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


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 117258 by Dwaipayan Shikari last updated on 10/Oct/20

$\frac{4}{3} . \frac{9}{8} . \frac{49}{48} . \frac{121}{120} . \frac{169}{168} . \frac{289}{288} . \frac{529}{528} . \frac{831}{830} . . . . . \infty$
	Answered by AbduraufKodiriy last updated on 10/Oct/20
	$− { ((𝛇(2)=(1/1^2 )+(1/2^2 )+(1/3^2 )+...)),(((1/2^2 )𝛇(2)=(1/2^2 )+(1/4^2 )+(1/6^2 )+...)) :}⇒ (1−(1/2^2 ))𝛇(2)=(1/1^2 )+(1/3^2 )+(1/5^2 )+(1/7^2 )+(1/9^2 )+... ⇒ ⇒ − { (((1−(1/2^2 ))𝛇(2)=(1/1^2 )+(1/3^2 )+(1/5^2 )+...)),(((1/3^2 )(1−(1/2^2 ))𝛇(2)=(1/3^2 )+(1/9^2 )+(1/(15^2 ))+...)) :}⇒ ⇒ (1−(1/3^2 ))(1−(1/2^2 ))𝛇(2)=(1/1^2 )+(1/5^2 )+(1/7^2 )+(1/(11^2 ))+(1/(13^2 ))+... .............................. 𝛇(2)(1−(1/2^2 ))(1−(1/3^2 ))(1−(1/5^2 ))...=1 (π^2 /6)=(2^2 /(2^2 −1))∙(3^2 /(3^2 −1))∙(5^2 /(5^2 −1))∙(7^2 /(7^2 −1))∙... (π^2 /6)=(4/3)∙(9/8)∙((25)/(24))∙((49)/(48))∙... ⇒ ((4π^2 )/(25))=(4/3)∙(9/8)∙((49)/(48))∙((121)/(120))∙...$
	$- {\begin{cases} ζ (2) = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . \\ \frac{1}{2^{2}} ζ (2) = \frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . . \end{cases} \Rightarrow (1 - \frac{1}{2^{2}}) ζ (2) = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + \frac{1}{9^{2}} + . . . \Rightarrow$ $\Rightarrow - {\begin{cases} (1 - \frac{1}{2^{2}}) ζ (2) = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . \\ \frac{1}{3^{2}} (1 - \frac{1}{2^{2}}) ζ (2) = \frac{1}{3^{2}} + \frac{1}{9^{2}} + \frac{1}{15^{2}} + . . . \end{cases} \Rightarrow$ $\Rightarrow (1 - \frac{1}{3^{2}}) (1 - \frac{1}{2^{2}}) ζ (2) = \frac{1}{1^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + \frac{1}{11^{2}} + \frac{1}{13^{2}} + . . .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $ζ (2) (1 - \frac{1}{2^{2}}) (1 - \frac{1}{3^{2}}) (1 - \frac{1}{5^{2}}) . . . = 1$ $\frac{π^{2}}{6} = \frac{2^{2}}{2^{2} - 1} \cdot \frac{3^{2}}{3^{2} - 1} \cdot \frac{5^{2}}{5^{2} - 1} \cdot \frac{7^{2}}{7^{2} - 1} \cdot . . .$ $\frac{π^{2}}{6} = \frac{4}{3} \cdot \frac{9}{8} \cdot \frac{25}{24} \cdot \frac{49}{48} \cdot . . . \Rightarrow \frac{4 π^{2}}{25} = \frac{4}{3} \cdot \frac{9}{8} \cdot \frac{49}{48} \cdot \frac{121}{120} \cdot . . .$
	Commented by Dwaipayan Shikari last updated on 10/Oct/20

	$G r e a t s i r!$ $S o g e n e r a l l y$ $ζ (s) = \underset{p}{\prod^{\infty}} {(1 - \frac{1}{p^{s}})}^{- 1} (p = p r i m e N u m b e r s)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum