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Question Number 113641 by ZiYangLee last updated on 14/Sep/20

$$\mathrm{Prove}\mathrm{that}\mathrm{there}\mathrm{exists}M>0\mathrm{such}\mathrm{that}$$ $$\mathrm{for}\mathrm{any}\mathrm{positive}\mathrm{integers}n,\mathrm{we}\mathrm{have}$$ $$\sqrt{1+\sqrt{2+\sqrt{...+\sqrt{n+1}}}}⩽M$$

Commented bymr W last updated on 14/Sep/20

$$A_n=\sqrt{1+\sqrt{2+\sqrt{3+...\sqrt{n}}}}$$ $$A_n>\sqrt{1+\sqrt{1+\sqrt{1+...\sqrt{1}}}}=C$$ $$1+C=C^2$$ $$C^2-C-1=0$$ $$C=\frac{1+\sqrt{5}}{2}$$ $$A_n<\sqrt{n+\sqrt{n+\sqrt{n+...\sqrt{n}}}}=D$$ $$n+D=D^2$$ $$D^2-D-n=0$$ $$D=\frac{1+\sqrt{1+4n}}{2}$$ $$\frac{1+\sqrt{5}}{2}<A_n<\frac{1+\sqrt{1+4n}}{2}$$ $$withM=\lceil \frac{1+\sqrt{1+4n}}{2}\rceil whichalwaysexists$$ $$A_n<M$$

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