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Given 2 positive numbers $a,b>0$. Let us define
$$a_{0}=a\quad ,b_{0}=b\quad,a_{n+1}=\sqrt{a_{n}b_{n}}\qquad,b_{n+1}=\frac{a_{n}+b_{n}}{2}$$
prove that both sequences convergence and $\lim(a_{n})_{n\to \infty}=\lim(b_{n})_{n\to \infty}$

Now I know that $$a_{n+1}=\sqrt{a_{n}b_{n}}\le\frac{a_{n}+b_{n}}{2}=b_{n+1}$$ from AM-GM inequality so I tried to prove that $a_n$ in increasing and $b_n$ is decreasing using induction but it's too complicated there is another way to show that without induction or maybe I'm not doing it right?

Kenta S
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math_addict
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1 Answers1

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First of all, usual notation is $a_n$ is the arithmetic mean and $b_n$ is geometric mean. But I will stick with your notation.

Let us show $b_n\geq a_n$ for all $n$. \begin{equation} b_{n+1} \geq a_{n+1} \iff \frac{b_n+a_n}{2}\geq \sqrt{b_na_n} \iff (b_n-a_n)^2\geq 0 \end{equation} which shows that $b_{n+1}\geq a_{n+1}$ for all $n$, and without loss of generality you can assume $b\geq a$ and say $b_n\geq a_n$ for all $n$.
Next, we show that $b_n$ is decreasing and $a_n$ is increasing. \begin{equation} b_{n+1} \leq b_n \iff \frac{a_n+b_n}{2} \leq b_n \iff a_n\leq b_n \end{equation} and \begin{equation} a_n \leq a_{n+1} \iff a_n \leq \sqrt{a_nb_n}\iff a_n\leq b_n \end{equation} both are true. Therefore; \begin{equation} a_n \leq a_{n+1} \leq b_{n+1} \leq b_n\label{bdd}\tag{1} \end{equation} Lastly, we need to show that they converge to the same limit. \begin{equation} 0 \leq b_{n+1}-a_{n+1}\leq \frac{b_n+a_n}{2} - \sqrt{b_na_n} = \frac{1}{2}\left(\sqrt{b_n}-\sqrt{a_n}\right)^2 = \frac{1}{2}\frac{b_n-a_n}{\left(\sqrt{b_n}+\sqrt{a_n}\right)^2}(b_n-a_n) \end{equation} On noting that; \begin{equation} b_n-a_n \leq \left(\sqrt{b_n} + \sqrt{a_n}\right)^2 \end{equation} We get; \begin{equation} b_{n+1} - a_{n+1} \leq \frac{1}{2}(b_n-a_n) \leq 2^{-n}(b-a) \end{equation} Hence $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}b_n$ exists and equal.


Note: [\ref{bdd}] shows that limits exists, since they are bounded monotone sequence. Last results shows they are converging to same limit.

Let me also add my reference:
Borwein, Jonathan M., and Peter B. Borwein.
Pi and the AGM : a study in analytic number theory and computational complexity.
New York: Wiley, 1987.

Atbey
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