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Question Number 32290 by abdo imad last updated on 22/Mar/18

let give u_0 =1 and u_(n+1) =(√(1+(√u_n )))  prove that u_n  is  increasing .

$${let}\:{give}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}+\mathrm{1}} =\sqrt{\mathrm{1}+\sqrt{{u}_{{n}} }}\:\:{prove}\:{that}\:{u}_{{n}} \:{is} \\ $$$${increasing}\:. \\ $$

Commented by prof Abdo imad last updated on 04/Apr/18

first let prove that  u_n >0  for n=0  u_0 =1>0 let suppose u_n >0  ⇒1+(√(u_n  ))>0  ⇒ (√(1+(√u_n )  ))  >0 ⇒ u_(n+1)  >0  we have u_(n+1)  =f(u_n ) /f(x) =(√(1+(√x) )) .  f^′ (x) = ((1/(2(√x)))/(2(√(1+(√x) ))))) =  (1/(4(√x) (√(1+(√x)))))  >0 ⇒ f is  increazing  ⇒ (u_n ) is increazing .

$${first}\:{let}\:{prove}\:{that}\:\:{u}_{{n}} >\mathrm{0}\:\:{for}\:{n}=\mathrm{0} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}>\mathrm{0}\:{let}\:{suppose}\:{u}_{{n}} >\mathrm{0}\:\:\Rightarrow\mathrm{1}+\sqrt{{u}_{{n}} \:}>\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{\mathrm{1}+\sqrt{{u}_{{n}} }\:\:}\:\:>\mathrm{0}\:\Rightarrow\:{u}_{{n}+\mathrm{1}} \:>\mathrm{0} \\ $$$${we}\:{have}\:{u}_{{n}+\mathrm{1}} \:={f}\left({u}_{{n}} \right)\:/{f}\left({x}\right)\:=\sqrt{\mathrm{1}+\sqrt{{x}}\:}\:. \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\mathrm{2}\sqrt{\left.\mathrm{1}+\sqrt{{x}}\:\right)}}\:=\:\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}\:\sqrt{\mathrm{1}+\sqrt{{x}}}}\:\:>\mathrm{0}\:\Rightarrow\:{f}\:{is} \\ $$$${increazing}\:\:\Rightarrow\:\left({u}_{{n}} \right)\:{is}\:{increazing}\:. \\ $$$$ \\ $$

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