Prove $\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$ Given the sequence $x_1 = 0$, $x_{n+1} = \sqrt{2+x_n}$, proove:
$$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$$
I have used the relations
$$2\cos({\frac x 2}) = \sqrt{2 + 2\cos(x)}$$
and
$$2\sin({\frac x 2}) = \sqrt{2 - 2\cos(x)}$$
Observe:
$x_1 = 0 = 2\cos(\alpha) \iff \alpha = \frac \pi 2$
I then hoped we can set $x_n = 2\cos(\frac{\alpha}{2^{n-1}})$ but if we can do this, why is that?
I assummed we can, so:
$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \lim_{n\to\infty} 2^n \sqrt{2-2\cos(\frac{\alpha}{2^{n-1}})} = \lim_{n\to\infty} 2^n \sqrt{4 \sin^2(\frac{\alpha}{2^n})}$
We know $$\alpha = \frac \pi 2$$
By substituting:
$ = \lim_{n\to\infty} 2^{n+1} \sin(\frac{\pi}{2^{n+1}})$
Again, by substituting
$$\frac{\pi}{2^{n+1}} = \frac 1 m$$
$= \lim_{m\to\infty} \pi m \sin(\frac 1 m)= \pi$ 
Is this proof correct, and if not, how to prove it?
The even more important question is the why is that? part mentioned above.
 A: $\cos(x) = \cos(2\cdot \frac{x}{2})= \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = -1+ 2\cos^2(\frac{x}{2})$ and hence $\cos(\frac{x}{2})= \sqrt{\frac{1}{2}+\frac{1}{2}\cos(x)} = \frac{1}{2}\sqrt{2+2\cos(x)}$. That is the inductive step for the calculation.
Geometrically, you can check that $2^n \sqrt{2-x_n}$ is half the perimeter of a regular $2^{n+1}$-gon which is inscribed in the unit circle. Which for $n \to \infty$ approaches half the circumference of a unit circle, which is $\pi$.
Added: About your question what happens when you choose a different starting value $x_1 \in [-2,2]$ (note that outside this interval, at least one of the square roots in the calculation becomes unreal, and accordingly, the geometric interpretation breaks down).


*

*In the calculation: You now get $\alpha = \displaystyle \arccos\left(\frac{x_1}{2}\right)$. The rest of the computation stays literally the same, just at the end you get the limit $\lim_{n \to \infty} 2^{n+1}\cdot \sin\displaystyle\left(\frac{\alpha}{2^n}\right) = 2\alpha$.

*Geometrically: You get a polygonal chain of $2^n$ pieces, each of which is a chord within a circular segment of arc length $\displaystyle\frac{\alpha}{2^{n-1}}$. So the total arc in which this polygonal chain is inscribed has arc length $2\alpha$. As $n$ grows, the polygonal chain approximates the arc which gives the geometric reason for the limit being just $2\alpha$.


(Archimedes would be so happy with this.)
