A nice answer!

Let $R = 1$ and label $C$ and $D$ as follows:

Then $PD = DO = CD = \cfrac{1}{2}$ , and by the Pythagorean Theorem on $\triangle ADO, AD = \sqrt{AO^2 + DO^2} = \sqrt{1^2 + \bigg(\cfrac{1}{2}\bigg)^2} = \cfrac{\sqrt{5}}{2}$ .

Then $AC = AD - CD = \cfrac{\sqrt{5}}{2} - \cfrac{1}{2} = \cfrac{\sqrt{5} - 1}{2}$ .

The sum of the radii of all the red circles is $S = \cfrac{1}{2} AC + CD = \cfrac{1}{2} \cdot \cfrac{\sqrt{5} - 1}{2} + \cfrac{1}{2} = \cfrac{\sqrt{5} + 1}{4}$ .

Therefore, $\cfrac{2S}{R} = \cfrac{2 \cdot \frac{\sqrt{5} + 1}{4}}{1} = \cfrac{\sqrt{5} + 1}{2} = \phi \approx \boxed{1.618}$ .

$Aw_{0} = \sqrt{5}R_{1} \implies \sqrt{5} = \dfrac{R_{1} + R_{2}}{R_{1} - R_{2}}$ $\implies (\sqrt{5} - 1)R_{1} = (\sqrt{5} + 1)R_{2} \implies$

$R_{2} = \dfrac{\sqrt{5} - 1}{\sqrt{5} + 1}R_{1}$ and $\sqrt{5} = \dfrac{R_{2} + R_{3}}{R_{2} - R_{3}} \implies R_{3} = \dfrac{\sqrt{5} - 1}{\sqrt{5} + 1}R_{2} = (\dfrac{\sqrt{5} - 1}{\sqrt{5} + 1})^2R_{1}$

In General $R_{n} = (\dfrac{\phi - 1}{\phi})^{n - 1} R_{1} \implies S = R_{1}\sum_{n = 1}^{\infty} R_{n} = \phi R_{1} = \dfrac{\phi}{2}R \implies$

$\dfrac{2S}{R} = \phi \approx \boxed{1.61803398875}$