Question Number 165081 by mathocean1 last updated on 25/Jan/22
$${Given}\:{a};\:{b}\:>\mathrm{0}\:{and}\:\forall\:{n}\:\in\:\mathbb{N},\:{u}_{{n}} ;{v}_{{n}} >\mathrm{0}\:. \\ $$$$\:\begin{cases}{{u}_{\mathrm{0}} ={a}}\\{{u}_{{n}+\mathrm{1}} =\sqrt{{u}_{{n}} {v}_{{n}} }}\end{cases}\:{and}\:\begin{cases}{{v}_{\mathrm{0}} ={b}}\\{{v}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{n}} +{v}_{{n}} \right).}\end{cases} \\ $$$${Show}\:{that}\:{the}\:{sequences}\:{u}_{{n}} \:{and}\:{u}_{{n}} \\ $$$${are}\:{convergent}\:{and}\:{have}\:{the}\:{same} \\ $$$${limit}\:{l}\:\left({l}\:{is}\:{called}\:{the}\:{arithmetico}−{geometric}\:{limit}\right). \\ $$
Answered by aleks041103 last updated on 27/Jan/22
$${we}\:{know} \\ $$$$\sqrt{{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right) \\ $$$$\Rightarrow\mathrm{0}<{u}_{\mathrm{1}} \leqslant{v}_{\mathrm{1}} \\ $$$${also} \\ $$$$\sqrt{{u}_{{k}} {v}_{{k}} }={u}_{{k}+\mathrm{1}} \leqslant{v}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{u}_{{k}} \right) \\ $$$${suppose}\:{u}_{{k}} \leqslant{v}_{{k}} \left({true}\:{if}\:{v}_{{k}} ,{u}_{{k}} \geqslant\mathrm{0}\right) \\ $$$$\Rightarrow{u}_{{k}+\mathrm{1}} =\:\sqrt{{v}_{{k}} {u}_{{k}} }\geqslant\sqrt{{u}_{{k}} {u}_{{k}} }={u}_{{k}} \\ $$$$\Rightarrow{u}_{{k}+\mathrm{1}} \geqslant{u}_{{k}} \\ $$$$\Rightarrow{u}_{{k}} \geqslant{u}_{\mathrm{1}} >\mathrm{0}\Rightarrow{v}_{{k}} ,{u}_{{k}} >\mathrm{0} \\ $$$$\Rightarrow{u}_{{k}} \leqslant{v}_{{k}} \:{is}\:{true} \\ $$$$\Rightarrow{u}_{{k}+\mathrm{1}} =\sqrt{{v}_{{k}} {u}_{{k}} }\leqslant\sqrt{{v}_{{k}} {v}_{{k}} }={v}_{{k}} \\ $$$$\Rightarrow{u}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \\ $$$${Now} \\ $$$${v}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{u}_{{k}} \right)\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{k}} +{u}_{{k}} \right)={u}_{{k}} \\ $$$$\Rightarrow{v}_{{k}+\mathrm{1}} \geqslant{u}_{{k}} \\ $$$${v}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{u}_{{k}} \right)\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{v}_{{k}} \right)={v}_{{k}} \\ $$$$\Rightarrow{v}_{{k}+\mathrm{1}} \geqslant{v}_{{k}} \\ $$$$ \\ $$$${From}\:{all}\:{of}\:{this}\:{we}\:{have}: \\ $$$${u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{u}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \leqslant{v}_{\mathrm{1}} \\ $$$$\Rightarrow{u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{v}_{\mathrm{1}} \Rightarrow\left\{{u}_{{n}} \right\}\:{is}\:{bounded}. \\ $$$${also}\:{u}_{{k}+\mathrm{1}} \geqslant{u}_{{k}} \Rightarrow\left\{{u}_{{n}} \right\}\:{is}\:{increasing} \\ $$$$\Rightarrow\left\{{u}_{{n}} \right\}\:{converges}. \\ $$$${u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{v}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \leqslant{v}_{\mathrm{1}} \\ $$$$\Rightarrow{u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{v}_{\mathrm{1}} \Rightarrow\left\{{v}_{{n}} \right\}\:{is}\:{bounded}. \\ $$$${also}\:{v}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \Rightarrow\left\{{v}_{{n}} \right\}\:{is}\:{decreasing} \\ $$$$\Rightarrow\left\{{v}_{{n}} \right\}\:{converges}. \\ $$$$ \\ $$$${u}_{{k}+\mathrm{1}} =\sqrt{{u}_{{k}} {v}_{{k}} } \\ $$$$\Rightarrow\underset{{k}\rightarrow\infty} {{lim}}\frac{{u}_{{k}+\mathrm{1}} }{{u}_{{k}} }=\sqrt{\frac{\underset{{k}\rightarrow\infty} {{lim}v}_{{k}} }{\underset{{k}\rightarrow\infty} {{lim}u}_{{k}} }} \\ $$$${since}\:\left\{{u}_{{k}} \right\}\:{converges},\:\underset{{k}\rightarrow\infty} {{lim}}\frac{{u}_{{k}+\mathrm{1}} }{{u}_{{k}} }=\mathrm{1} \\ $$$$\Rightarrow\underset{{k}\rightarrow\infty} {{lim}v}_{{k}} =\underset{{k}\rightarrow\infty} {{lim}u}_{{k}} ={l} \\ $$
Commented by mathocean1 last updated on 27/Jan/22
$${thank}\:{you}\:{very}\:{much}. \\ $$