Rational approximation of $\pi$ by recursion

Question

I wonder if the following result is already known and may be considered as interesting : let $\mathcal{C}$ be the real algebra of continuous functions $f:\left[0,1\right]\to\mathbf{R}$, $T:\mathcal{C}\to \mathcal{C}$ the map defined by $$ T\left(f\right)\left(x\right)=f\left(x\right)\left(2-\left(1+x^2\right)f\left(x\right)\right) $$ $\left(f_n\right)$ the sequence of $\mathcal{C}$ defined by $$ f_n= \begin{cases} T\left(f_{n-1}\right) & \text{if } n>0\\ 1/2 & \text{if } n=0 \end{cases} $$ and $\left(I_n\right)$ the real sequence defined by $I_n=\int_0^1 f_n\left(x\right) \, dx$. One has $$ \forall n\in\mathbf{N} \qquad 0 \leq \frac{\pi}{4} - I_n \leq 2^{-2^n} $$ and $$ \forall n\in\mathbf{N} \qquad I_n = \frac{1}{2} \sum_{k=0}^{2^n-1} \frac{2^k}{\left(2k+1\right)\binom{2k}{k}} $$

Answer 1

Too long for a comment.

Using the Gaussian hypergeometric function $$I_n = \frac{1}{2} \sum_{k=0}^{2^n-1} \frac{2^k}{\left(2k+1\right)\binom{2k}{k}}=\frac \pi 4-J_n$$ $$J_n=\frac{2^{2^n-1}}{\left(2^{n+1}+1\right) \binom{2^{n+1}}{2^n}}\,\,\, _2F_1\left(1,2^n+1;2^n+\frac{3}{2};\frac{1}{2}\right)$$ $$J_n < \frac{2^{2^n}}{\left(2^{n+1}+1\right) \binom{2^{n+1}}{2^n}}$$ $$\log \big[-\log (J_n)\big]=n \log (2)-\frac{2}{5}+\frac{3}{5 n}+O\left(\frac{1}{n^2}\right)$$ $$\log \left(-\log \left(\frac{J_{n+1}}{J_n}\right)\right) \sim n\log(2)-\frac 13$$

Edit

The first $I_n$ form the sequence $$\left\{\frac{1}{2},\frac{2}{3},\frac{16}{21},\frac{11776}{15015}, \frac{208481288192}{265447707525},\frac{3374092439600157908008 96}{429602792188525533252675}\right\}$$