Π_(k=2) ^∞ (((k^3 −1)/(k^3 +1))) =?


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Question Number 156924 by cortano last updated on 17/Oct/21

$\underset{k = 2}{\prod^{\infty}} (\frac{k^{3} - 1}{k^{3} + 1}) = ?$
	Answered by puissant last updated on 17/Oct/21
	$P=Π_(k=2) ^∞ (((k^3 −1)/(k^3 +1))) = lim_(n→∞) Π_(k=2) ^n (((k^3 −1)/(k^3 +1))) = lim_(n→∞) Π_(k=2) ^n ((((k−1)(k^2 +k+1))/((k+1)(k^2 −k+1)))) =lim Π_(k=2) ^n (((k−1)/(k+1)))×lim Π_(k=2) ^n (((k^2 +k+1)/(k^2 −k+1)))_(n→∞) =lim_(n→∞) {(1/3)•(2/4)•(3/5)•...•((n−2)/n)•((n−1)/(n+1))}×lim_(n→∞) {(7/3)•((13)/7)•((21)/(13))•...•((n^2 +n+1)/(n^2 −n+1))} =lim_(n→∞) {(2/(n(n+1)))}×lim_(n→∞) {((n^2 +n+1)/3)}= (2/3)lim_(n→∞) {((n^2 (1+(1/n)+(1/n^2 )))/(n^2 (1+(1/n))))}=(2/3).. ∴∵ P = Π_(k=2) ^∞ (((k^3 −1)/(k^3 +1)))=(2/3)... ......................Le puissant..................$
	$P = \underset{k = 2}{\prod^{\infty}} (\frac{k^{3} - 1}{k^{3} + 1}) = \lim_{n \to \infty} \underset{k = 2}{\prod^{n}} (\frac{k^{3} - 1}{k^{3} + 1})$ $= \lim_{n \to \infty} \underset{k = 2}{\prod^{n}} (\frac{(k - 1) (k^{2} + k + 1)}{(k + 1) (k^{2} - k + 1)}) \underset{n \to \infty}{= \lim \underset{k = 2}{\prod^{n}} (\frac{k - 1}{k + 1}) \times \lim \underset{k = 2}{\prod^{n}} (\frac{k^{2} + k + 1}{k^{2} - k + 1})}$ $= \lim_{n \to \infty} {\frac{1}{3} ∙ \frac{2}{4} ∙ \frac{3}{5} ∙ . . . ∙ \frac{n - 2}{n} ∙ \frac{n - 1}{n + 1}} \times \lim_{n \to \infty} {\frac{7}{3} ∙ \frac{13}{7} ∙ \frac{21}{13} ∙ . . . ∙ \frac{n^{2} + n + 1}{n^{2} - n + 1}}$ $= \lim_{n \to \infty} {\frac{2}{n (n + 1)}} \times \lim_{n \to \infty} {\frac{n^{2} + n + 1}{3}} = \frac{2}{3} \lim_{n \to \infty} {\frac{n^{2} (1 + \frac{1}{n} + \frac{1}{n^{2}})}{n^{2} (1 + \frac{1}{n})}} = \frac{2}{3} . .$ $∴∵ P = \underset{k = 2}{\prod^{\infty}} (\frac{k^{3} - 1}{k^{3} + 1}) = \frac{2}{3} . . .$ $. . . . . . . . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . . . . .$
	Commented by mnjuly1970 last updated on 17/Oct/21

	$v e r y n i c e s i r p u i s s a n t . .$
	Commented by puissant last updated on 17/Oct/21

	$t h a n k s s i r M n j u l y . . .$
	Commented by Tawa11 last updated on 17/Oct/21

	$great sir$
	Commented by cortano last updated on 18/Oct/21

	$y e s g a v e k u d o s$
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