Previous in Relation and Functions Next in Relation and Functions
Question Number 51982 by maxmathsup by imad last updated on 01/Jan/19
$${let}\:{u}_{{n}+\mathrm{1}} =\sqrt{\sum_{{k}=\mathrm{1}} ^{{n}} \:{u}_{{k}} }\:\:\:\:\:\:\:{with}\:{n}>\mathrm{0}\:\:\:{and}\:{u}_{\mathrm{1}} =\mathrm{1} \\ $$ $$\left.\mathrm{1}\right){calculate}\:{u}_{\mathrm{2}} ,{u}_{\mathrm{3}} ,{u}_{\mathrm{4}} {and}\:{u}_{\mathrm{5}} \\ $$ $$\left.\mathrm{2}\right){prove}\:{that}\:\:\forall{n}\geqslant\mathrm{2}\:\:\:\:\:{u}_{{n}+} ^{\mathrm{2}} ={u}_{{n}} ^{\mathrm{2}} \:+{u}_{{n}} \\ $$ $$\left.\mathrm{3}\right){study}\:{the}\:{variation}\:{of}\:{u}_{{n}} \\ $$ $$\left.\mathrm{4}\right){prove}\:{that}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} =+\infty \\ $$ $$\left.\mathrm{5}\right){prove}\:{that}\:{u}_{{n}+\mathrm{1}} \sim{u}_{{n}} \:\:\left({n}\rightarrow+\infty\right) \\ $$ $$\left.\mathrm{6}\right){let}\:{v}_{{n}} ={u}_{{n}+\mathrm{1}} −{u}_{{n}} \:\:{prove}\:{that}\:\left({v}_{{n}} \right)\:{converges}\:{and}\:{find}\:{its}\:{limit}. \\ $$