It is easy to show that, for two fixed real numbers $\alpha, \beta > 0$, the sequences given by $a_ 1 = \frac{\alpha + \beta }{2}$, $ g_1 = \sqrt{\alpha\beta}$, and $a_{n+1} = \frac{a_n + g_n}{2}$, $g_{n+1} = \sqrt{a_ng_n}$ are both convergent and that $\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} g_n$. Let's call this limit $\operatorname{agm}(\alpha, \beta)$. I am convinced that the number $\operatorname{agm}(1,2)$ is irrational, but I don't know how to prove it.

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    $\begingroup$ What is $b_n$ in $a_{n + 1} = \frac{a_n + b_n}2$? Should it be $g_n$? $\endgroup$ – LSpice Apr 16 at 19:31
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    $\begingroup$ It boils down to understand if some complete elliptic integral of the first kind is irrational. In fact, $$\operatorname{agm}(1, \, 2)=\frac{\pi}{4} \cdot 3 K\left(-\frac{1}{3}\right)^{-1}.$$ Numerical computations seem to suggest so. $\endgroup$ – Francesco Polizzi Apr 16 at 19:34
  • $\begingroup$ I agree with Francisco. It would be truly remarkable if $$ \frac{\pi}{4} \cdot 3 K\left(-\frac{1}{3}\right)^{-1}\qquad\text{is rational} \tag1$$ Note $K(1/3) = K(-1/3)$. And $(1)$ holds if and only if $K(1/3)$ is a rational multiple of $\pi$. $\endgroup$ – Gerald Edgar Apr 16 at 20:35
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    $\begingroup$ For some $k$ called singular values, the complete elliptic integral $K(k)$ can be expressed using $\pi$ and the $\Gamma$ function at rational arguments. These $k$ are algebraic numbers but the following table does not include any rational $k$: mathworld.wolfram.com/EllipticLambdaFunction.html $\endgroup$ – François Brunault Apr 16 at 23:11
  • $\begingroup$ Perhaps studying the symmetries of the considered elliptic integral could help determine the structure of $\mathbb{Q}(K(1/p))$ for $p$ a prime number. $\endgroup$ – Sylvain JULIEN Apr 17 at 11:34

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