The arithmetic-geometric mean,
$a_{k+1}=\frac{a_k+b_k}{2} \quad b_{k+1}=\sqrt{a_k b_k}$
is one of the celebrated discoveries of Gauss, who found out that it is equivalent to computing a (complete) elliptic integral (which is a special case of the Gauss hypergeometric function ${}_2 F_1$).
I have been wondering if nth-order generalizations of the iteration,
$a_{k+1}=\frac{a_k+b_k}{n} \quad b_{k+1}=\sqrt[n]{a_k b_k}$
have ever been systematically studied. I've seen this paper by Borwein, but have had trouble searching for other papers. In particular, I'm interested if the coupled sequences also have a common limit, and if so, whether the limit is expressible as a hypergeometric function (or generalizations like those of Appell or Lauricella).
Another possible generalization I thought involves $n$ variables and makes use of the elementary symmetric polynomials. To use $n=4$ as an example:
$a_{k+1}=\frac{a_k+b_k+c_k+d_k}{4}$
$b_{k+1}=\sqrt{\frac{a_k b_k+a_k c_k+a_k d_k+b_k c_k+b_k d_k+c_k d_k}{3}}$
$c_{k+1}=\sqrt[3]{\frac{{a_k b_k c_k}+{a_k b_k d_k}+{a_k c_k d_k}+{b_k c_k d_k}}{2}}$
$d_{k+1}=\sqrt[4]{a_k b_k c_k d_k}$
Would these four sequences (and in general the $n$ sequences) tend to a common limit $F(a_0,b_0,c_0,d_0,\dots)$ like in the $n=2$ case, and if so, are they expressible in terms of known functions?
EDIT
Taking into account Darsh Ranjan's comments, I realized that what I should be looking at instead is the generalization whose denominators are binomial coefficients (thus, the general form $\sqrt[j]{\frac{e_j}{\binom{n}{j}}}$, for $j=1\dots n$ where $e_j$ is the jth elementary symmetric polynomial). The case $n=4$ now looks like
$a_{k+1}=\frac{a_k+b_k+c_k+d_k}{4}$
$b_{k+1}=\sqrt{\frac{a_k b_k+a_k c_k+a_k d_k+b_k c_k+b_k d_k+c_k d_k}{6}}$
$c_{k+1}=\sqrt[3]{\frac{{a_k b_k c_k}+{a_k b_k d_k}+{a_k c_k d_k}+{b_k c_k d_k}}{4}}$
$d_{k+1}=\sqrt[4]{a_k b_k c_k d_k}$
So, still the same question: is there a common limit, and if so, is the limit expressible in terms of known functions?
Best Answer
You want Maclaurin's inequality. Given $n$ positive numbers $a_1, a_2,\dots,a_n$, write $$ (x+a_1)(x+a_2)\cdots(x+a_n) = x^n + S_1x^{n-1} + \cdots + S_{n-1}x + S_n, $$ so $S_i$ is the $i$-th elementary symmetric function of $a_1,\dots,a_n$. For $i = 1,\dots,n$, set $A_i = S_i/\binom{n}{i}$. When $n = 2$, $A_1 = (a_1+a_2)/2$ and $A_2 = a_1a_2$. Maclaurin's inequality is that $$ A_1 \geq \sqrt{A_2} \geq \sqrt[3]{A_3} \geq \cdots \geq \sqrt[n]{A_n}, $$ where the inequality signs are all strict unless $a_1,\dots,a_n$ are all equal. The inequality of the outer terms, $A_1 \geq \sqrt[n]{A_n}$, is the arithmetic-geometric mean inequality for $n$ positive numbers.
From a list of $n$ positive numbers $a_1,\dots,a_n$ we have produced another list of $n$ positive numbers $A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n}$. The construction can be repeated.
Theorem: All the terms in the list tend to the same limit.
Off the top of my head I can't recall a reference where this is proved. It was studied by Meissel in 1875 for $n = 3$.
For example, if we start with the three numbers 1, 2, 3 then after 4 iterations the three numbers we get all look like 1.9099262335 to 10 digits after the decimal point.
[Edit: Here is a proof of the common limit, based on Will Sawin's first comment below to my answer. Order the numbers $a_1,\dots,a_n$ so that $a_1 \geq \cdots \geq a_n > 0$. By Maclaurin's inequality (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will bound $A_1 - \sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $\sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above,
$$ A_1 = \frac{a_1 + \cdots + a_n}{n} \leq \frac{(n-1)a_1 + a_n}{n} = a_1 - \frac{a_1 - a_n}{n} $$ and to bound $A_n$ from below we write $A_n = a_1\cdots a_n \geq a_n^n$, so $$ \sqrt[n]{A_n} \geq a_n. $$ Therefore $$ 0 \leq A_1 - \sqrt[n]{A_n} \leq \left(a_1 - \frac{a_1 - a_n}{n}\right) - a_n = \left(1 - \frac{1}{n}\right)(a_1 - a_n). $$ Start from an $n$-tuple $(a_1,a_2,\dots,a_n)$ which is ordered so that $a_1 \geq \cdots \geq a_n > 0$ and construct the $n$-tuple $(A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n})$ and keep repeating this, which produces a sequence of $n$-tuples $(a_1^{(k)},a_2^{(k)},\dots,a_n^{(k)})$ for $k = 0,1,2,\dots$, where $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} \geq a_1^{(1)} \geq a_1^{(2)} \geq \cdots > 0$. Therefore the sequence $a_1^{(k)}$ converges as $k \rightarrow \infty$. (The limit is positive because the sequence of last coordinates $a_n^{(k)}$ is non-decreasing and $a_1^{(k)} \geq a_n^{(k)} \geq a_n^{(0)} = a_n$ for all $k$.) The above calculation shows $$ 0 \leq a_1^{(k)} - a_n^{(k)} \leq \left(1 - \frac{1}{n}\right)(a_1^{(k-1)} - a_n^{(k-1)}), $$ so $0 \leq a_1^{(k)} - a_n^{(k)} \leq (1 - 1/n)^k(a_1 - a_n)$. Letting $k \rightarrow \infty$ we see the sequence of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},\dots$ converges to the limit of the sequence of first coordinates $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},\dots$. Since $a_1^{(k)} \geq a_i^{(k)} \geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},\dots$ converges to the same limit. ]