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Question Number 204334 by universe last updated on 13/Feb/24

Answered by Frix last updated on 13/Feb/24

$$n\to \mathrm{\infty }\Rightarrow a_n=\sqrt{2a_n}\Rightarrow a_n=2$$

Answered by MM42 last updated on 14/Feb/24

$$step1:$$$$a_{n+2}^2-a_{n+1}^2=a_{n+1}-a_{n-1}=\frac{a_{n+1}^2-a_{n-1}^2}{a_{n+1}+a_{n-1}}=\frac{a_{n-1}-a_{n-2}}{a_{n+1}+a_{n-1}}$$$$=\frac{a_{n-1}^2-a_{n-2}^2}{(a_{b+1}+a_{n-1})(a_{n-1}+a_{n-2})}$$$$=\frac{a_{n-2}-a_{n-3}}{(a_{n+1}+a_{n-1})(a_{n-1}+a_{n-2})}=...$$$$=\frac{a_2-a_1}{(a_{n+1}+a_{n-1})(a_{n-1}+a_{n-2})...(a_3+a_2)}>0$$$$\Rightarrow \left\{a_n\right\}\nearrow$$$$step2:$$$$a_1=0<2,a_2=1<2,a_3=1<2$$$$a_4=\sqrt{2}<2,a_5=\sqrt{2+\sqrt{2}}<2$$$$\mathrm{\forall }m<n\to a_m<2$$$$a_n^2=a_{n-2}+a_{n-1}<4\Rightarrow a_n<2✓$$$$\Rightarrow \left\{a_n\right\}isconvergent$$$$let{a}_n\to a$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}{a}_{n+2}={lim}_{n\to \mathrm{\infty }}\sqrt{a_n+a_{n+1}}$$$$\Rightarrow a=\sqrt{2a}\Rightarrow a=2✓$$$$$$

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