# 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?

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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 $0$.

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Thanks for showing me that the sequence converges to 0, but what I really need to know is why $0$ is the supremum of the set(Before proving this convergence). This is because I would like to use the monotone convergence theorem to prove that $x_n$ converges. –  user38268 Aug 27 '11 at 23:34
@D B: I wonder why you would want to do such a thing. OK, how about if you just leave off the last sentence and instead argue that $|x_n/x_{n-1}|\le1/2$ implies that the supremum cannot be non-zero (since otherwise you'd jump across it once you get near it)? Then you've used monotone convergence instead of the comparison with a geometric sequence. –  joriki Aug 28 '11 at 4:15
Hmm you're right perhaps showing that the supremum is zero is not so easy, better to show it converges to 0 straight away. –  user38268 Aug 28 '11 at 5:43
I don't see how $\left|\frac{x_n}{x_{n-1}}\right|$ being less than 1 means that its absolute value is dominated by $x_12^{-1}$. –  user38268 Aug 28 '11 at 6:14
Sorry, it should have been $x_12^{-(n-1)}$ (neither $x_12^n$ as I wrote nor $x_12^{-1}$ as you wrote). I only added the $<1$ line to indicate that the factor $1/2$ of the geometric sequence is less than $1$, so the geometric sequence itself converges. That $x_n$ is dominated by the geometric sequence $x_12^{-(n-1)}$ follows from the penultimate inequality $|x_n/x_{n-1}|\le1/2$, since we start with $x_1$ and in each step we reduce the absolute value at least by a factor of $1/2$, and $x_12^{-(n-1)}$ also starts with $x_1$ and reduces its absolute value by a factor of $1/2$ in each step. –  joriki Aug 28 '11 at 6:40

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, $|b_n^2-a_n^2|\le4^{-n}|b_0^2-a_0^2|$.

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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.

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