study-the-sequence-u-0-1-and-u-n-1-1-1-u-n-2-

Question Number 52671 by maxmathsup by imad last updated on 11/Jan/19

s t u d y t h e s e q u e n c e u_{0} = 1 a n d u_{n + 1} = \frac{1}{1 + u_{n}^{2}}

Commented by maxmathsup by imad last updated on 11/Mar/19

i t s c l e a r t h a t u_{n} > 0 \forall n l e t p r o v e t h a t 0 < u_{n} ⩽ 1

n = 0 \Rightarrow 0 < u_{o} ⩽ 1 (t r u e) l e t s u p p o s e 0 < u_{n} ⩽ 1 \Rightarrow u_{n + 1} - 1 = \frac{1}{1 + u_{n}^{2}} - 1

= - \frac{u_{n}^{2}}{1 + u_{n}^{2}} ⩽ 0 \Rightarrow 0 < u_{n + 1} ⩽ 1 w e h a v e

u_{n + 1} = f (u_{n}) w i t h f (x) = \frac{1}{1 + x^{2}} (w e c a n t a k e x > 0) \Rightarrow f^{^{'}} (x) = - \frac{2 x}{1 + x^{2}} < 0 \Rightarrow

f i s d e c r e s i n g

x 0 + \infty

f^{^{'}} (x) -

f (x) 1 d e c r e 0 i s f h a v e a f i x e d p o i n t

f (x) = x \Rightarrow \frac{1}{1 + x^{2}} = x \Rightarrow 1 = x + x^{3} \Rightarrow x^{3} + x - 1 = 0

l e t p (x) = x^{3} + x - 1 w e h a v e p^{^{'}} (x) = 3 x^{2} + 1 > 0 \Rightarrow p i s i n c r e s s i n g

p (0) = - 1 < 0 a n d p (1) = 1 > 0 \Rightarrow \exists! α_{0} \in] 0, 1 [/ p (α_{0}) = 0 a n d α i s t h e f i x e d

p o i n t \Rightarrow α_{0} = {l i m}_{n \to + \infty} u_{n}