study-the-sequence-U-n-1-u-n-1-2-with-u-0-1-2-and-determine-lim-n-U-n-

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} = \frac{1 + u_{n - 1}}{2}

u_{n} < 1 \dots 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