Finding the area of this figure inside a circle? Let's say I have a unit circle and I draw a triangular sector with an angle $\pi/2$. Next to it, I draw a sector with an angle $\pi/8$, next $\pi/18$, so that the $n^{\text{th}}$ sector has an angle of $\pi/(2n^2)$. Now I am only looking at the area of the triangle inside each sector. What is the area of the triangles if I were to draw infinitely many of them, as shown in the figure below? 
I was only able to find an approximation for this. I got $f(1)/2 + \int_1^\infty f(x)dx$, where $f(x)=(1/2)\sin(\pi/(2x^2)) \approx 1.95$. How would you find the exact area of this?

 A: As you seem to know, the area of an isosceles triangle with common side length 1 and angle between them $\theta$ is $\sin(\theta)/2$.  Thus, your area is
$$\sum _{i=1}^{\infty } \frac{1}{2} \sin \left(\frac{\pi }{2 i^2}\right),$$
which is just a bit bigger than 1.  I don't see any easy reason to think that this is expressible in closed form; Mathematica, at least, can't seem to do so.
A: The union of the sectors should be a sector of angle
$$\sum_{n=1}^\infty\frac{\pi}{2n^2}=\frac{\pi}{2}\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi}{2}\cdot\frac{\pi^2}{6}=\frac{\pi^3}{12}$$
and the area of a sector of angle $\theta$ in a circle of radius $r$ is $\frac{\theta}{2\pi}\cdot\pi r^2$, so the area of the union of the sectors is
$$\frac{\pi^3/12}{2\pi}\cdot\pi=\frac{\pi^3}{24}$$
A: Area of a sector with angle $\theta$ is $\frac{\theta}{2}$ (Can you see this?).
Now, in your case, the area is given by
$ A = \sum_{x=1}^{x=\infty}\frac{\pi}{4x^{2}} = \frac{\pi}{4}\sum_{x=1}^{\infty}\frac{1}{x^{2}}$
Do you know the value of the last sum? It is $\frac{\pi^{2}}{6}$.
A: The resulting shape is a sector of the circle.  It would be helpful to find the central angle of this sector.  This angle is equal to the following sum:
$$\theta=\frac{\pi}{2}+\frac{\pi}{8}+\frac{\pi}{18}+\ldots=\frac{\pi}{2}\left(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\ldots\right)$$
Now here is a rather astonishing fact: the term inside the parentheses is equal to $\pi^2/6$.  This means we'd like to find the area of the sector of a unit circle ($r=1$) with central angle $\theta=\pi^3/12$.  The area is then given by:
$$A=\frac{1}{2}\cdot r\cdot\theta=\frac{1}{2}\cdot 1\cdot\frac{\pi^3}{12}=\frac{\pi^3}{24}$$
