Nested Tetrahedrons and Spheres

Prior to doing the problem note that the ratio $\dfrac{V_{s}}{V_{T}} = \dfrac{V_{s}(n)}{V_{T}(n)} \:\ \forall n \in \mathbb{N}$ ,
so that $\dfrac{V_{s}}{V_{T}} = \dfrac{V_{s}(1)}{V_{T}(1)} = \dfrac{3\sqrt{3}}{2}\pi$ , but since the problem was about constructing a nested sequence of inscribed and circumscribed spheres and regular tetrahedrons, I provide the construction below.

$V_{s}(1) = \dfrac{4}{3}\pi R_{1}^3$ and $V_{T}(1) = \dfrac{a_{1}^3}{6\sqrt{2}}$

$R_{1}^2 = \dfrac{2}{3}a_{1}^2 - 2\sqrt{\dfrac{2}{3}}a_{1}R_{1} + R_{1}^2 + \dfrac{a_{1}^2}{3} \implies a_{1}^2 - 2\sqrt{\dfrac{2}{3}}a_{1}R_{1} =0 \implies a_{1}(a_{1} - 2\sqrt{\dfrac{2}{3}}R_{1}) = 0$

$a_{1} \neq 0 \implies a_{1} = 2\sqrt{\dfrac{2}{3}}R_{1}$

$\implies V_{1}(T) = \dfrac{8}{9\sqrt{3}}R_{1}^3 = \dfrac{4}{3}\pi R^3(\dfrac{2}{3\sqrt{3}\pi}) = \dfrac{2}{3\sqrt{3}\pi}V_{1}(s)$

$\dfrac{a_{1}^2}{3} + R_{2}^2 = \dfrac{2}{3}a_{1}^2 - 2\sqrt{\dfrac{2}{3}}a_{1}R_{2} + R_{2}^2 \implies \dfrac{a_{1}^2}{3} - 2\sqrt{\dfrac{2}{3}}a_{1}R_{2} = 0 \implies a_{1}(\dfrac{a_{1}}{3} - 2\sqrt{\dfrac{2}{3}}R_{2}) = 0$

$a_{1} \neq 0 \implies R_{2} = \dfrac{\sqrt{3}}{6\sqrt{2}}a_{1} =$ $\dfrac{\sqrt{3}}{6\sqrt{2}}(2\sqrt{\dfrac{2}{3}})R_{1}$ = $\dfrac{R_{1}}{3}$

$\implies V_{s}(2) = \dfrac{4\pi}{3}(\dfrac{R_{1}}{3})^3 = \dfrac{1}{27}V_{s}(1)$

$\therefore a_{1} = 2\sqrt{\dfrac{2}{3}}R_{1}, R_{2} = \dfrac{R_{1}}{3}$ and $a_{2} = 2\sqrt{\dfrac{2}{3}}R_{2} = 2\sqrt{\dfrac{2}{3}}\dfrac{R_{1}}{3}, R_{3} = \dfrac{R_{2}}{3} = \dfrac{R_{1}}{3^2},$ $a_{3} = 2\sqrt{\dfrac{2}{3}}R_{3} = 2\sqrt{\dfrac{2}{3}}\dfrac{R_{1}}{3^2}$

In General:

$R_{n} = (\dfrac{1}{3})^{n - 1}R_{1}$ and $a_{n} = 2\sqrt{\dfrac{2}{3}}(\dfrac{1}{3})^{n - 1}R_{1}$

$\implies V_{s}(n) = (\dfrac{1}{27})^{n - 1}V_{s}(1)$ and $V_{T}(n) = \dfrac{2}{3\sqrt{3}\pi}(\dfrac{1}{27})^{n - 1}V_{s}(1)$

and,

$V_{s} = \sum_{n = 1}^{\infty} V_{s}(n) = V_{s}(1)\sum_{n = 1}^{\infty} (\dfrac{1}{27})^{n - 1} = \dfrac{27}{26}V_{s}(1)$

and

$V_{T} = \sum_{n = 1}^{\infty} V_{T}(n) = \dfrac{2}{3\sqrt{3}\pi}V_{s}(1)\sum_{n = 1}^{\infty} (\dfrac{1}{27})^{n - 1} = \dfrac{2}{3\sqrt{3}\pi}(\dfrac{27}{26})V_{s}(1)$

$\implies \dfrac{V_{s}}{V_{T}} = \dfrac{3\sqrt{3}}{2}\pi = \dfrac{a\sqrt{a}}{b}\pi \implies a + b = \boxed{5}$ .