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.
]
Best Answer
There are two possible ways to attack this problem
Both K and K' can be expressed in terms of the Theta function as described here http://mathworld.wolfram.com/EllipticModulus.html. If you compute $\Theta_3$, you can get both at the same time.
The other way is to observe that both K and K' are expressible in terms of the hypergeometric function $_2F_1(\frac{1}{2}, \frac{1}{2} ; 1; m)$. They are solutions of the same self-adjoint Gauss hypergeometric differential equation (since the equation is invariant under the transformation (m $\rightarrow$ 1-m))
$(k^3 - k)\frac{d^2y}{dk^2} + (3k^2 -1)\frac{dy}{dk} + ky = 0 $
By virtue of this fact, both K and K' are connected. You will find the following series expansion for K'(k) derived in Borwein's book Pi and AGM Section 1.3
$ K'(k) = \frac{2}{\pi} log \frac {4}{k} K(k) - 2 [(\frac{1}{2})^2(\frac{1}{1.2}k^2 + (\frac{1.3}{2.4})^2(\frac{1}{1.2} + \frac{1}{3.4})k^4 + (\frac{1.3.5}{2.4.6})^2(\frac{1}{1.2} + \frac{1}{3.4} + \frac{1}{5.6})k^6 $.....(infinite series) + ]
You may also find Chapter 5 of Armitage and Eberlein's book on Elliptic Functions useful.
EDIT1: I put in the complete series expansion for K'(k).
Regarding the computation of E(k), E(k) and K(k) are connected by the differential equation $ \frac{dK}{dk} = \frac{E - (1-k^2)K}{k(1-k^2)} $ which is how the Legendre relation you mention above comes about.
Again Borwein has the solution for this problem(buy the book!). Exercise 3 in Sec 1.4 has the formula based on the quartic AGM iteration $ E(k) = K(k)[1 - \sum_{n=0}^{\infty} 4^n [\alpha_n^4 - (\frac{\alpha_n^2+\beta_n^2}{2})^2 ] $ where
and $ a_n, b_n$ and $c_n $ satisfy the AGM relation