Question Number 136031 by mathmax by abdo last updated on 18/Mar/21
$$\mathrm{study}\:\mathrm{the}\:\mathrm{sequence}\:\:\mathrm{U}_{\mathrm{n}} =\sqrt{\frac{\mathrm{1}+\mathrm{u}_{\mathrm{n}−\mathrm{1}} }{\mathrm{2}}} \\ $$$$\mathrm{with}\:\mathrm{u}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{determine}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} \\ $$
Answered by mindispower last updated on 18/Mar/21
$${u}_{{n}} ^{\mathrm{2}} =\frac{\mathrm{1}+{u}_{{n}−\mathrm{1}} }{\mathrm{2}} \\ $$$${u}_{{n}} <\mathrm{1}\:…{easy}\:{to}\:{see} \\ $$$${u}_{\mathrm{0}} =\mathrm{0}.\mathrm{5}\leqslant\mathrm{1} \\ $$$${suppose}\:{u}_{{n}} <\mathrm{1}\:,\forall{n} \\ $$$${u}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+{u}_{{n}} }{\mathrm{2}}}<\sqrt{\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}\Leftrightarrow{u}_{{n}+\mathrm{1}} \leqslant\mathrm{1} \\ $$$${by}\:{reccursion}\:{u}_{{n}} \leqslant\mathrm{1} \\ $$$${u}_{{n}} ^{\mathrm{2}} −{u}_{{n}−\mathrm{1}} <{u}_{{n}} −{u}_{{n}−\mathrm{1}} \\ $$$${u}_{{n}} ^{\mathrm{2}} −{u}_{{n}−\mathrm{1}} =\frac{\mathrm{1}−{u}_{{n}−\mathrm{1}} }{\mathrm{2}_{} }>\mathrm{0},\Rightarrow{u}_{{n}} −{u}_{{n}−\mathrm{1}} >\mathrm{0} \\ $$$${u}_{{n}} {increase}\:{withe}\:{bounded}\:\mathrm{0}<{u}_{{n}} <\mathrm{1} \\ $$$$\Rightarrow{u}_{{n}} \:\:{conveege} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} ={l}\Rightarrow{l}=\sqrt{\frac{\mathrm{1}+{l}}{\mathrm{2}}}\Leftrightarrow\mathrm{2}{l}^{\mathrm{2}} −{l}−\mathrm{1} \\ $$$${l}\in\left\{\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$${l}=\mathrm{1}\:,{u}_{{n}} \in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\left[\right.\right. \\ $$
Commented by mathmax by abdo last updated on 18/Mar/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mind} \\ $$
Commented by mindispower last updated on 20/Mar/21
$${withe}\:{pleasur} \\ $$
Answered by mathmax by abdo last updated on 18/Mar/21
$$\mathrm{we}\:\mathrm{have}\:\mathrm{u}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)\:\Rightarrow\mathrm{u}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{2}}}=\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)\:\mathrm{let} \\ $$$$\mathrm{suppose}\:\mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}} }\right)\:\Rightarrow\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}} }\right)}{\mathrm{2}}} \\ $$$$=\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\right)\:\Rightarrow\forall\mathrm{n}\geqslant\mathrm{1}\:\:\mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\frac{\pi}{\mathrm{3}.\mathrm{2}^{\mathrm{n}} }\right)\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1} \\ $$