The Arithmetic-Geometric-Harmonic Mean

Write $g = GM(x,y)$ . Then $g_1 = g$ and $a_1h_1 = g^2$ . Suppose $n \in \mathbb{N}$ and that $g_n = g$ and $a_nh_n = g^2$ . Then $\begin{aligned} a_{n+1} & =\; \tfrac13(a_n + g + h_n) \\ g_{n+1} & = \; \sqrt[3]{a_n \times g \times h_n} \; = \; g \\ h_{n+1} & = \; \frac{3a_ngh_n}{a_ng + a_nh_n + gh_n} \; = \; \frac{3g^3}{g(a_n + g + h_n)} \; = \; \frac{3g^2}{a_n + g + h_n} \end{aligned}$ so that $g_{n+1} = g$ and $a_{n+1}h_{n+1} = g^2$ .

Thus we deduce that $g_n = g$ and that $a_nh_n = g^2$ for all $n \ge 1$ . Now we know that $a_n > g_n = g$ for all $n$ . Moreover $a_n - a_{n+1} \; = \; \tfrac13(2a_n - g - h_n) \; = \; \tfrac13\big(2a_n - g - \tfrac{g^2}{a_n}\big) \; = \; \tfrac{1}{3a_n}(a_n - g)(2a_n + g) \; > \; 0$ so we know that $(a_n)$ is a decreasing sequence, and hence converges to some $\alpha \ge g$ . Since $a_{n+1} \; = \; \tfrac13\big(a_n + g + \tfrac{g^2}{a_n}\big)$ we deduce that $\alpha \; =\; \tfrac13\big(\alpha + g + \tfrac{g^2}{\alpha}\big)$ which implies that $2\alpha^2 - g\alpha- g^2 = 0$ and hence, since $\alpha \ge g$ , that $\alpha = g$ . Thus $\lim_{n \to \infty}a_n = g$ . Since $h_n = \tfrac{g^2}{a_n}$ , we deduce that $\lim_{n\to\infty}h_n = g$ as well. Thus we deduce that $AGHM(x,y) = g = GM(x,y)$ .

It is worth noting that $a_{n+1} - g \; = \; \tfrac{1}{3a_n}(a_n - g)^2$ and so the rate of convergence of $a_n$ to $g$ is quadratic.

Note : This is not a full solution, but more of an outline on how one can obtain the solution.

This question was inspired after reading up on the arithmetic-geometric mean and the geometric-harmonic mean (both really interesting means, by the way).

Let us first prove that $\text{AGHM}(x,y)$ exists for all positive reals $x,y$ . Without loss of generality, assume $x\le y$ .

Denote $a=\displaystyle\lim_{n\to\infty}a_n,g=\displaystyle\lim_{n\to\infty}g_n,h=\displaystyle\lim_{n\to\infty}h_n$ . Since $a_{n+1}\le a_{n}$ and $h_{n+1}\ge h_{n}$ for all $n\ge 1$ and both $a_n,h_n$ are bounded, $a$ and $h$ both exist. (We're going to show later that $g_n=g_{n+1}$ for all $n$ and thus $g$ also exists.)

Note that by the AM-GM-HM inequality, $h\le g\le a$ . We know that $3a_n=a_{n-1}+g_{n-1}+h_{n-1}$ . Take $n\to\infty$ to get the result that $2a=g+h$ . Since $a\ge g\ge h$ , this necessarily means that $a=g=h$ . Thus, $\text{AGHM}(x,y)$ exists.

Now, it's easy to show that $a_1h_1=g_1^2$ . Then, assuming that $a_nh_n=g_n^2$ for some $n$ , one can also show that $a_{n+1}h_{n+1}=g_{n+1}^2$ . Knowing that $a_nh_n=g_n^2$ for all $n\ge 1$ , it is easy to demonstrate that $a_nh_n=a_{n+1}h_{n+1}$ and thus $g_{n}=g_{n+1}$ for all $n\ge 1$ .

Thus, $\text{AGHM}(x,y)=\text{GM}(x,y)$ .