We divide the unit circle into equal sectors and inscribe circles within each sector. Assuming we make smaller and smaller sectors, what does the total area of the inscribed circles converges to?
I think I have solved it, but since I'm not a mathematician, I wonder if this is correct.
I began by asking: What is the total area of all the circles in a single sector?
We will start with the unit circle as the outer perimeter, with $\theta$ being half of the subtended angle, and $r_0$ being the radius of the first ring of circles, $r_1$ being the radius of the second ring of circles, and so on.
Since the unit circle has a radius of 1, we can immediately deduce the following: $$r_{0} = ({\sin{\theta} \over 1+\sin{\theta}})$$ and we also know that: $$r_{n+1} = r_{n} ({1-\sin{\theta} \over 1+\sin{\theta}})$$ that means that: $$r_{1} = ({\sin{\theta} \over 1+\sin{\theta}}) ({1-\sin{\theta} \over 1+\sin{\theta}}) = {(\sin{\theta})({1-\sin{\theta})} \over (1+\sin{\theta})^2}$$ and: $$r_{2} = {(\sin{\theta})({1-\sin{\theta})^2} \over (1+\sin{\theta})^3}$$ so generally we can say that: $$r_{n} = {(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}}$$ the area of a single circle in a sector is: $$A_{n} = \pi r_{n}^2 = \pi ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2$$ so the area of a single sector of circles will be: $$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 $$ and since we're dealing with two or more sectors, we can say that: $$ 0 < \theta \leq {\pi \over 2} $$ so this means the sum of areas of all the circles inscribed in a sector converges to: $$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 = {\pi \sin{\theta} \over 4}$$
I then continued with: What is the total area of all the circles in all the sectors?
Assuming we have $k$ sectors, then $\theta = {\pi \over k}$ and our sum would be: $$A_{k} = {k A_{s}} = {\pi k \sin{\pi \over k} \over 4}$$ and if we take the limit of $k$, we will get: $$\lim_{k \to \infty} {\pi k \sin{\pi \over k} \over 4} = {\pi^2 \over 4} \tag*{$\blacksquare$}$$
Is this valid?