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Question Number 159403 by LEKOUMA last updated on 16/Nov/21

$$U_{n+1}=\frac{1}{2}(u_n+\frac{a}{u_n})with{u}_1>0,a>0$$ $$Provethat\frac{u_{n+1}}{u_n}⩽1$$

Commented bymathmax by abdo last updated on 18/Nov/21

$$\mathrm{i}\mathrm{think}\frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}⩾1...$$

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