It's All Squares 3 !

$\dfrac{2a_{1}}{1 - a_{1}} = \tan(60^{\circ}) = \sqrt{3} \implies a_{1} = \dfrac{\sqrt{3}}{2 + \sqrt{3}}$

and

$\dfrac{2a_{2}}{a_{1} - a_{2}} = \tan(60^{\circ}) = \sqrt{3} \implies a_{2} = \dfrac{\sqrt{3}a_{1}}{2 + \sqrt{3}} = (\dfrac{\sqrt{3}}{2 + \sqrt{3}})^2$

$\dfrac{2a_{3}}{a_{2} - a_{3}} = \sqrt{3} \implies a_{3} = \dfrac{\sqrt{3}a_{2}}{2 + \sqrt{3}} = (\dfrac{\sqrt{3}}{2 + \sqrt{3}})^3$

In General for each positive integer $n$ we have: $a_{n} = (\dfrac{\sqrt{3}}{2 + \sqrt{3}})^n$

$\implies$

$S = \sum_{n = 1}^{\infty} a_{n}^2 = (\dfrac{\sqrt{3}}{2 + \sqrt{3}})^2(\dfrac{2 + \sqrt{3}}{2}) =$

$\dfrac{3}{2}(\dfrac{1}{2 + \sqrt{3}}) = \dfrac{\sqrt{3}}{2}(\dfrac{\sqrt{3}}{2 + \sqrt{3}}) =$

$\dfrac{\sqrt{3}}{2}a_{1} = height_{\triangle{ABC}} * a_{1} \implies$ $\dfrac{S}{a_{1}} = height_{\triangle{ABC}} = \dfrac{\sqrt{3}}{2} = \dfrac{\alpha}{\beta}$

$\implies \alpha + \beta = \boxed{5}$ .