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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

DXT
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1 Answers1

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$$ \eqalign{ & \underbrace {\sqrt 2 }_1 = 2\cos \left( {\pi /4} \right) \cr & \underbrace {\sqrt {2 + \sqrt 2 } }_2 = \sqrt {2 + 2\cos \left( {\pi /4} \right)} = \cr & = \sqrt {2\left( {1 + \cos \left( {\pi /4} \right)} \right)} = \sqrt {2\left( {1 + \cos ^2 \left( {\pi /8} \right) - \sin ^2 \left( {\pi /8} \right)} \right)} = \cr & = \sqrt {4\cos ^2 \left( {\pi /8} \right)} = 2\cos \left( {\pi /8} \right) \cr & \quad \quad \vdots \cr & \underbrace {\sqrt {2 + \sqrt { \cdots + \sqrt 2 } } }_n = 2\cos \left( {\pi /2^{n + 1} } \right) \cr} $$

Then probably you are looking for the value of $\lim_ {n \to \infty}{(2-\sqrt{2+\sqrt{\cdots}})4^n}$.

G Cab
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    You can also mention the asymptotic $\cos x\sim 1-x^2/2$ – Szeto Oct 16 '18 at 09:59
  • @szeto: you are right, and it is applicable if there is a typo in the expression of which to take the limit and it's leading term is $2$ instead of $\sqrt{2}$ – G Cab Oct 16 '18 at 10:24