Question Number 103260 by 675480065 last updated on 13/Jul/20
Answered by Rio Michael last updated on 13/Jul/20
$$\:\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{1}}\\{{u}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}{u}_{{n}} {v}_{{n}} }{{u}_{{n}} +\:{v}_{{n}} }}\end{cases}\:\mathrm{and}\:\begin{cases}{{v}_{\mathrm{0}} \:=\:\mathrm{2}}\\{{v}_{{n}+\mathrm{1}\:} =\:\frac{{u}_{{n}} \:+\:{v}_{{n}} }{\mathrm{2}}}\end{cases}\:\:\forall\:{n}\:\in\:\mathbb{N} \\ $$$$\left(\mathrm{a}\right)\:{n}\:\in\:\mathbb{N}\:\Rightarrow\:\mathrm{2}{u}_{{n}} {v}_{{n}} \:>\:\mathrm{0}\:\mathrm{and}{u}_{{n}} \:+\:{v}_{{n}} \:>\:\mathrm{0}\:\mathrm{since}\:{u}_{{n}} \:\mathrm{and}\:{v}_{{n}} \:>\:\mathrm{0} \\ $$$$\mathrm{hence}\:{u}_{{n}} \:\mathrm{and}\:{v}_{{n}} \:\mathrm{are}\:\mathrm{strickly}\:\mathrm{positive}\:\mathrm{sequences}. \\ $$$$\left(\mathrm{b}\right)\:{w}_{{n}} \:=\:{v}_{{n}} −{u}_{{n}} \:\Rightarrow\:{w}_{{n}+\mathrm{1}} \:=\:{v}_{{n}+\mathrm{1}} −{u}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:{w}_{{n}+\mathrm{1}} \:=\:\frac{{u}_{{n}} \:+\:{v}_{{n}} }{\mathrm{2}}−\frac{\mathrm{2}{u}_{{n}} {v}_{{n}} }{{u}_{{n}} +\:{v}_{{n}} }\:=\:\frac{{u}_{{n}} ^{\mathrm{2}} \:+\:\mathrm{2}{u}_{{n}} {v}_{{n}} \:+\:{v}_{{n}} ^{\mathrm{2}} \:−\mathrm{4}{u}_{{n}} {v}_{{n}} }{\mathrm{2}\left({u}_{{n}} \:+\:{v}_{{n}} \right)}\:=\:\frac{{u}_{{n}} ^{\mathrm{2}} \:\:−\mathrm{2}{u}_{{n}} {v}_{{n}} +\:{v}_{{n}} ^{\mathrm{2}} }{\mathrm{2}\left({u}_{{n}} \:+\:{v}_{{n}} \right)} \\ $$$$\Rightarrow\:{w}_{{n}+\mathrm{1}} \:=\:\frac{\left({u}_{{n}} −{v}_{{n}} \right)^{\mathrm{2}} }{\mathrm{2}\left({u}_{{n}\:} +\:{v}_{{n}} \right)}\:\Rightarrow\:{w}_{{n}+\mathrm{1}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{n}} −{u}_{{n}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{w}_{{n}} \\ $$$$\mathrm{hence}\:\mathrm{0}\:\leqslant\:{w}_{{n}+\mathrm{1}} \leqslant\:\frac{\mathrm{1}}{\mathrm{2}}{w}_{{n}} \\ $$$$\left(\mathrm{c}\right)\underset{{n}\rightarrow\infty} {\mathrm{lim}}{w}_{{n}} \:=\:\mathrm{0} \\ $$$$ \\ $$