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).


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Question Number 165081 by mathocean1 last updated on 25/Jan/22
$Given a; b >0 and ∀ n ∈ N, u_n ;v_n >0 . { ((u_0 =a)),((u_(n+1) =(√(u_n v_n )))) :} and { ((v_0 =b)),((v_(n+1) =(1/2)(u_n +v_n ).)) :} Show that the sequences u_n and u_n are convergent and have the same limit l (l is called the arithmetico−geometric limit).$
$G i v e n a; b > 0 a n d \forall n \in N, u_{n}; v_{n} > 0 .$ ${\begin{cases} u_{0} = a \\ u_{n + 1} = \sqrt{u_{n} v_{n}} \end{cases} a n d {\begin{cases} v_{0} = b \\ v_{n + 1} = \frac{1}{2} (u_{n} + v_{n}) . \end{cases}$ $S h o w t h a t t h e s e q u e n c e s u_{n} a n d u_{n}$ $a r e c o n v e r g e n t a n d h a v e t h e s a m e$ $l i m i t l (l i s c a l l e d t h e a r i t h m e t i c o - g e o m e t r i c l i m i t) .$
	Answered by aleks041103 last updated on 27/Jan/22
	$we know (√(ab))≤(1/2)(a+b) ⇒0<u_1 ≤v_1 also (√(u_k v_k ))=u_(k+1) ≤v_(k+1) =(1/2)(v_k +u_k ) suppose u_k ≤v_k (true if v_k ,u_k ≥0) ⇒u_(k+1) = (√(v_k u_k ))≥(√(u_k u_k ))=u_k ⇒u_(k+1) ≥u_k ⇒u_k ≥u_1 >0⇒v_k ,u_k >0 ⇒u_k ≤v_k is true ⇒u_(k+1) =(√(v_k u_k ))≤(√(v_k v_k ))=v_k ⇒u_(k+1) ≤v_k Now v_(k+1) =(1/2)(v_k +u_k )≥(1/2)(u_k +u_k )=u_k ⇒v_(k+1) ≥u_k v_(k+1) =(1/2)(v_k +u_k )≤(1/2)(v_k +v_k )=v_k ⇒v_(k+1) ≥v_k From all of this we have: u_1 ≤u_k ≤u_(k+1) ≤v_k ≤v_1 ⇒u_1 ≤u_k ≤v_1 ⇒{u_n } is bounded. also u_(k+1) ≥u_k ⇒{u_n } is increasing ⇒{u_n } converges. u_1 ≤u_k ≤v_(k+1) ≤v_k ≤v_1 ⇒u_1 ≤u_k ≤v_1 ⇒{v_n } is bounded. also v_(k+1) ≤v_k ⇒{v_n } is decreasing ⇒{v_n } converges. u_(k+1) =(√(u_k v_k )) ⇒lim_(k→∞) (u_(k+1) /u_k )=(√((lim_(k→∞) v_k )/(lim_(k→∞) u_k ))) since {u_k } converges, lim_(k→∞) (u_(k+1) /u_k )=1 ⇒lim_(k→∞) v_k =lim_(k→∞) u_k =l$
	$w e k n o w$ $\sqrt{a b} ⩽ \frac{1}{2} (a + b)$ $\Rightarrow 0 < u_{1} ⩽ v_{1}$ $a l s o$ $\sqrt{u_{k} v_{k}} = u_{k + 1} ⩽ v_{k + 1} = \frac{1}{2} (v_{k} + u_{k})$ $s u p p o s e u_{k} ⩽ v_{k} (t r u e i f v_{k}, u_{k} ⩾ 0)$ $\Rightarrow u_{k + 1} = \sqrt{v_{k} u_{k}} ⩾ \sqrt{u_{k} u_{k}} = u_{k}$ $\Rightarrow u_{k + 1} ⩾ u_{k}$ $\Rightarrow u_{k} ⩾ u_{1} > 0 \Rightarrow v_{k}, u_{k} > 0$ $\Rightarrow u_{k} ⩽ v_{k} i s t r u e$ $\Rightarrow u_{k + 1} = \sqrt{v_{k} u_{k}} ⩽ \sqrt{v_{k} v_{k}} = v_{k}$ $\Rightarrow u_{k + 1} ⩽ v_{k}$ $N o w$ $v_{k + 1} = \frac{1}{2} (v_{k} + u_{k}) ⩾ \frac{1}{2} (u_{k} + u_{k}) = u_{k}$ $\Rightarrow v_{k + 1} ⩾ u_{k}$ $v_{k + 1} = \frac{1}{2} (v_{k} + u_{k}) ⩽ \frac{1}{2} (v_{k} + v_{k}) = v_{k}$ $\Rightarrow v_{k + 1} ⩾ v_{k}$ $F r o m a l l o f t h i s w e h a v e :$ $u_{1} ⩽ u_{k} ⩽ u_{k + 1} ⩽ v_{k} ⩽ v_{1}$ $\Rightarrow u_{1} ⩽ u_{k} ⩽ v_{1} \Rightarrow {u_{n}} i s b o u n d e d .$ $a l s o u_{k + 1} ⩾ u_{k} \Rightarrow {u_{n}} i s i n c r e a s i n g$ $\Rightarrow {u_{n}} c o n v e r g e s .$ $u_{1} ⩽ u_{k} ⩽ v_{k + 1} ⩽ v_{k} ⩽ v_{1}$ $\Rightarrow u_{1} ⩽ u_{k} ⩽ v_{1} \Rightarrow {v_{n}} i s b o u n d e d .$ $a l s o v_{k + 1} ⩽ v_{k} \Rightarrow {v_{n}} i s d e c r e a s i n g$ $\Rightarrow {v_{n}} c o n v e r g e s .$ $u_{k + 1} = \sqrt{u_{k} v_{k}}$ $Double subscripts: use braces to clarify$ $s i n c e {u_{k}} c o n v e r g e s, \underset{k \to \infty}{l i m} \frac{u_{k + 1}}{u_{k}} = 1$ $Double subscripts: use braces to clarify$
	Commented bymathocean1 last updated on 27/Jan/22

	$t h a n k y o u v e r y m u c h .$
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