How Might Geometry Be Useful Here?

Suppose exactly $2^n$ identical line segments are inscribed in a semicircle of radius $1$ (like the 4 green lines in the picture above). Then I say $s_n$ is the length of each of the identical line segments.

If $s_n$ is the length of each line segment, then their total length is $2^n s_n$ . As $n$ approaches infinity, the green lines approximate the circumference of the semicircle better and better, and hence their total length gets closer and closer to $\pi$ . Thus $2^n s_n \to \boxed{\pi}$ .

All we need to do is prove the formula for $s_n$ : $s_n = 2\sin{\left(\dfrac{\pi}{2^{n+1}}\right)}$ (I leave it as an exercise to show that this is the correct formula for the length of each line segment).

This can be proved by induction:

$s_0 = 2 = 2\sin{\left(\dfrac{\pi}{2^{1}}\right)}$ , so the base case works.

Now suppose that for some $k \in \mathbb{N}$ , we have $s_k = 2\sin{\left(\dfrac{\pi}{2^{k+1}}\right)}$ .

Let $r = \dfrac{\pi}{2^{k+2}}$ . Then, $\cos(2r) = 1 - 2\sin^2{r}$ $\sin(r) = \pm\sqrt{\dfrac{1-\cos(2r)}{2}}$ $= \pm\sqrt{\dfrac{1 \mp \sqrt{1-\sin^2(2r)}}{2}}$ Since $0 < r < 2r \le \dfrac{\pi}{2}$ , then we must have $\sin(r) > 0$ and $\cos(2r) \ge 0$ . Hence, $\sin(r) = \sqrt{\dfrac{1 - \sqrt{1-\sin^2(2r)}}{2}}$ $2\sin{(r)} = 2\sqrt{\dfrac{1 - \sqrt{1-\sin^2(2r)}}{2}}$ $2\sin(r) = \sqrt{2 - 2\sqrt{1-\sin^2(2r)}} = \sqrt{2 - \sqrt{4-4\sin^2(2r)}}$ $2\sin\left( \dfrac{\pi}{2^{k+2}}\right) = \sqrt{2 - \sqrt{4-s_k^2}} = s_{k+1}$

Hence the induction is complete. Thus $s_n = 2\sin{\left(\dfrac{\pi}{2^{n+1}}\right)}$ for all $n \in \mathbb{N}$ .

Let $s_n = 2\sin(\theta)$

$s_{n+1} = \sqrt{2-\sqrt{4-s_n^{2}}}$

$\implies s_{n+1} = \sqrt{2-2\sqrt{1-(\frac{s_n}{2})^{2}}}$

Substituting $s_n = 2\sin(\theta)$ :

$s_{n+1} = \sqrt{2-2\sqrt{1-sin^{2}(\theta)}}$

$\implies s_{n+1} = \sqrt{2-2\sqrt{cos^{2}(\theta)}}$

$\implies s_{n+1} = \sqrt{2-2cos(\theta)}$

$\implies s_{n+1} = \sqrt{2(1-cos(\theta))}$

$\implies s_{n+1} = \sqrt{2(2sin^{2}(\frac{\theta}{2}))}$

$\implies s_{n+1} = 2\sin(\frac{\theta}{2})$

If we choose $s_0 = 2\sin(\theta)$ , it can shown via induction that $s_n = 2\sin(\frac{\theta}{2^{n}})$

$\because s_0 = 2 \implies \theta = \frac{\pi}{2}$

Finally, $\lim_{n\to\infty}2^{n}s_n = \lim_{n\to\infty}2^{n}(2\sin(\frac{\theta}{2^{n}}))$

$=\lim_{n\to\infty}2\theta(\frac{sin(\frac{\theta}{2^{n}})}{\frac{\theta}{2^{n}}})$ $=2\theta$ $=2\frac{\pi}{2}$

$=\boxed{\pi}$