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Question Number 208445 by alcohol last updated on 16/Jun/24

u_{n + 1} = u_{n} - u_{n}^{3}, u_{0} \in] 0, 1 [

v_{n} = \frac{1}{u_{n + 1}^{2}} - \frac{1}{u_{n}^{2}} = f (u_{n}^{2}); f (x) = \frac{2 - x}{{(1 - x)}^{2}}

v_{n} c o n v e r g e s t o 2, v_{n} i s d e c r e a s i n g

. s h o w t h a t v_{n} ⩾ 2

x_{n} = \frac{1}{n + 1} \underset{m = 0}{\sum^{m}} (v_{m})

. s h o w t h a t x_{0} ⩾ x_{n} ⩾ v_{n}

. s h o w t h a t x_{n} i s d e c r e a s i n g a n d \lim_{n \to \infty} x_{n} = l ⩾ 2

. s h o w t h a t 2 x_{n + 1} - x_{n} ⩽ v_{n + 1} a n d d e d u c e l

. e x p r e s s x_{n + 1} - x_{n} i n t e r m s o f u_{n}

. d e d u c e \lim_{n \to \infty} {n u}_{n}^{2}

Answered by Berbere last updated on 16/Jun/24

u_{n + 1} = u_{n} - u_{n}^{3} = g (u_{n})

g (x) = x - x^{3} \Rightarrow g^{'} = 1 - 3 x^{2}

g (0) = g (1) \Rightarrow g (x) \in [0, g (\frac{1}{\sqrt{3}})] = [0 \frac{2}{3 \sqrt{3}}] \subset [0, 1]

u_{n} = g^{n} (u_{0}) \subset [0, 1] \Rightarrow \forall n \in N U_{n} \in [0, 1]

u_{n + 1} - u_{n} = - u_{n}^{3} < 0 \Rightarrow u_{n} d e c r e a s e B o u n d e d C v

\lim_{n \to \infty} u_{n} = l; s o l u t i o n o f g (x) = x \Rightarrow l = 0

v_{n} = \frac{1}{u_{n + 1}^{2}} - \frac{1}{u_{n}^{2}} = \frac{1}{u_{n}^{2} {(1 - u_{n}^{2})}^{2}} - \frac{1}{u_{n}^{2}}

u_{n}^{2} = w_{n} \Rightarrow v_{n} = \frac{1}{w_{n}} [\frac{1 - {(1 - w_{n})}^{2}}{{(1 - w_{n})}^{2}}] = \frac{2 w_{n} - w_{n}^{2}}{w_{n} {(1 - w_{n})}^{2}} = \frac{2 - w_{n}}{{(1 - w_{n})}^{2}}

= f (w_{n}) = f (u_{n}^{2})

\lim_{n \to \infty} V_{n} = \lim_{n \to \infty} f (u_{n}^{2}) = f (\lim_{n \to \infty} u_{n}^{2}) = f (0) = 2

b y c o n t i n u i t y f i n 2

f (x) = \frac{2 - x}{{(1 - x)}^{2}} = \frac{- {(1 - x)}^{2} + 2 (1 - x) (2 - x)}{{(1 - x)}^{4}} =

\frac{(1 - x) (- (1 - x) + 2 (2 - x)}{{(1 - x)}^{4}} = \frac{(1 - x) (3 - x)}{{(1 - x)}^{4}} > 0

f (u_{n}^{2}) \in] 2, \infty [\Rightarrow v_{n} > 2

u_{n + 1} < u_{n} \Rightarrow f (u_{n + 1}^{2}) < f (u_{n}^{2}) \Rightarrow v_{n + 1} < v_{n}; v_{n} d e c r e a s e

x_{n} = \frac{1}{n + 1} \underset{k = 0}{\sum^{n}} v_{m}; \forall k \in [0, n + 1]

v_{n} ⩽ v_{k} ⩽ v_{0} \overset{v_{n} d e c r e a s e}{\Rightarrow} v_{n} ⩽ x_{n} ⩽ v_{0}

x_{n + 1} - x_{n} = \frac{1}{n + 2} \underset{k = 0}{\sum^{n + 1}} v_{k} - \frac{1}{n + 1} \underset{k = 0}{\sum^{n}} v_{k} = \frac{(n + 1) Σ v_{k} - (n + 2) Σ v_{k}}{(n + 2) (n + 1)}

= \frac{(n + 1) v_{n + 1} - \underset{k = 0}{\sum^{n}} v_{k}}{(n + 1) (n + 2)} = \frac{Σ (v_{n + 1} - v_{k})}{(n + 1) (n + 2)} < 0

\forall k \in [0, n] v_{k} > v_{n + 1} v_{n} d e c r e a s e

x_{n} d e c r e a s e b o u n d e d C v l; x_{n} ⩾ v_{n} \Rightarrow l ⩾ 2

2 x_{n + 1} - x_{n} = \frac{2}{n + 2} \underset{k = 0}{\sum^{n + 1}} v_{k} - \frac{1}{n + 1} \underset{k = 0}{\sum^{n}} v_{k}

= \frac{2 (n + 1) \sum^{n + 1} v_{k} - (n + 2) \underset{k = 0}{\sum^{n}} v_{k}}{(n + 2) (n + 1)} = \frac{2 v_{n + 1} + 2 {n v}_{n + 1} + n \underset{k = 0}{\sum^{n}} v_{k}}{}

⩾ \frac{(2 + 2 n) v_{n + 1} + n (n + 1) v_{n + 1}}{(n + 1) (n + 2)} ⩾ v_{n + 1}

n o t 2 x_{n + 1} - x_{n} < v_{n + 1}

l = 2 b y c e s a r o

x_{n} = \frac{1}{n + 1} (\underset{k = 0}{\sum^{n}} \frac{1}{u_{k + 1}^{2}} - \frac{1}{u_{k}^{2}}) = \frac{1}{n + 1} (\frac{1}{u_{n + 1}^{2}} - \frac{1}{u_{0}^{2}})

\Rightarrow \lim_{n \to \infty} x_{n} = 2 = \lim_{n \to \infty} \frac{1}{(n + 1) u_{n + 1}^{2}}

\Rightarrow \lim_{n \to \infty} (n + 1) u_{n + 1}^{2} = \frac{1}{2}

w_{n} = (n + 1) u_{n + 1}^{2} c v \Rightarrow \lim_{n \to \infty} W_{n - 1} = \lim_{n \to \infty} W_{n} = \frac{1}{2}

{n U}_{n}^{2} \to \frac{1}{2}

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