Let $a_0 = 1$ and $b_0 = x \ge 1$. Let $$ a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n+1} b_n})/2. $$

Numeric computation suggests that regardless of the choice of $x$, $a_n$ and $b_n$ always converge to the same value. Can we prove this?

Moreover, if we assume this is true and define $f(x) = \lim_{n \to \infty} a_n$, then numeric computation shows that $$ f'(1) \approx 0.57142857142857 \approx 4/7 \\ f''(1) \approx -8/49 \\ f^{(3)}(1) \approx 1056/4459 \\ f^{(4)}(1) \approx -65664/111475. $$ There seems to be some patterns here. Can we actually find the limit with these clues?

I was actually trying to check this recursion $$ a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n} b_n})/2. $$ I made a mistaked in my code and computed the one at the top. See The Computer as Crucible, pp 130.


1 Answer 1


We have $1\le a_0\le b_0\le x$.

We can prove that $1\le a_n\le a_{n+1}\le b_{n+1}\le b_n\le x$ by mathematical induction.

$\displaystyle a_1=\frac{1+\sqrt{x}}{2}$ and $\displaystyle b_1=\frac{x+\sqrt{a_1x}}{2}$.

Obviously, $a_1\ge 1$. Hence $\displaystyle b_1-a_1=\frac{(x-1)+(\sqrt{a_1}-1)\sqrt{x}}{2}\ge0$

Note that



$$b_1\le\frac{x+\sqrt{x\cdot x}}{2}=x$$

Therefore, $1\le a_0\le a_{1}\le b_{1}\le b_0\le x$.

Suppose that $1\le a_k\le a_{k+1}\le b_{k+1}\le b_k\le x$. Then






\begin{align} b_{k+2}&\le\frac{b_{k+1}+\frac{b_{k+2}+b_{k+1}}{2}}{2}\\ 4b_{k+2}&\le3b_{k+2}+b_{k+1}\\ b_{k+2}&\le b_{k+1} \end{align}

This complete the induction proof.

So, $\{a_n\}$ is increasing and bounded above by $x$. $\{b_n\}$ is decreasing and bounded below by $1$. Both the sequences converge.

Let $\displaystyle \lim_{n\in\infty} a_n=a$ and $\displaystyle \lim_{n\in\infty} b_n=b$. Then we have


This implies that $a=b$.

Unsolved: How to find the value of the common limit?


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