This is from a problem in Spivak's Calculus. It is in the chapter on least upper bounds, though this particular item of a problem is related to geometry.
11b) Suppose $P$ is a regular polygon inscribed inside a circle. If $P'$ is the inscribed regular polygon with twice as many sides, show that the difference between the area of the circle and the area of the $P'$ is less than half the difference between the area of the circle and the area of $P$.
This is easy to show using a geometric argument:
We want to show that
$$R_2 < \frac{1}{2}(R_1+R_2)$$
$$R_2<R_1$$
This is clearly true because $R_2<R_2+R_3=R_1$.
I'd like an analytical proof. Below I will simply show what I did, but it's a path that didn't lead me to an answer, feel free to ignore it.
Here is what I thought was leading to the correct conclusion but actually not.
$A_1=\frac{b_1h_1}{2}$
$A_2=\frac{b_2h_2}{2}$
$\sin{\left (\frac{\alpha}{2} \right )}=\frac{b_1}{r}$
$b_1 = r \sin{\left (\frac{\alpha}{2} \right )}\tag{1}$
$\cos{\left (\frac{\alpha}{2} \right )} = \frac{h_1}{r}$
$$h_1 = r\cos{\left (\frac{\alpha}{2} \right )}\tag{2}$$
Therefore, $$A_1 = \frac{r^2 \sin{\left (\frac{\alpha}{2} \right )} \cos{\left (\frac{\alpha}{2} \right )}}{2}\tag{3}$$
$\sin{\beta}=\frac{b_1}{b_2}$
$$b_1 = b_2 \sin{\beta}\tag{4}$$
From $(1)$ and $(4)$
$$r \sin{\left (\frac{\alpha}{2} \right )} = b_2\sin{\beta}$$ $$b_2 = \frac{r \sin{\left (\frac{\alpha}{2} \right )}}{\sin{\beta}}\tag{5}$$
$\sin{\beta}=\frac{h_2}{r}$
$$h_2 = r \sin{\beta}\tag{6}$$
From $(5)$ and $(6)$
$$A_2 = \frac{r^2 \sin{\left (\frac{\alpha}{2} \right )}}{2}\tag{7}$$
Therefore,
$$A_1=A_2 \cos{\left (\frac{\alpha}{2} \right )}$$
But an inscribed polygon has $n \geq 3$ sides.
$$\alpha = \frac{2\pi}{n}, n \geq 3$$
$$0 < \alpha < \frac{2\pi}{3}$$
$$0 < \frac{\alpha}{2} < \frac{\pi}{3}$$
$$\frac{1}{2} \leq \cos{\left (\frac{\alpha}{2} \right )} < 1$$
$$\frac{A_2}{2} \leq A_2 \cos{\left (\frac{\alpha}{2} \right )} < A_2$$
$$\frac{A_2}{2} \leq A_1 < A_2$$
So we know that $A_1$ is more than half as large as $A_2$, and smaller than $A_2$
$$-A_2 < -A_1 \leq -\frac{A_2}{2}$$
$$\pi r^2-A_2 < \pi r^2-A_1 \leq \pi r^2-\frac{A_2}{2}\tag{8}$$
$(8)$ doesn't seem quite right.
