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Show that if $a_1>b_1>0,a_{n+1}=\sqrt{a_nb_n}$ and $b_{n+1}=\frac{a_n+b_n}{2}$, then $a_n$ and $b_n$ both converge to a common limit.

It seems a bit intuitive but I am not able to get it in writing. I tried using Cauchy's test but it isn't helping(introducing more variables). Some help please. Thanks.


marked as duplicate by Arnaud D., Eevee Trainer, YiFan, Leucippus, Lord Shark the Unknown Mar 15 at 5:37

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One has $$a_1>b_2>b_3>b_4>\cdots>a_4>a_3>a_2>b_1.$$ So both $(a_n)$ and $(b_n)$ converge, to $A$ and $B$ say. Then $b_{n+1}=(a_n+b_n)/2\to (A+B)/2$. But $b_{n+1}\to B$.

The common limit is the arithmetic-geometric mean (AGM).


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