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 
 A: Incomplete answer (sorry, going out with my children). It's definitely monotone.
$$\color{red}{M_{n+1}} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)=
\frac{2^{n+1}}{M_n}\left(\sqrt{2^{2n}+ M^2_{n}}-2^n\right)=\\
\frac{2^{n+1}}{M_n}\left(\frac{M_n^2}{\sqrt{2^{2n}+ M^2_{n}}+2^n}\right)\color{red}{\leq}
\frac{2^{n+1}}{M_n}\left(\frac{M_n^2}{2^{n+1}}\right)=\color{red}{M_n}$$
and $M_n>0$ should be easy to show.
A: 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 
$$M_n = 2^n \tan\left(\frac{π}{2^{n+2}}\right)\to \frac{π}{4}$$
A: Define $a_n = \dfrac{M_n}{2^n} \ (n \in \mathbb{N}_+)$, then$$
a_{n + 1} = \frac{\sqrt{1 + a_n^2} - 1}{a_n} > 0.
$$
Define $b_n = \sqrt{1 + a_n^2} \ (n \in \mathbb{N}_+)$, then$$
b_n > 1 \Longrightarrow a_n = \sqrt{b_n^2 - 1},
$$
and\begin{align*}
&\mathrel{\phantom{\Longrightarrow}}\sqrt{b_{n + 1}^2 - 1} = \frac{b_n - 1}{\sqrt{b_n^2 - 1}} = \sqrt{\frac{b_n - 1}{b_n + 1}}\\
&\Longrightarrow b_{n + 1}^2 = \frac{b_n - 1}{b_n + 1} + 1 = \frac{2b_n}{b_n + 1}.
\end{align*}
Define $c_n = \dfrac{1}{b_n} \ (n \in \mathbb{N}_+)$, then$$
c_{n + 1}^2 = \frac{1 + c_n}{2}.
$$
Because $c_1 = \dfrac{2}{\sqrt{5}} < 1$, denote $θ = \arccos \dfrac{2}{\sqrt{5}}$, then by induction on $n$, there is $0 < c_n < 1$. Define $θ_n = \arccos c_n \ (n \in \mathbb{N}_+)$, then$$
\cos θ_{n + 1} = \sqrt{\frac{1 + \cos θ_n}{2}} = \cos \frac{θ_n}{2} \Longrightarrow θ_{n + 1} = \frac{θ_n}{2}.
$$
Therefore, $θ_n = \dfrac{θ}{2^{n - 1}} \ (n \in \mathbb{N}_+)$, which implies$$
M_n = 2^n \tan \frac{θ}{2^{n - 1}}.
$$
A: Convergence :
$$
M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)
$$
$$
=>(\frac{M_{n+1}M_n}{2^{2n+1}}+1)^2=1+ 2^{-2n}M^2_{n}
$$
On Solving,
$$
(\frac{M_{n+1}}{2^{n+1}})^2+\frac{M_{n+1}}{M_n}=1
$$
As $(\frac{M_{n+1}}{2^{n+1}})^2 > 0$ ($M_n>0$ for all $n\geq1$). So, $(\frac{M_{n+1}}{2^{n+1}})^2+\frac{M_{n+1}}{M_n}>\frac{M_{n+1}}{M_n}$ which implies $1>\frac{M_{n+1}}{M_n}$. Hence, we get $M_n>M_{n+1}$.
Closed Form:
I am not able to find the closed form. I also try it with Wolfram Alpha but it doesn't work.
