$$ a_0 = \frac{1}{\sqrt 2} $$
$$ a_{n+1} = \frac{( \sqrt {1 - a_n^2} -1)^2}{a_n^2} $$
$$ \lim_{n \to \infty} \frac{4^{\frac{1}{2^n}}}{a_{n+1}^{\frac{1}{2^n}}} = \exp(\pi) $$
How did Gelfond find this nice result ? And how to prove it ? The 4 is trivial , but the rest is not. Is this related to trigonometry ? Is this related to continued fractions ?
Are there analogues known for cube roots ?
Notice a proof alone might not explain how he found the result.
This is the Gauss-Legendre algorithm non homogeneous version for the elliptic modulus k. That is, let $\, a_n = k(q_n) = (\theta_2(q_n)/\theta_3(q_n))^2 \,$ where $\, q_n = \exp(-\pi 2^n). \,$ We have $\, a_{n+1} = 4 q_n + O(q_n^3). \,$
Legendre used the following: $\, k_{n+1} = (1 - k'_n)/(1 + k'_n) = k_n^2/(1 + k'_n)^2 = (1 - k'_n)^2/k_n^2 \,$ where $\, 1 = k_n^2 + k_n'^2. \,$ The AGM of Gauss is $\, a_{n+1} \!=\! \frac12(a_n \!+\! b_n), $ $ b_{n+1} \!=\! \sqrt{a_n b_n}, \, $ $ c_{n+1} \!=\! \frac12(a_n\!-\!b_n). \,$ The connection is $\, k_n = c_n/a_n, \, k_n' = b_n/a_n. \,$ For more details, you can read the standard book Pi and the AGM by Borwein and Borwein.
By the way, the recursion as written $\, a_{n+1} = (\sqrt {1 - a_n^2} -1)^2/a_n^2, \,$ is not good from a numerical point of view. The subtraction of two numbers nearly equal causes drastic loss of significance. The equivalent recursion $\, a_{n+1} = a_n^2 / (1 + \sqrt {1 - a_n^2})^2 \,$ does not suffer from significance loss.