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

$$studythesequence{u}_0=1and{u}_{n+1}=\frac{1}{1+u_n^2}$$

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

$$itsclearthat{u}_n>0\mathrm{\forall }nletprovethat0<u_n⩽1$$$$n=0\Rightarrow 0<u_o⩽1\left(true\right)letsuppose0<u_n⩽1\Rightarrow u_{n+1}-1=\frac{1}{1+u_n^2}-1$$$$=-\frac{u_n^2}{1+u_n^2}⩽0\Rightarrow 0<u_{n+1}⩽1wehave$$$$u_{n+1}=f\left(u_n\right)withf\left(x\right)=\frac{1}{1+x^2}(wecantakex>0)\Rightarrow f^{{}^{\prime }}\left(x\right)=-\frac{2x}{1+x^2}<0\Rightarrow$$$$fisdecresing$$$$x0+\mathrm{\infty }$$$$f^{{}^{\prime }}\left(x\right)-$$$$f\left(x\right)1decre0isfhaveafixedpoint$$$$f\left(x\right)=x\Rightarrow \frac{1}{1+x^2}=x\Rightarrow 1=x+x^3\Rightarrow x^3+x-1=0$$$$letp\left(x\right)=x^3+x-1wehave{p}^{{}^{\prime }}\left(x\right)=3x^2+1>0\Rightarrow pisincressing$$$$p\left(0\right)=-1<0andp\left(1\right)=1>0\Rightarrow \mathrm{\exists }!{\alpha }_0\in ]0,1[/p\left({\alpha }_0\right)=0and\alpha isthefixed$$$$point\Rightarrow {\alpha }_0={lim}_{n\to +\mathrm{\infty }}{u}_n$$

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