Is this series for Pi correct?

Question

The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.

Here is the series I got using Pythagorean’s theorem , is it correct?

$$\begin{align} A&=3r^{2} + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=r\sqrt{2-\sqrt{3}} ,\\x_{n+1}&=\sqrt{2r^{2}-2r\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}} \end{align}$$

So-for-a-unit-circle

$$\begin{align} \pi&=3 + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{1-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=\sqrt{2-\sqrt{3}} ,\\x_{n+1}&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}} \end{align}$$

Answer 1

Numerical calculation suggests that you have made an error (or that I have, of course.)

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
    return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
    return 1 - sqrt(1-x*x/4)

for _ in range(50):
    xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

produces the output

answer= 3.0117993877991496
11.999999999999849

Have you dropped a factor of $12$ before the sum, perhaps?

Answer 2

Note that, in $x_{n+1} =\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}} $ if $x_n = 2\sin(t) $ then

$\begin{array}\\ x_{n+1} &=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}\\ &=\sqrt{2-2\sqrt{1-\sin^2(t)}}\\ &=\sqrt{2-2\cos(t)}\\ &=2\sqrt{\dfrac{1-\cos(t)}{2}}\\ &=2\sin(\dfrac{t}{2})\\ \end{array} $