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Question Number 117258 by Dwaipayan Shikari last updated on 10/Oct/20

(4/3).(9/8).((49)/(48)).((121)/(120)).((169)/(168)).((289)/(288)).((529)/(528)).((831)/(830)).....∞

$$\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{9}}{\mathrm{8}}.\frac{\mathrm{49}}{\mathrm{48}}.\frac{\mathrm{121}}{\mathrm{120}}.\frac{\mathrm{169}}{\mathrm{168}}.\frac{\mathrm{289}}{\mathrm{288}}.\frac{\mathrm{529}}{\mathrm{528}}.\frac{\mathrm{831}}{\mathrm{830}}.....\infty \\ $$

Answered by AbduraufKodiriy last updated on 10/Oct/20

− { ((𝛇(2)=(1/1^2 )+(1/2^2 )+(1/3^2 )+...)),(((1/2^2 )𝛇(2)=(1/2^2 )+(1/4^2 )+(1/6^2 )+...)) :}⇒ (1−(1/2^2 ))𝛇(2)=(1/1^2 )+(1/3^2 )+(1/5^2 )+(1/7^2 )+(1/9^2 )+... ⇒ ⇒ − { (((1−(1/2^2 ))𝛇(2)=(1/1^2 )+(1/3^2 )+(1/5^2 )+...)),(((1/3^2 )(1−(1/2^2 ))𝛇(2)=(1/3^2 )+(1/9^2 )+(1/(15^2 ))+...)) :}⇒ ⇒ (1−(1/3^2 ))(1−(1/2^2 ))𝛇(2)=(1/1^2 )+(1/5^2 )+(1/7^2 )+(1/(11^2 ))+(1/(13^2 ))+... .............................. 𝛇(2)(1−(1/2^2 ))(1−(1/3^2 ))(1−(1/5^2 ))...=1 (π^2 /6)=(2^2 /(2^2 −1))∙(3^2 /(3^2 −1))∙(5^2 /(5^2 −1))∙(7^2 /(7^2 −1))∙... (π^2 /6)=(4/3)∙(9/8)∙((25)/(24))∙((49)/(48))∙... ⇒ ((4π^2 )/(25))=(4/3)∙(9/8)∙((49)/(48))∙((121)/(120))∙...

$$−\begin{cases}{\boldsymbol{\zeta}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...}\\{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\boldsymbol{\zeta}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+...}\end{cases}\Rightarrow\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\boldsymbol{\zeta}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }+...\:\Rightarrow \\ $$$$\Rightarrow\:−\begin{cases}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\boldsymbol{\zeta}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+...}\\{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\boldsymbol{\zeta}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{15}^{\mathrm{2}} }+...}\end{cases}\Rightarrow \\ $$$$\Rightarrow\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\boldsymbol{\zeta}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{13}^{\mathrm{2}} }+... \\ $$$$.............................. \\ $$$$\boldsymbol{\zeta}\left(\mathrm{2}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\right)...=\mathrm{1} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}\centerdot\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}\centerdot\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} −\mathrm{1}}\centerdot\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{7}^{\mathrm{2}} −\mathrm{1}}\centerdot... \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{9}}{\mathrm{8}}\centerdot\frac{\mathrm{25}}{\mathrm{24}}\centerdot\frac{\mathrm{49}}{\mathrm{48}}\centerdot...\:\Rightarrow\:\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{25}}=\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{9}}{\mathrm{8}}\centerdot\frac{\mathrm{49}}{\mathrm{48}}\centerdot\frac{\mathrm{121}}{\mathrm{120}}\centerdot... \\ $$

Commented by Dwaipayan Shikari last updated on 10/Oct/20

Great sir! So generally ζ(s)=Π_p ^∞ (1−(1/p^s ))^(−1) (p=prime Numbers)

$${Great}\:{sir}! \\ $$$${So}\:{generally} \\ $$$$\zeta\left({s}\right)=\underset{{p}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }\right)^{−\mathrm{1}} \left({p}={prime}\:{Numbers}\right) \\ $$

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