\right)!}$$$$\mathrm{For}k=3:$$$$\zeta \left(6\right)=\frac{{(-1)}^2{B}_6{\left(2\pi \right)}^6}{2\times 6!}=\frac{2}{45}{B}_6{\pi }^6$$$$B_6=\frac{1}{42}$$$$\zeta \left(6\right)=\frac{2}{45}\times \frac{1}{42}{\pi }^6=\frac{{\pi }^6}{945}$$

Answered by Dwaipayan Shikari last updated on 26/Dec/20

$$\zeta \left(2n\right)=\frac{{(-1)}^{n+1}{\left(2\pi \right)}^{2n}}{2\left(2n\right)!}{B}_{2n}$$$$\zeta \left(6\right)=\frac{{\left(2\pi \right)}^6}{2.6!}.\frac{1}{42}=\frac{{\pi }^6}{15.63}=\frac{{\pi }^6}{945}$$$$\zeta \left(8\right)=\frac{{\pi }^8}{9450}$$$$\zeta \left(10\right)=\frac{{\pi }^{10}}{93555}$$$$....$$

Commented by Dwaipayan Shikari last updated on 26/Dec/20

$$\frac{sin\pi x}{\pi x}=\prod ^{\mathrm{\infty }}(1-\frac{x^2}{n^2})$$$$log\left(sin\pi x\right)-log\left(\pi x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}log(1-\frac{x^2}{n^2})$$$$\pi \frac{cos\pi x}{sin\pi x}-\frac{\pi }{\pi x}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{-\frac{2x}{n^2}}{1-\frac{x^2}{n^2}}\Rightarrow \pi cot\left(\pi x\right)-\frac{1}{x}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2x}{x^2-n^2}$$$$\Rightarrow \pi xcot\left(\pi x\right)=1-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\frac{x^2}{n^2}}{1-\frac{x^2}{n^2}}\Rightarrow \pi xcot\left(\pi x\right)=1-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{k=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{x}{n}\right)}^{2k}$$$$\Rightarrow \pi xcot\left(\pi x\right)=1-2\underset{k=1}{\sum ^{\mathrm{\infty }}}\zeta \left(2k\right)x^{2k}\Rightarrow \pi xi\left(\frac{e^{\pi xi}+e^{-\pi xi}}{e^{\pi xi}-e^{-\pi xi}}\right)=1-2\sum ^{\mathrm{\infty }}\zeta \left(2k\right)x^{2k}$$$$\Rightarrow \pi xi+\frac{2\pi xi}{e^{2\pi xi}-1}=1-2\underset{k⩾1}{\sum ^{\mathrm{\infty }}}\zeta \left(2k\right)x^{2k}\Rightarrow 1-\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{\beta }_{2k}}{2(2k!)}{\left(2\pi ix\right)}^{2k}$$$$\zeta \left(2k\right)x^{2k}=\frac{{\beta }_{2k}}{2(2k!)}{\left(2\pi ix\right)}^{2k}\Rightarrow \zeta \left(2k\right)={(-1)}^{k+1}\frac{{\beta }_{2k}{\left(2\pi \right)}^{2k}}{2(2k!)}$$$$$$

Commented by mnjuly1970 last updated on 26/Dec/20

$$verynicemrpayan$$$$solutionwithexplanation..$$

Answered by liberty last updated on 26/Dec/20

$$viaFouriersineseries$$$$b_n=\frac{2}{\pi }{\int }_0^{\pi }(\pi x-x^2)\sin \left(nx\right)dx=\frac{4}{\pi }.\frac{1-{(-1)}^{n+1}}{n^3}$$$$b_n=\{\begin{array}{l}0;ifnisodd\\ \frac{8}{\pi n^3};ifneven\end{array}$$$$writtingn=2k-1forsomepositiveintegerk$$$$weget\pi x-x^2\sim \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{8\sin \left((2k-1)x\right)}{\pi {(2k-1)}^3}$$$$\frac{2}{\pi }{\int }_0^{\pi }{(\pi x-x^2)}^{2}dx=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{8}{\pi {(2k-1)}^3}\right)}^2$$$$simplifyingthisyields$$$$\frac{{\pi }^4}{15}=\frac{64}{{\pi }^2}\underset{k=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{2k-1}\right)}^{6}and\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2k-1)}^6}=\frac{{\pi }^6}{960}$$$$however$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2k-1)}^6}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^6}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2n}=(1-\frac{1}{2^6})\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^6}$$$$=\frac{63}{64}\zeta \left(6\right)$$$$\zeta \left(6\right)=\frac{64}{63}.\frac{{\pi }^6}{960}=\frac{{\pi }^6}{945}.$$

Commented by mnjuly1970 last updated on 26/Dec/20

$$verynicebravomrliberty$$$$excellent...$$

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