Arithmetic–geometric mean and elliptic integrals

Remarks

Proof idea: by the inequality of arithmetic and geometric means, $b_n \le b_{n+1} \le a_{n+1} \le a_n$ for $n \ge 1$, so both sequences are monotone and bounded. Since $b_{n+1} \ge b_n$, one gets $a_{n+1} - b_{n+1} \le a_{n+1} - b_n = (a_n - b_n)/2$, so the difference tends to $0$ and the two limits agree. For part 2, call the integral $I(a, b)$; a change of variables found by Gauss shows that $I(a, b) = I(a_1, b_1)$. Hence $I(a, b) = I(a_n, b_n)$ for all $n$, and letting $n \to \infty$ gives $I(M, M) = \pi / (2M)$. Part 3 is part 2 with $a = 1$ and $b = k'$, since $\cos^2\theta + k'^2\sin^2\theta = 1 - k^2\sin^2\theta$.

Example: $M(1, \sqrt2) \approx 1.19814$, and $1/M(1, \sqrt 2)$ is Gauss's constant, equal to $\varpi/\pi$ where $\varpi$ is the lemniscate constant.

Going deeper

For a student

The convergence is quadratic: once $a_n$ and $b_n$ are close, the number of correct digits roughly doubles at each step, because $a_{n+1} - b_{n+1} = (\sqrt{a_n} - \sqrt{b_n})^2/2$. This makes the formula $K(k) = \pi/(2M(1, k'))$ the standard way to evaluate $K$ numerically. Its invariance step $I(a, b) = I(a_1, b_1)$ is a form of Landen's transformation. Combined with Legendre's relation it gives the Gauss–Legendre algorithm for $\pi$.