TITLE: Limit of $a_{k+1}=\dfrac{a_k+b_k}{2}$, $b_{k+1}=\sqrt{a_kb_k}$?
QUESTION [7 upvotes]: Let $a,b>0$
and let $a_0=a$, $b_0=b $,
$a_{k+1}=\dfrac{a_k+b_k} 2$,$b_{k+1}=\sqrt{a_kb_k}$ 
 $\quad k\geq0$.
This converges  to a number between a and b. Also $a_k>b_k$ for $k\geq1$ (AM-GM inequality). Can we find the limit explicitly in terms of $a$ and $b$?

REPLY [4 votes]: The limit $L$ is known as arithmetic-geometric mean, see MathWorld entry, and can be expressed in terms of elliptic integrals:
$$ L = \frac{(a+b)\pi}{4K((a-b)/(a+b))}$$
where $K$ is a complete elliptic integral. This goes back to Legendre and Gauss.