Ford Circles

Let us denote $C(p,n)$ the area of a circle, where p and n are coprime.

$C(p,n)=\pi.\frac{1}{4.n^4}$

Since $C(p,n)$ is the same for every p, coprime to n, we get that all of the area is

$\displaystyle\sum_{n=1}^{\infty} \frac{\pi}{4}.\frac{\varphi (n)}{n^4}=\frac{\pi}{4}\sum_{n=1}^{\infty} \frac{\varphi (n)}{n^4}$ .

Now this is a formal Dirichlet series where $a(n)=\varphi(n)$ .

We know that

$D(a,s).D(b,s)=D(a•b,s)$ ,

where a•b is the convolution of the functions a and b, and the D stands for Dirichlet series.

Let $a(n)=\varphi(n)$ and $b(n)=1$ .

Then $D(b,s)=\zeta(s)$

Also, the convolution of a and b becomes

$\displaystyle \sum_{d|n}\varphi(d).b(\frac{n}{d})= \sum_{d|n}\varphi(d)=n$ , so

$D(a,s).D(b,s)=D(a•b,s)$

$D(a,s).\zeta(s)=\zeta(s-1)$

$D(a,s)=\frac{\zeta(s-1)}{\zeta(s)}$

Now we need to plug in s=4 and our answer is

$\boxed{\frac{\pi}{4}.\frac{\zeta(3)}{\zeta(4)}}$