Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 165081 by mathocean1 last updated on 25/Jan/22

Given a; b >0 and ∀ n ∈ N, u_n ;v_n >0 .    { ((u_0 =a)),((u_(n+1) =(√(u_n v_n )))) :} and  { ((v_0 =b)),((v_(n+1) =(1/2)(u_n +v_n ).)) :}  Show that the sequences u_n  and u_n   are convergent and have the same  limit l (l is called the arithmetico−geometric limit).

$${Given}\:{a};\:{b}\:>\mathrm{0}\:{and}\:\forall\:{n}\:\in\:\mathbb{N},\:{u}_{{n}} ;{v}_{{n}} >\mathrm{0}\:. \\ $$ $$\:\begin{cases}{{u}_{\mathrm{0}} ={a}}\\{{u}_{{n}+\mathrm{1}} =\sqrt{{u}_{{n}} {v}_{{n}} }}\end{cases}\:{and}\:\begin{cases}{{v}_{\mathrm{0}} ={b}}\\{{v}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{n}} +{v}_{{n}} \right).}\end{cases} \\ $$ $${Show}\:{that}\:{the}\:{sequences}\:{u}_{{n}} \:{and}\:{u}_{{n}} \\ $$ $${are}\:{convergent}\:{and}\:{have}\:{the}\:{same} \\ $$ $${limit}\:{l}\:\left({l}\:{is}\:{called}\:{the}\:{arithmetico}−{geometric}\:{limit}\right). \\ $$

Answered by aleks041103 last updated on 27/Jan/22

we know  (√(ab))≤(1/2)(a+b)  ⇒0<u_1 ≤v_1   also  (√(u_k v_k ))=u_(k+1) ≤v_(k+1) =(1/2)(v_k +u_k )  suppose u_k ≤v_k (true if v_k ,u_k ≥0)  ⇒u_(k+1) = (√(v_k u_k ))≥(√(u_k u_k ))=u_k   ⇒u_(k+1) ≥u_k   ⇒u_k ≥u_1 >0⇒v_k ,u_k >0  ⇒u_k ≤v_k  is true  ⇒u_(k+1) =(√(v_k u_k ))≤(√(v_k v_k ))=v_k   ⇒u_(k+1) ≤v_k   Now  v_(k+1) =(1/2)(v_k +u_k )≥(1/2)(u_k +u_k )=u_k   ⇒v_(k+1) ≥u_k   v_(k+1) =(1/2)(v_k +u_k )≤(1/2)(v_k +v_k )=v_k   ⇒v_(k+1) ≥v_k     From all of this we have:  u_1 ≤u_k ≤u_(k+1) ≤v_k ≤v_1   ⇒u_1 ≤u_k ≤v_1 ⇒{u_n } is bounded.  also u_(k+1) ≥u_k ⇒{u_n } is increasing  ⇒{u_n } converges.  u_1 ≤u_k ≤v_(k+1) ≤v_k ≤v_1   ⇒u_1 ≤u_k ≤v_1 ⇒{v_n } is bounded.  also v_(k+1) ≤v_k ⇒{v_n } is decreasing  ⇒{v_n } converges.    u_(k+1) =(√(u_k v_k ))  ⇒lim_(k→∞) (u_(k+1) /u_k )=(√((lim_(k→∞) v_k )/(lim_(k→∞) u_k )))  since {u_k } converges, lim_(k→∞) (u_(k+1) /u_k )=1  ⇒lim_(k→∞) v_k =lim_(k→∞) u_k =l

