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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).

Givena;b>0andnN,un;vn>0. {u0=aun+1=unvnand{v0=bvn+1=12(un+vn). Showthatthesequencesunandun areconvergentandhavethesame limitl(liscalledthearithmeticogeometriclimit).

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

weknow ab12(a+b) 0<u1v1 also ukvk=uk+1vk+1=12(vk+uk) supposeukvk(trueifvk,uk0) uk+1=vkukukuk=uk uk+1uk uku1>0vk,uk>0 ukvkistrue uk+1=vkukvkvk=vk uk+1vk Now vk+1=12(vk+uk)12(uk+uk)=uk vk+1uk vk+1=12(vk+uk)12(vk+vk)=vk vk+1vk Fromallofthiswehave: u1ukuk+1vkv1 u1ukv1{un}isbounded. alsouk+1uk{un}isincreasing {un}converges. u1ukvk+1vkv1 u1ukv1{vn}isbounded. alsovk+1vk{vn}isdecreasing {vn}converges. uk+1=ukvk Double subscripts: use braces to clarify since{uk}converges,limkuk+1uk=1 Double subscripts: use braces to clarify

Commented bymathocean1 last updated on 27/Jan/22

thank you very much.

thankyouverymuch.

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