Showing $\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$ $$\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$$
We cannot use the Herschfeld's Convergence Theorem because you can see that it's negative. I have tried some generalization but nothing good. I have looked for some algorithm on the web like Landau's algorithm but we speak about and infinite nested radicals so I was wondering myself if there is not a recursive formula existing. 
So if you have some papers or a proof, I will be thankful.
Any helps is greatly appreciated
Thanks a lot.
 A: I suppose the right-hand side is defined (that is, interpreted) as $\lim\limits_{n\to\infty}a_n(1)$, where $$a_1(x)=x,\qquad a_{n+1}(x)=a_n\big(\sqrt{1-2^{-n}x}\big).$$ Now observe that $a_n(1-2^{-n-1})=3/4$ for each $n$; to prove it, use induction and $$a_{n+1}(1-2^{-n-2})=a_n\left(\sqrt{1-2^{-n}(1-2^{-n-2})}\right)=a_n(1-2^{-n-1}).$$ Thus, it remains to prove $\lim\limits_{n\to\infty}\big(a_n(1)-a_n(1-2^{-n-1})\big)=0$. This can be done using the mean value theorem: we have $a_n(1)-a_n(1-2^{-n-1})=2^{-n-1}a_n'(x_n)$ for some $x_n\in(1-2^{-n-1},1)$, and the (very crude) estimate $$x\in(0,1)\implies|a_n'(x)|\leqslant 1\quad\color{gray}{\left[\impliedby a_{n+1}'(x)=-\frac{a_n'\big(\sqrt{1-2^{-n}x}\big)}{2^{n+1}\sqrt{1-2^{-n}x}}\right]}$$ suffices (of course, a much more precise result can be obtained this way).
A: Induction at work:
$$\frac{3}{4}=\sqrt{1-\frac{7}{16}}=
\sqrt{1-\color{blue}{\frac{1}{2}}\cdot\color{red}{\frac{7}{8}}}=\\
\sqrt{1-\frac{1}{2}\sqrt{1-\frac{15}{64}}}=
\sqrt{1-\frac{1}{2}\sqrt{1-\color{blue}{\frac{1}{4}}\cdot\color{red}{\frac{15}{16}}}}=\\
\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{\frac{225}{256}}}}=
\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\color{blue}{\frac{1}{8}}\cdot\color{red}{\frac{31}{32}}}}}$$
Do you see the pattern?
At every step (in blue/red) we have something like $\color{blue}{\frac{1}{2^{n-2}}}\cdot\color{red}{\frac{2^n-1}{2^n}}$, then
$$\color{blue}{\frac{1}{2^{n-2}}}\cdot\color{red}{\frac{2^n-1}{2^n}}=
\color{blue}{\frac{1}{2^{n-2}}}\sqrt{\frac{2^{2n}-2^{n+1}+1}{2^{2n}}}=\\
\color{blue}{\frac{1}{2^{n-2}}}\sqrt{1-\color{blue}{\frac{1}{2^{n-1}}}\cdot\color{red}{\frac{2^{n+1}-1}{2^{n+1}}}}$$
and the result follows ...
A: Let $a_{2n}=\sqrt{1-\dfrac12\sqrt{...\sqrt{1-\dfrac1{2^{2n}}}}}$
$a_{2n+2}=\sqrt{1-\dfrac12\sqrt{...\sqrt{1-\dfrac1{2^{2n}}\sqrt{1-\dfrac1{2^{2n+1}}\sqrt{1-\dfrac1{2^{2n+2}}}}}}}$
$\sqrt{1-\dfrac1{2^{2n+2}}}\lt 1 \implies \sqrt{1-\dfrac1{2^{2n+1}}\sqrt{1-\dfrac1{2^{2n+2}}}}\gt\sqrt{1-\dfrac1{2^{2n+1}}}\implies \sqrt{1-\dfrac1{2^{2n}}\sqrt{1-\dfrac1{2^{2n+1}}\sqrt{1-\dfrac1{2^{2n+2}}}}}\lt \sqrt{1-\dfrac1{2^{2n}}\sqrt{1-\dfrac1{2^{2n+1}}}}$
The minus signs balance out, so $a_{2n+1}\gt a_{2n}, a_{2n+2}\gt a_{2n}$
 On the other hand it is clear that $a_n\lt 1 \;\forall n\in \Bbb N, $ so $\lim_{n\to \infty}a_n$ exists.
