In a Python program while trying to compute some expressions, I've encountered a recursive sequence $\{k_i\}$
for any $0 \le k_0 \le 1$ we have: $$ k_i = \begin{cases} k = k_0 \quad \text{if} \quad i = 0 \\ \dfrac{2\sqrt{k_{i-1}}}{1+k_{i-1}} \quad \text{for i > 0} \end{cases} $$
Is it possible to convert this recursive expression to an explicit formula where I could only give $i$ as input and get back $k_i$ as output
I appreciate any help.
It's possible and related to the AGM. You decide if it's okay. Let $$ k_0 \!=\! k'(q) \!:=\! \left(\frac{\theta_4(0,q)}{\theta_3(0,q)}\right)^2 \!=\! 1 - 8q + 32q^2 + \dots \tag{1} $$ where $\,\theta_3(0,q)\,$ and $\,\theta_4(0,q)\,$ are Jacobi theta functions and where $\,k'(q)\,$ is the generating function of OEIS sequence A139820. Then it is known that $$ k_n := k'(q^{2^n}) = \frac{2\sqrt{k_{n-1}}}{1+k_{n-1}}. \tag{2} $$ Using the inverse function of $\,k'\,$ let $\, q := k'^{(-1)}(k_0),\,$ equation $(2)$ is an explicit formula for $\,k_n.$
This is a numerical example. Let $\,q = .01\,$ and $\,k_0 \approx 0.92310649897852255... = k'(q).\,$ Then $\,k_1 \approx 0.99920031990402559... = k'(q^2).\,$ Next $\,k_2 \approx 0.99999992000000319... = k'(q^4).\,$ Next $\,k_3 \approx 0.99999999999999920... = k'(q^8).\,$ In general $\,k_n = k'(q^{2^n}).\,$