Geometry problem involving infinite number of circles What is the sum of the areas of the grey circles? I have not made any progress so far.


This is an alternate approach to derive the area of the circles
using Descartes four circle theorem.
WOLOG, assume $R = 1$. Let us call


*

*the outer circle (with radius $r_a = 1$) as $C_a$.

*the inner green circle (with radius $r_b = \frac12$) as $C_b$.

*the $1^{st}$ gray circle (the largest one at bottom) as $C_0$.

*the $2^{nd}$ gray circle (the one above $C_b$ and $C_0$) as $C_1$.

*in general, for any $k > 0$, we will call the gray circle above $C_b$ and $C_{k-1}$ as $C_k$.

*We can reflect the figure in question vertically. For $k < 0$, we will let $C_k$ be the mirror image of the circle  $C_{-(1+k)}$.


The key is for any $k\in\mathbb{Z}$, the four circles $C_a, C_b, C_k$ and $C_{k\pm 1}$ are kissing each other. Let $r_k$ be the radius of $C_k$. We can apply Descartes four circle theorem to obtain:
$$\left(\frac{1}{r_{k\pm 1}} + \frac{1}{r_k} + \frac{1}{r_b} - \frac{1}{r_a} \right)^2
= 2 \left(\frac{1}{r_{k\pm 1}^2} + \frac{1}{r_k^2} + \frac{1}{r_b^2} + \frac{1}{r_a^2}\right)\tag{*1}$$
Since $r_a = 1$ and $r_b = \frac12$, this implies $\displaystyle\;\frac{1}{r_{k \pm 1}}\;$ are the two roots of quadratic equation
$$\left(\rho + \frac{1}{r_k} + 2 - 1 \right)^2
= 2 \left(\rho^2 + \frac{1}{r_k^2} + 2^2 + 1^2 \right)
$$
and hence
$$\frac{1}{r_{k+1}} + \frac{1}{r_{k-1}} = 2\left(\frac{1}{r_k} + 1\right)
\quad\iff\quad
\frac{1}{r_{k+1}} + \frac{1}{r_{k-1}} - \frac{2}{r_k} = 2$$
The RHS is an inhomogeneous linear recurrence relation on $\displaystyle\;\frac{1}{r_k}\;$ 
with constant term. Since the characteristic polynomial $(\lambda-1)^2$ has a double root at $1$, we know its solution must have the form $\displaystyle\;\frac{1}{r_k} = k^2 + \lambda k + \mu\;$.  By symmetry, 
$$r_k = r_{-(1+k)}\quad\implies\quad \lambda = 1 \quad\implies\quad
\frac{1}{r_k} = k(k+1) + \mu$$
To fix $\mu$, apply $(*1)$ to the case $C_a, C_b, C_0$ and $C_{-1}$, we get
$$\left(2\mu + 1\right)^2 = 2\left(2\mu^2 + 5\right)
\quad\implies\quad \mu = \frac{9}{4}
$$
From this, we get 
$$r_k = \frac{4}{4k^2 + 4k + 9}
\quad\implies\quad
\text{Area} = \sum_{k=0}^\infty \frac{16\pi}{(4k^2+4k+9)^2}$$
reproducing what's in J.J's answer. 
Update
To evaluate the sum, we will use following infinite product expansion of $\cosh(\pi x)$,
$$\cosh\pi x = \prod_{k=0}^\infty \left( 1 + \frac{x^2}{(k+\frac12)^2}\right)
$$
Taking logarithm, differentiate w.r.t $x$ and divide by $2x$, we get
$$\sum_{k=0}^\infty \frac{1}{(k+\frac12)^2 + x^2} = \frac{\pi}{2x} \tanh\pi x$$
Differentiate w.r.t $x$ and divide by $-2x$ once more, we get
$$\sum_{k=0}^\infty \frac{1}{((k+\frac12)^2 + x^2)^2} 
= \frac{\pi}{4x^3}\tanh(\pi x) - \frac{\pi^2}{4x^2\cosh^2(\pi x)}
$$
With this, we find
$$\text{Area} 
=  \sum_{k=0}^\infty \frac{\pi}{((k+\frac12)^2 + 2)^2}
=  \frac{\pi^2}{16}\left[\sqrt{2}\tanh(\pi\sqrt{2}) - \frac{2\pi}{\cosh^2(\pi\sqrt{2})}\right]
$$
Once again, this agrees with J.J's result.
