Supremum of a Set containing terms of a sequence I am trying to prove that the sequence defined by $x_n = a_n - b_n$ converges to zero where $a_n$ and $b_n$ are sequences such that
$a_n = \sqrt{a_{n-1}b_{n-1}}$ and $ b_n = \frac{a_{n-1} + b_{n-1}}{2}$.
By definition $a_1 = \sqrt{ab}$ and $b_1 = \frac{a + b}{2}$ where $a,b$ are real numbers with $a > b > 0$.
Now I have proven that $x_n$ is strictly increasing, so I just need to show that the supremum of the set of all $x_n's$ is zero. How can I do this?
I tried using contradiction namely to show that $\forall \epsilon > 0$, $\exists n$ such that $-\epsilon < x_n < 0$, but to even produce such an $n$ is tough.
If we complete the square we find that $x_n = -\frac{1}{2}(\sqrt{a_{n-1}} - \sqrt{b_{n-1}})^2 \leq 0.$
That's all I've got at the moment. How can I find the supremum of this?
 A: Using the expression you derived for $x_n$,
$$
\begin{eqnarray}
\left|\frac{x_n}{x_{n-1}}\right|
&=&
\frac{\left|-\frac12(\sqrt{a_{n-1}} - \sqrt{b_{n-1}})^2\right|}{\left|a_{n-1}-b_{n-1}\right|}
\\
&=&
\frac{\left|-\frac12(\sqrt{a_{n-1}} - \sqrt{b_{n-1}})^2\right|}{\left|(\sqrt{a_{n-1}} - \sqrt{b_{n-1}})(\sqrt{a_{n-1}} + \sqrt{b_{n-1}})\right|}
\\
&=&
\frac12\frac{\left|(\sqrt{a_{n-1}} - \sqrt{b_{n-1}})\right|}{(\sqrt{a_{n-1}} + \sqrt{b_{n-1}})}
\\
&\le&\frac12
\\
&<&1\;.
\end{eqnarray}
$$
Thus the absolute value is dominated by $x_12^{-(n-1)}$ and therefore converges to $0$.
A: To show that $x_n\le0$,
$$
\begin{align}
x_n&=a_n-b_n\\
   &=\frac{a_n^2-b_n^2}{a_n+b_n}\\
   &=\frac{a_{n-1}b_{n-1}-\frac{1}{4}(a_{n-1}+b_{n-1})^2}{a_n+b_n}\\
   &=\frac{-\frac{1}{4}(a_{n-1}-b_{n-1})^2}{a_n+b_n}\\
   &\le0
\end{align}
$$
But instead of  showing that $a_n-b_n\to0$, it is simpler to show that $b_n^2-a_n^2\to0$:
$$
\begin{align}
|b_n^2-a_n^2|&=\left|\frac{1}{4}(a_{n-1}+b_{n-1})^2-a_{n-1}b_{n-1}\right|\\
           &=\frac{1}{4}|b_{n-1}-a_{n-1}|^2\tag{1}
\end{align}
$$
Divide $(1)$ by $|b_{n-1}^2-a_{n-1}^2|$:
$$
\begin{align}
\left|\frac{b_n^2-a_n^2}{b_{n-1}^2-a_{n-1}^2}\right|
&=\frac{1}{4}\left|\frac{(b_{n-1}-a_{n-1})^2}{b_{n-1}^2-a_{n-1}^2}\right|\\
&=\frac{1}{4}\left|\frac{b_{n-1}-a_{n-1}}{b_{n-1}+a_{n-1}}\right|\\
&\le\frac{1}{4}\tag{2}
\end{align}
$$
because $a_n,b_n\ge0$.
Therefore, $|b_n^2-a_n^2|\le4^{-n}|b_0^2-a_0^2|$.
A: On account of the AGM inequality you should define $a_n:=(a_{n-1}+b_{n-1})/2$ and $b_n:=\sqrt{a_{n-1}b_{n-1}}$; then $a_n\geq b_n$ for all $n$. From the recursion formula you immediately obtain
$$a_n - b_n={(a_{n-1}-b_{n-1})^2\over 2(\sqrt{a_{n-1}}+\sqrt{b_{n-1}})^2 }\leq{1\over2}(a_{n-1}-b_{n-1})\ .$$
The middle term of this inequality shows that the convergence of $a_n$ and $b_n$ to some number $\mu$ (the so called arithmetic-geometric mean of the given $a$, $b$) is even quadratic.
