Deriving the closed form of $M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$ I have the sequence, let $M_0=1$
$$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$$
Which I would like first to study the convergence and fine the closed form. 
 I failed  to show that $M_n$ is bounded and  monotone. 
This could be easy if have the explicit expression of it. 

Question:
  Is there a closed a form of this sequence? does anyone has an idea?

FYI In the book it is mentioned that this sequence is used to approximate the area of the unit circle. may be some else has a more clever explanation to this connection 

Answer thanks to @Masacroso comment.

$$M_n = 2^n \tan\left(\frac{π}{2^{n+2}}\right)\to \frac{π}{4}$$
  and 
  $$\lim_{n\to\infty}M_n = \lim_{n\to\infty} \frac{π}{4} \frac{\tan\left(\frac{π}{2^{n+2}}\right)}{\frac{π}{2^{n+2}}}=\frac{π}{4}\lim_{x\to 0}  \frac{\tan\left(x\right)}{x} =\frac{π}{4}$$

One can show that $M_n\ge 0,$ let $X_n:=\frac{M_n}{2^n}$ then  $$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)\implies X_{n+1}=\frac{1}{X_n}(\sqrt{1+X_n^2}-1)$$
There exists $a_n\in[0, \frac{π}{2}]$ such that, $$X_n=\tan(a_n)$$
hence we have, 
$$\begin{align}\tan(a_{n+1})= X_{n+1}&=\frac{1}{X_n}\left(\sqrt{1+X_n^2}-1\right) \\&=\frac{1}{\tan(a_n)}\left(\sqrt{1+\tan^2(a_n)}-1\right)\\&= \frac{1}{\tan(a_n)}\left(\frac{1}{\cos(a_n)}-1\right) \\&= \frac{1-\cos(a_n)}{\sin(a_n)} = \frac{1-\cos^2(\frac{a_n}{2})+\sin^2(\frac{a_n}{2})}{2\cos(\frac{a_n}{2})\sin(\frac{a_n}{2})} \ \\&= \frac{1-\cos(a_n)}{\sin(a_n)} = \frac{2\sin^2(\frac{a_n}{2})}{2\cos(\frac{a_n}{2})\sin(\frac{a_n}{2})} \\&=\color{blue}{\tan(\frac{a_n}{2})}\end{align}$$
Finally we have 
$$a_{n+1}= \frac{a_n}{2}\implies a_n=\frac{a_0}{2^{n}} $$
But we have $a_0 = π/4$ therefore we end up, with 
$$M_n = 2^n \tan\left(\frac{π}{2^{n+2}}\right)\to \frac{π}{4}$$
