Circles in Geometric Progression?

For this problem we will use the identity $\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}$ where $x$ and $y$ are the radii of the two larger circles and $z$ is the radius of the smaller circle that is tangent to the two larger circles and the line which is tangent to both the larger circles. This can be easily proved geometrically.

Let $r_n$ be the radius of circle $C_n$ .

With this identity, $\frac{1}{\sqrt{r_1}}+\frac{1}{\sqrt{r_2}} = \frac{1}{\sqrt{r_3}}$ therefore $1 + 1 = \frac{1}{\sqrt{r_3}}$ so $r_3 = \frac{1}{4}$

Similarly, $\frac{1}{\sqrt{r_{n-1}}}+\frac{1}{\sqrt{r_2}} = \frac{1}{\sqrt{r_{n}}}$ for all $n>3$ . (Note $\frac{1}{\sqrt{r_2}} = 1$ )

$\frac{1}{\sqrt{r_{3}}}+ \frac{1}{\sqrt{r_2}} = \frac{1}{\sqrt{r_{4}}}$

$2+ 1 = \frac{1}{\sqrt{r_{4}}}$ so $r_4 = \frac{1}{9}$

You can see that this pattern conitues forever so that $r_n = \frac{1}{(n-1)^2}$ .

Therefore $A_n = \pi (\frac{1}{(n-1)^2})^2$

$\displaystyle \sum_{n=2}^{\infty} A_n = \pi(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}...) = \pi\zeta(4) = \frac{\pi ^5}{90}$ where $\zeta(s)$ is defined as the Riemann-Zeta function.

So $\frac{b}{a} = \boxed{18}$