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Question Number 117990 by Dwaipayan Shikari last updated on 14/Oct/20

(2.3.5.7.9.11.13.17......∞)×(√(6/(3.8.24.48.80.120.168.288.....∞)))

$$\left(\mathrm{2}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}.\mathrm{11}.\mathrm{13}.\mathrm{17}......\infty\right)×\sqrt{\frac{\mathrm{6}}{\mathrm{3}.\mathrm{8}.\mathrm{24}.\mathrm{48}.\mathrm{80}.\mathrm{120}.\mathrm{168}.\mathrm{288}.....\infty}} \\ $$

Answered by Dwaipayan Shikari last updated on 14/Oct/20

(√6).(√((2^2 .3^2 .5^2 .7^2 .9^2 .....)/((2^2 −1)(3^2 −1)(5^2 −1)(7^2 −1)...)))  (√6).(√(1/((1−(1/2^2 ))(1−(1/3^2 ))(1−(1/5^2 )).....)))=(√6).(√(π^2 /6))=π  ζ(s)=Π_p ^∞ (1−(1/p^s ))^(−1)   ζ(2)=((2^2 .3^2 .5^2 .7^2 .9^2 .....∞)/((2^2 −1)(3^2 −1)....∞))=(π^2 /6)

$$\sqrt{\mathrm{6}}.\sqrt{\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} .\mathrm{5}^{\mathrm{2}} .\mathrm{7}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} .....}{\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{5}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{7}^{\mathrm{2}} −\mathrm{1}\right)...}} \\ $$$$\sqrt{\mathrm{6}}.\sqrt{\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\right).....}}=\sqrt{\mathrm{6}}.\sqrt{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}}=\pi \\ $$$$\zeta\left({s}\right)=\underset{{p}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }\right)^{−\mathrm{1}} \\ $$$$\zeta\left(\mathrm{2}\right)=\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} .\mathrm{5}^{\mathrm{2}} .\mathrm{7}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} .....\infty}{\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)....\infty}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$

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