The arithmetic–geometric mean

Ordering. For $x,y>0$, $\sqrt{xy}\le\frac{x+y}2$. So $b_{n+1}\le a_{n+1}$ for all $n\ge0$, i.e. $b_n\le a_n$ for all $n\ge1$.

Monotonicity. For $n\ge1$:
- $a_{n+1}=\frac{a_n+b_n}2\le a_n$, so $(a_n)_{n\ge1}$ is decreasing;
- $b_{n+1}=\sqrt{a_nb_n}\ge\sqrt{b_n^2}=b_n$, so $(b_n)_{n\ge1}$ is increasing.

Contracting gap. For $x\ge y>0$,
$$\frac{x+y}2-\sqrt{xy}=\frac{(\sqrt x-\sqrt y)^2}2\le\frac{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}2=\frac{x-y}2.$$
So $0\le a_{n+1}-b_{n+1}\le\frac12(a_n-b_n)$, hence $a_n-b_n\le2^{-n}(a_1-b_1)\to0$.

Conclusion. The sequences are adjacent, so they converge to a common limit $M(1,2)$, the arithmetic–geometric mean of $1$ and $2$.

Numerics.
- $n=1$: $a_1=1.5$, $b_1=1.41421$
- $n=2$: $a_2=1.45711$, $b_2=1.45648$
- $n=3$: $a_3=1.4567910$, $b_3=1.4567910$

So $M(1,2)\approx1.4567910$. Convergence is extremely fast (the number of correct digits doubles).