Question about arithmetic–geometric mean [duplicate]

If $a_1=b_1$, the two sequences are constant, and the proof is trivial. Otherwise, suppose $a_1<b_1$ or $b_1<a_1$. We have the property that for any two numbers $a<b$, $$a<GM<AM<b,$$ where $GM$ is the geometric mean, and $AM$ is the arithmetic mean. So, we must have $$a_1<a_2<b_2<b_1$$ or $$b_1<a_2<b_2<a_1.$$ In any case, for $n>1$ $$a_n<a_{n+1}<b_{n+1}<b_n$$ That shows us that $a_n$ is increasing and $b_n$ is decreasing. To apply the Cantor's lemma we must only show that $b_n-a_n\rightarrow 0$. To show that, we realize that $b_{n+1}-a_{n+1}<\frac{b_n-a_n}{2}$, since $b_{n+1}$ is at the middle of the points $a_n$ and $b_n$.