= Solution
The <arithmetic-geometric mean inequality> gives
$$
0<a_n\leq\sqrt{a_nb_n}=a_{n+1}\leq b_{n+1}=\frac{a_n+b_n}{2}\leq b_n.
$$
By induction these inequalities hold at every step, so $(a_n)$ is increasing and bounded above by $b$, while $(b_n)$ is decreasing and bounded below by $a$. The <bounded monotone sequence theorem> gives limits $\alpha,\beta$ with $0<a\leq\alpha\leq\beta\leq b$. Taking limits in the <arithmetic mean> recurrence gives $\beta=(\alpha+\beta)/2$, whence \b[$\boxed{\alpha=\beta}$]. Thus both <sequences> converge to the same positive limit, the <arithmetic-geometric mean iteration>'s limit. No elementary closed formula for that limit is needed.
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