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 (Wayback Machine) 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?