$${we}\:{know} \\ $$ $$\sqrt{{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right) \\ $$ $$\Rightarrow\mathrm{0}<{u}_{\mathrm{1}} \leqslant{v}_{\mathrm{1}} \\ $$ $${also} \\ $$ $$\sqrt{{u}_{{k}} {v}_{{k}} }={u}_{{k}+\mathrm{1}} \leqslant{v}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{u}_{{k}} \right) \\ $$ $${suppose}\:{u}_{{k}} \leqslant{v}_{{k}} \left({true}\:{if}\:{v}_{{k}} ,{u}_{{k}} \geqslant\mathrm{0}\right) \\ $$ $$\Rightarrow{u}_{{k}+\mathrm{1}} =\:\sqrt{{v}_{{k}} {u}_{{k}} }\geqslant\sqrt{{u}_{{k}} {u}_{{k}} }={u}_{{k}} \\ $$ $$\Rightarrow{u}_{{k}+\mathrm{1}} \geqslant{u}_{{k}} \\ $$ $$\Rightarrow{u}_{{k}} \geqslant{u}_{\mathrm{1}} >\mathrm{0}\Rightarrow{v}_{{k}} ,{u}_{{k}} >\mathrm{0} \\ $$ $$\Rightarrow{u}_{{k}} \leqslant{v}_{{k}} \:{is}\:{true} \\ $$ $$\Rightarrow{u}_{{k}+\mathrm{1}} =\sqrt{{v}_{{k}} {u}_{{k}} }\leqslant\sqrt{{v}_{{k}} {v}_{{k}} }={v}_{{k}} \\ $$ $$\Rightarrow{u}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \\ $$ $${Now} \\ $$ $${v}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{u}_{{k}} \right)\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{k}} +{u}_{{k}} \right)={u}_{{k}} \\ $$ $$\Rightarrow{v}_{{k}+\mathrm{1}} \geqslant{u}_{{k}} \\ $$ $${v}_{{k}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{u}_{{k}} \right)\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({v}_{{k}} +{v}_{{k}} \right)={v}_{{k}} \\ $$ $$\Rightarrow{v}_{{k}+\mathrm{1}} \geqslant{v}_{{k}} \\ $$ $$ \\ $$ $${From}\:{all}\:{of}\:{this}\:{we}\:{have}: \\ $$ $${u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{u}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \leqslant{v}_{\mathrm{1}} \\ $$ $$\Rightarrow{u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{v}_{\mathrm{1}} \Rightarrow\left\{{u}_{{n}} \right\}\:{is}\:{bounded}. \\ $$ $${also}\:{u}_{{k}+\mathrm{1}} \geqslant{u}_{{k}} \Rightarrow\left\{{u}_{{n}} \right\}\:{is}\:{increasing} \\ $$ $$\Rightarrow\left\{{u}_{{n}} \right\}\:{converges}. \\ $$ $${u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{v}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \leqslant{v}_{\mathrm{1}} \\ $$ $$\Rightarrow{u}_{\mathrm{1}} \leqslant{u}_{{k}} \leqslant{v}_{\mathrm{1}} \Rightarrow\left\{{v}_{{n}} \right\}\:{is}\:{bounded}. \\ $$ $${also}\:{v}_{{k}+\mathrm{1}} \leqslant{v}_{{k}} \Rightarrow\left\{{v}_{{n}} \right\}\:{is}\:{decreasing} \\ $$ $$\Rightarrow\left\{{v}_{{n}} \right\}\:{converges}. \\ $$ $$ \\ $$ $${u}_{{k}+\mathrm{1}} =\sqrt{{u}_{{k}} {v}_{{k}} } \\ $$ $$\Rightarrow\underset{{k}\rightarrow\infty} {{lim}}\frac{{u}_{{k}+\mathrm{1}} }{{u}_{{k}} }=\sqrt{\frac{\underset{{k}\rightarrow\infty} {{lim}v}_{{k}} }{\underset{{k}\rightarrow\infty} {{lim}u}_{{k}} }} \\ $$ $${since}\:\left\{{u}_{{k}} \right\}\:{converges},\:\underset{{k}\rightarrow\infty} {{lim}}\frac{{u}_{{k}+\mathrm{1}} }{{u}_{{k}} }=\mathrm{1} \\ $$ $$\Rightarrow\underset{{k}\rightarrow\infty} {{lim}v}_{{k}} =\underset{{k}\rightarrow\infty} {{lim}u}_{{k}} ={l} \\ $$

Commented bymathocean1 last updated on 27/Jan/22

thank you very much.

$${thank}\:{you}\:{very}\:{much}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com