<s>[INST] Finding value of infinite series limit 
Finding value of $\displaystyle \lim_{n\rightarrow \infty}\frac{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+........+\sqrt{2}}}}}_{\bf{n\; times}}}{4^{-n}}$

Try: I am trying to convert it into $\cos$ ine series sum
Let $$\displaystyle \sqrt{2+2\cos \theta } = 2\cos \frac{\theta}{2}$$ and 
$$\displaystyle \sqrt{2+\sqrt{2+2\cos \theta}} = 2\cos \frac{\theta}{4}$$
$$\displaystyle \sqrt{2+\sqrt{2+\sqrt{2+2\cos \theta}}} = 2\cos \frac{\theta}{8}$$
could some help me how i write $$\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+........+\sqrt{2}}}}}_{\bf{n\; times}}$$ into cosine series form. thanks
 [/INST] $$
\eqalign{
  & \underbrace {\sqrt 2 }_1 = 2\cos \left( {\pi /4} \right)  \cr 
  & \underbrace {\sqrt {2 + \sqrt 2 } }_2 = \sqrt {2 + 2\cos \left( {\pi /4} \right)}  =   \cr 
  &  = \sqrt {2\left( {1 + \cos \left( {\pi /4} \right)} \right)}  = \sqrt {2\left( {1 + \cos ^2 \left( {\pi /8} \right) - \sin ^2 \left( {\pi /8} \right)} \right)}  =   \cr 
  &  = \sqrt {4\cos ^2 \left( {\pi /8} \right)}  = 2\cos \left( {\pi /8} \right)  \cr 
  & \quad \quad  \vdots   \cr 
  & \underbrace {\sqrt {2 + \sqrt { \cdots  + \sqrt 2 } } }_n = 2\cos \left( {\pi /2^{n + 1} } \right) \cr} 
$$
Then probably you are looking for the value of $\lim_ {n \to \infty}{(2-\sqrt{2+\sqrt{\cdots}})4^n}$.
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