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Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$

Hint:

Use the following formulas: $$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$ $$\sin\left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}$$

Any idea how to solve this problem?

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Note that as $x_1 \le 2$, $x_n \le 2 \forall n$. This can be shown inductively.

Thus, we may write $x_n = 2 \cos \theta_n$ for some $\theta_n \in [0, \pi/2]$, with $\theta_1 = \pi/2$

Now, $2 \cos \theta_{n+1} = \sqrt{2(1+ \cos \theta_n)} = 2 \cos \frac{\theta_n}2 \implies \theta_{n+1} = \frac{\theta_n}{2} = \frac{\theta_1}{2^n} = \frac{\pi}{2^{n+1}}$.

This implies that $2^n\sqrt{2 - x_n} = 2^{n+1} \sin \frac{\pi}{2^{n+1}}$. Finish off by using $\frac{\sin x}{x} \to 1$ as $x \to 0$.

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By induction and the formula from the hint, we can easily get $x_n=2\cos(\frac{\pi}{2^n})$. Then $$\lim_{n\rightarrow\infty}2^n\sqrt{2-x_n}=\lim_{n\rightarrow\infty}2^n\sqrt{2}\sqrt{1-\cos(\frac{\pi}{2^n})}=\lim_{n\rightarrow\infty}2^{n+1}\sin\left(\frac{\pi}{2^{n+1}}\right)=\pi$$

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By mathematical induction it can be proved that $x_n=2\cos(\frac{\pi}{2^n})$. for base of induction $x_0=0=2\cos(\frac{\pi}{2})$. Also we have $x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(\frac{\pi}{2^n})}=2\sqrt{\frac{1+\cos(\frac{\pi}{2^n})}{2}}=2\cos(\frac{\pi}{2^{n+1}})$. Therefore $\sqrt{2-x_n}=2\sqrt{\frac{1-\cos(\frac{\pi}{2^n})}{2}}=2\sin(\frac{\pi}{2^{n+1}})$, so $\lim_{n\rightarrow \infty}2^n\sqrt{2-x_n}=\lim_{n\rightarrow \infty}2^{n+1}\sin(\frac{\pi}{2^{n+1}})=\lim_{n\rightarrow \infty}\pi( \frac{\sin(\frac{\pi}{2^{n+1}})}{\frac{\pi}{2^{n+1}}})=\pi$

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