<s>[INST] 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.
 [/INST] 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 ...
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