study the sequence U_n =(√((1+u_(n−1) )/2)) with u_0 =(1/2) and determine lim_(n→+∞) U


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Question Number 136031 by mathmax by abdo last updated on 18/Mar/21

$study the sequence U_{n} = \sqrt{\frac{1 + u_{n - 1}}{2}}$ $with u_{0} = \frac{1}{2} and determine \lim_{n \to + \infty} U_{n}$
	Answered by mindispower last updated on 18/Mar/21
	$u_n ^2 =((1+u_(n−1) )/2) u_n <1 ...easy to see u_0 =0.5≤1 suppose u_n <1 ,∀n u_(n+1) =(√((1+u_n )/2))<(√((1+1)/2))=1⇔u_(n+1) ≤1 by reccursion u_n ≤1 u_n ^2 −u_(n−1) <u_n −u_(n−1) u_n ^2 −u_(n−1) =((1−u_(n−1) )/2_ )>0,⇒u_n −u_(n−1) >0 u_n increase withe bounded 0<u_n <1 ⇒u_n conveege lim_(n→∞) u_n =l⇒l=(√((1+l)/2))⇔2l^2 −l−1 l∈{1,−(1/2)} l=1 ,u_n ∈[(1/2),1[$
	$u_{n}^{2} = \frac{1 + u_{n - 1}}{2}$ $u_{n} < 1 . . . e a s y t o s e e$ $u_{0} = 0.5 ⩽ 1$ $s u p p o s e u_{n} < 1, \forall n$ $u_{n + 1} = \sqrt{\frac{1 + u_{n}}{2}} < \sqrt{\frac{1 + 1}{2}} = 1 \Leftrightarrow u_{n + 1} ⩽ 1$ $b y r e c c u r s i o n u_{n} ⩽ 1$ $u_{n}^{2} - u_{n - 1} < u_{n} - u_{n - 1}$ $u_{n}^{2} - u_{n - 1} = \frac{1 - u_{n - 1}}{2_{}} > 0, \Rightarrow u_{n} - u_{n - 1} > 0$ $u_{n} i n c r e a s e w i t h e b o u n d e d 0 < u_{n} < 1$ $\Rightarrow u_{n} c o n v e e g e$ $\lim_{n \to \infty} u_{n} = l \Rightarrow l = \sqrt{\frac{1 + l}{2}} \Leftrightarrow 2 l^{2} - l - 1$ $l \in {1, - \frac{1}{2}}$ $l = 1, u_{n} \in [\frac{1}{2}, 1 [$
	Commented by mathmax by abdo last updated on 18/Mar/21

	$thank you sir mind$
	Commented by mindispower last updated on 20/Mar/21

	$w i t h e p l e a s u r$
	Answered by mathmax by abdo last updated on 18/Mar/21

	$we have u_{0} = \frac{1}{2} = \cos (\frac{π}{3}) \Rightarrow u_{1} = \sqrt{\frac{1 + \cos (\frac{π}{3})}{2}} = \cos (\frac{π}{6}) let$ $suppose u_{n} = \cos (\frac{π}{3 . 2^{n}}) \Rightarrow u_{n + 1} = \sqrt{\frac{1 + \cos (\frac{π}{3 . 2^{n}})}{2}}$ $= \cos (\frac{π}{3 . 2^{n + 1}}) \Rightarrow \forall n ⩾ 1 u_{n} = \cos (\frac{π}{3 . 2^{n}}) \Rightarrow \lim_{n \to + \infty} u_{n} = \cos (0) = 1$
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