Question Number 32290 by abdo imad last updated on 22/Mar/18
$${let}\:{give}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}+\mathrm{1}} =\sqrt{\mathrm{1}+\sqrt{{u}_{{n}} }}\:\:{prove}\:{that}\:{u}_{{n}} \:{is} \\ $$$${increasing}\:. \\ $$
Commented by prof Abdo imad last updated on 04/Apr/18
$${first}\:{let}\:{prove}\:{that}\:\:{u}_{{n}} >\mathrm{0}\:\:{for}\:{n}=\mathrm{0} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}>\mathrm{0}\:{let}\:{suppose}\:{u}_{{n}} >\mathrm{0}\:\:\Rightarrow\mathrm{1}+\sqrt{{u}_{{n}} \:}>\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{\mathrm{1}+\sqrt{{u}_{{n}} }\:\:}\:\:>\mathrm{0}\:\Rightarrow\:{u}_{{n}+\mathrm{1}} \:>\mathrm{0} \\ $$$${we}\:{have}\:{u}_{{n}+\mathrm{1}} \:={f}\left({u}_{{n}} \right)\:/{f}\left({x}\right)\:=\sqrt{\mathrm{1}+\sqrt{{x}}\:}\:. \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\mathrm{2}\sqrt{\left.\mathrm{1}+\sqrt{{x}}\:\right)}}\:=\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}\:\sqrt{\mathrm{1}+\sqrt{{x}}}}\:\:>\mathrm{0}\:\Rightarrow\:{f}\:{is} \\ $$$${increazing}\:\:\Rightarrow\:\left({u}_{{n}} \right)\:{is}\:{increazing}\:. \\ $$$$ \\ $$