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Question Number 173352 by cortano1 last updated on 10/Jul/22

	Answered by mr W last updated on 10/Jul/22
	$x_(n+1) ^2 =((2x_n )/(x_n +1)) ((1/x_(n+1) ))^2 =(1/2)(1+(1/x_n )) let a_n =(1/x_n ) a_(n+1) ^2 =(1/2)(1+a_n ) 2a_(n+1) ^2 −1=a_n { ?! 2cos^2 α−1=cos 2α} let a_n =cos θ_n 2 cos^2 θ_(n+1) −1=cos θ_n cos 2θ_(n+1) =cos θ_n ⇒2θ_(n+1) =θ_n ⇒θ_n =(θ_(n−1) /2)=(θ_(n−2) /2^2 )=...=(θ_1 /2^(n−1) ) x_1 =(1/a_1 )=(1/(cos θ_1 ))=(√2) ⇒cos θ_1 =(1/( (√2))) ⇒θ_1 =(π/4) ⇒θ_n =(π/(4×2^(n−1) ))=(π/2^(n+1) ) ⇒x_n =(1/a_n )=(1/(cos θ_n ))=(1/(cos (π/2^(n+1) ))) Π_(n=1) ^∞ x_n =lim_(n→∞) ((1/(cos (π/2^(n+1) ) cos (π/2^n ) ... cos (π/2^2 )))) =lim_(n→∞) (((2 sin (π/2^(n+1) ))/(2 sin (π/2^(n+1) ) cos (π/2^(n+1) ) cos (π/2^n ) ... cos (π/2^2 )))) =lim_(n→∞) (((2 sin (π/2^(n+1) ))/(sin (π/2^n ) cos (π/2^n ) ... cos (π/2^2 )))) =lim_(n→∞) (((2^2 sin (π/2^(n+1) ))/(sin (π/2^(n−1) ) cos (π/2^(n−1) ) ... cos (π/2^2 )))) ... =lim_(n→∞) (((2^(n−1) sin (π/2^(n+1) ))/(sin (π/2^2 ) cos (π/2^2 )))) =lim_(n→∞) (((2^n sin (π/2^(n+1) ))/(sin (π/2)))) =lim_(n→∞) (2^n sin (π/2^(n+1) )) =(π/2)lim_(n→∞) (((sin (π/2^(n+1) ))/(π/2^(n+1) ))) =(π/2)lim_(t→0) (((sin t)/t)) =(π/2) ✓$
	$x_{n + 1}^{2} = \frac{2 x_{n}}{x_{n} + 1}$ ${(\frac{1}{x_{n + 1}})}^{2} = \frac{1}{2} (1 + \frac{1}{x_{n}})$ $l e t a_{n} = \frac{1}{x_{n}}$ $a_{n + 1}^{2} = \frac{1}{2} (1 + a_{n})$ $2 a_{n + 1}^{2} - 1 = a_{n} {?! {2 \cos}^{2} α - 1 = \cos 2 α}$ $l e t a_{n} = \cos θ_{n}$ $2 \cos^{2} θ_{n + 1} - 1 = \cos θ_{n}$ $\cos 2 θ_{n + 1} = \cos θ_{n}$ $\Rightarrow 2 θ_{n + 1} = θ_{n}$ $\Rightarrow θ_{n} = \frac{θ_{n - 1}}{2} = \frac{θ_{n - 2}}{2^{2}} = . . . = \frac{θ_{1}}{2^{n - 1}}$ $x_{1} = \frac{1}{a_{1}} = \frac{1}{\cos θ_{1}} = \sqrt{2} \Rightarrow \cos θ_{1} = \frac{1}{\sqrt{2}} \Rightarrow θ_{1} = \frac{π}{4}$ $\Rightarrow θ_{n} = \frac{π}{4 \times 2^{n - 1}} = \frac{π}{2^{n + 1}}$ $\Rightarrow x_{n} = \frac{1}{a_{n}} = \frac{1}{\cos θ_{n}} = \frac{1}{\cos \frac{π}{2^{n + 1}}}$ $\underset{n = 1}{\prod^{\infty}} x_{n} = \lim_{n \to \infty} (\frac{1}{\cos \frac{π}{2^{n + 1}} \cos \frac{π}{2^{n}} . . . \cos \frac{π}{2^{2}}})$ $= \lim_{n \to \infty} (\frac{2 \sin \frac{π}{2^{n + 1}}}{2 \sin \frac{π}{2^{n + 1}} \cos \frac{π}{2^{n + 1}} \cos \frac{π}{2^{n}} . . . \cos \frac{π}{2^{2}}})$ $= \lim_{n \to \infty} (\frac{2 \sin \frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{n}} \cos \frac{π}{2^{n}} . . . \cos \frac{π}{2^{2}}})$ $= \lim_{n \to \infty} (\frac{2^{2} \sin \frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{n - 1}} \cos \frac{π}{2^{n - 1}} . . . \cos \frac{π}{2^{2}}})$ $. . .$ $= \lim_{n \to \infty} (\frac{2^{n - 1} \sin \frac{π}{2^{n + 1}}}{\sin \frac{π}{2^{2}} \cos \frac{π}{2^{2}}})$ $= \lim_{n \to \infty} (\frac{2^{n} \sin \frac{π}{2^{n + 1}}}{\sin \frac{π}{2}})$ $= \lim_{n \to \infty} (2^{n} \sin \frac{π}{2^{n + 1}})$ $= \frac{π}{2} \lim_{n \to \infty} (\frac{\sin \frac{π}{2^{n + 1}}}{\frac{π}{2^{n + 1}}})$ $= \frac{π}{2} \lim_{t \to 0} (\frac{\sin t}{t})$ $= \frac{π}{2} ✓$
	Commented by aleks041103 last updated on 10/Jul/22

	$T h a t i s n i c e . I s a w x_{n + 1} i n s t e a d o f x_{n} + 1$
	Commented by mr W last updated on 10/Jul/22

	$i s a w e x a c t l y t h e s a m e a s y o u a t t h e$ $b e g i n n i n g . l a t e r i r e a l i s e d s o m e t h i n g$ $m u s t b e w r o n g .$
	Commented by Tawa11 last updated on 11/Jul/22

	$Great sir$
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