Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\to\infty$, where $t_n=\sqrt{2+t_{n-1}}$ 
Possible Duplicate:
Why is this series of square root of twos equal $\pi$? 

Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\rightarrow\infty$, where $t_1=\sqrt{2}$, $t_2=\sqrt{2+\sqrt{2}}$, $t_3=\sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on.

$L_n = 2^{n+1} \sqrt{2-t_n}$. Then $L_n = 2^{n+1} \sqrt{2 - \sqrt{2+t_{n-1}}}$. Now $$2 - \sqrt{2+t_{n-1}} = \dfrac{4 - (2+t_{n-1})}{2 + \sqrt{2+t_{n-1}}} = \dfrac{2-t_{n-1}}{2 + t_n} = \dfrac{2-t_{n-1}}{t_{n+1}^2}$$
Hence, $$L_n = 2^{n+1} \sqrt{\dfrac{2-t_{n-1}}{t_{n+1}^2}} = \dfrac{2L_{n-1}}{t_{n+1}}$$
We have $L_0 = 2 \sqrt{2}$. Hence,
$$L_n = \dfrac{2}{t_{n+1}} \cdot \dfrac{2}{t_{n}} \cdot \dfrac{2}{t_{n-1}} \cdots \dfrac{2}{t_{2}} L_0 = \dfrac{2^n}{\displaystyle \prod_{k=2}^{n+1} t_k} 2 \sqrt{2} = 2 \dfrac{2^{n+1}}{\displaystyle \prod_{k=1}^{n+1} t_k}$$
Hence, we need to evaluate $$S_n = \dfrac{\displaystyle \prod_{k=1}^n t_k}{2^n}$$
Recall that $t_n \uparrow 2$. Let $t_ n = 2 \cos(\theta_n)$. We then have
$$\cos(\theta_{n+1}) = \dfrac{t_{n+1}}2 = \dfrac{\sqrt{2+t_n}}2 = \dfrac{\sqrt{2+2 \cos(\theta_n)}}2 = \cos(\theta_n/2)$$
Hence, we get that
$$\cos(\theta_{n+1}) = \cos(\theta_1/2^n)$$
Hence, $$S_n = \dfrac{\displaystyle \prod_{k=1}^n t_k}{2^n} = \prod_{k=0}^{n-1} \cos(\theta_1/2^k) = \dfrac{\displaystyle \sin(\theta_1/2^{n-1})\prod_{k=0}^{n-1} \cos(\theta_1/2^k)}{\sin(\theta_1/2^{n-1})} = \dfrac1{2^n} \dfrac{\sin(2\theta_1)}{\sin(\theta_1/2^{n-1})}$$
Note that $\cos(\theta_1) = \dfrac1{\sqrt{2}}$, which implies $\theta_1 = \dfrac{\pi}4$. Hence,
$$S_n = \dfrac1{2^n \sin(\pi/2^{n+1})}$$ Hence, $S_n \to \dfrac2{\pi}$. Hence, $$L_n \to \pi$$
