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Question Number 196828 by ERLY last updated on 01/Sep/23

Answered by Skabetix last updated on 01/Sep/23

$$2.a)U_{n+1}=\frac{1}{2}\times \frac{1}{U_n}$$$$Comme{U}_n>0\to U_n⩾\frac{1}{U_n}\to U_{n+1}⩽\frac{1}{2}{U}_n$$

Answered by a.lgnaoui last updated on 01/Sep/23

$$\bullet 1\mathrm{la}\mathrm{suute}\mathrm{est}\mathrm{vrai}\mathrm{pour}\mathrm{n}=0({\mathrm{u}}_0=1>0)$$$$\mathrm{pour}\mathrm{n}=1{\mathrm{u}}_1=\frac{{\mathrm{u}}_0}{2}=\frac{1}{2}>0$$$$\mathrm{suposons}\mathrm{qu}\mathrm{il}\mathrm{esr}\mathrm{vrai}\mathrm{et}\mathrm{montrons}$$$$\mathrm{qu}\mathrm{elle}\mathrm{est}\mathrm{vrai}\mathrm{pour}\mathbf{n}+1$$$${\mathrm{u}}_{\mathrm{n}+1}=\frac{{\mathrm{u}}_{\mathrm{n}}}{2+{\mathrm{u}}_{\mathrm{n}}^2}=\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{u}}_{\mathrm{n}}(\frac{2}{{\mathrm{u}}_{\mathrm{n}}}+{\mathrm{u}}_{\mathrm{n}})}=\frac{1}{{\mathrm{u}}_{\mathrm{n}}+\frac{2}{{\mathrm{u}}_{\mathrm{n}}}}$$$${\mathrm{u}}_{\mathrm{n}}>0\Rightarrow \frac{2}{{\mathrm{u}}_{\mathrm{n}}}>0\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}>0\mathrm{our}\mathrm{tout}\mathrm{n}\in \mathbb{N}$$$$\bullet 2\mathbf{a}){\mathrm{u}}_{\mathrm{n}+1}-\frac{1}{2}{\mathrm{u}}_{\mathrm{n}}=-\frac{-{\mathrm{u}}_{\mathrm{n}}^3}{2(2+{\mathrm{u}}_{\mathrm{n}}^2)}<0$$$$\mathrm{donc}{\mathbf{u}}_{\mathbf{n}+1}<\frac{{\mathbf{u}}_{\mathbf{n}}}{2}$$$$$$

Commented by a.lgnaoui last updated on 01/Sep/23

$$3\bullet \lim {\mathrm{u}}_{\mathrm{n}}(\mathrm{n}\to \mathrm{\infty })=\lim {\left(\frac{1}{2^{\mathrm{n}}}\right)}_{\mathrm{n}\to \mathrm{\infty }}=0$$$$\mathrm{de}\mathrm{plus}({\mathrm{u}}_{\mathrm{n}+1}-{\mathrm{u}}_{\mathrm{n}})<0\left(\mathrm{suite}\mathrm{decroissante}\right)$$$$\mathrm{donc}\mathrm{la}\mathrm{suitd}\left({\mathrm{u}}_{\mathrm{n}}\right)\mathrm{est}\mathrm{convervente}$$$$\mathrm{et}\mathrm{converge}\mathrm{vrrs}\left(0\right)$$$$$$

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