Building it up.

Since the answer we need is a ratio, for the ease of calculation, we can take $H_0=1$ and $R_0=1$ , therefore $V_{\text{cone}} = \dfrac 12 \pi H_0R_0^2 = \dfrac \pi 3$ .

Considering $r_1$ and $h_1$ , due to similar triangles $\dfrac {h_1}{R_0-r_1} = \dfrac {H_0}{R_0}$ $\implies \dfrac {h_1}{1-r_1} = \dfrac 11$ $\implies h_1 = 1-r_1$ . Then the $V_1$ is given by:

$\begin{aligned} V_1 & = \pi r_1^2h_1 = \pi r_1^2 (1-r_1) = \pi (r_1^2-r_1^3) & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }r_1 \\ \frac {dV_1}{dr_1} & = \pi (2r_1-3r_1^2) & \small \color{#3D99F6} \text{and} \\ \frac {d^2V_1}{dr_1^2} & = \pi (2-6r_1) \end{aligned}$

$\implies \dfrac {dV_1}{dr_1} = 0$ , when $r_1=0$ or $r_1=\dfrac 23$ . This means that $V_1$ is maximum when $r_1=\dfrac 23$ , since $\dfrac {d^2V_1}{dr_1^2} \bigg|_{r_1=\frac 23} < 0$ .

Similarly, we have $r_2 = \dfrac 23 r_1 = \left(\dfrac 23\right)^2 r_0$ , where $r_0=R_0=1$ , $r_3 = \left(\dfrac 23\right)^n r_0$ , $\implies r_n = \left(\dfrac 23\right)^n r_0$ . Similarly, $h_n = r_{n-1} - r_n$ . And $V_n = \pi r_n^2 h_n = \pi r_n^2 (r_{n-1}-r_n)$ . Therefore,

$\begin{aligned} V & = \sum_{n=1}^\infty V_n = \sum_{n=1}^\infty \pi r_n^2 \left(r_{n-1}-r_n\right) & \small \color{#3D99F6} \text{Since }r_n = \left(\frac 23\right)^n \\ & = \pi \sum_{n=1}^\infty \left(\frac 23\right)^{2n} \left(\left(\frac 23\right)^{n-1} - \left(\frac 23\right)^n\right) \\ & = \pi \sum_{n=1}^\infty \left(\frac 23\right)^{3n-1} \left(1 - \frac 23 \right) \\ & = \frac \pi 2 \sum_{n=1}^\infty \left(\frac 8{27}\right)^n \\ & = \frac \pi 2 \cdot \frac 8{27} \left(\frac 1{1-\frac 8{27}}\right) \\ & = \frac 4{19}\pi \end{aligned}$

Therefore, $\dfrac {V^*}{V_{\text{cone}}} = \dfrac {V_{\text{cone}}-V}{V_{\text{cone}}} = 1 - \dfrac {V}{V_{\text{cone}}} = 1 - \dfrac {\frac 4{19}}{\frac 13} = 1 - \dfrac {12}{19} = \dfrac 7{19}$ . Hence $\alpha + \beta = 7 + 19 = \boxed{26}$ .

Let the volume of the cone $V_{cone} = \dfrac{1}{3} \pi R_{0}^2 H_{0}$ and the volume of the cylinder $V_{1} = \pi r_{1}^2 h_{1}.$

From the geometry of the problem we have two similar triangles whose proportion is: $\dfrac{R_{0}}{R_{0} - r-{1}} = \dfrac{H_{0}}{h_{1}} \implies$

$h_{1} = \dfrac{H_{0}(R_{0} - r_{1})}{R_{0}} \implies V_{1} = \dfrac{H_{0} \pi}{R_{0}} * (R_{0}r_{1}^2 - r_{1}^3) \implies$

$\dfrac{dV_{1}}{dr_{1}} = \dfrac{H_{0} \pi }{R_{0}} * r_{1} * (2R_{0} - 3r_{1}) = 0, r_{1} \neq 0 \implies r_{1} = \dfrac{2R_{0}}{3}.$

$\dfrac{d^2V_{1}}{dr_{1}^2} = \dfrac{H_{0} \pi}{R_{0}} * (2R_{0} - 6r_{1})$ and $\dfrac{d^2V_{1}}{dr_{1}^2}|_{r_{1} = \frac{2R_{0}}{3}} = \dfrac{-2H_{0} \pi}{R_{0}} < 0 \implies$ we have a maximum at $r_{0} = \dfrac{2R_{0}}{3}$

and $\boxed{r_{1} = \dfrac{2R_{0}}{3} \implies h_{1} = \dfrac{H_{0}}{3}}$

$\implies$

$\boxed{r_{2} = \dfrac{2}{3} r_{1} = (\dfrac{2}{3})^2 R_{0}}, \:\ H_{1} = H_{0} - h_{1} = \dfrac{2}{3} H_{0} \implies \boxed{h_{2} = \dfrac{H_{1}}{r_{1}}(r_{1} - r_{2}) = \dfrac{2}{3^2} H_{0}}$

$\implies\boxed{r_{3} = \dfrac{2}{3} r_{2} = (\dfrac{2}{3})^3 R_{0}}, \:\ H_{2} = H_{1} - h_{2} = (\dfrac{2}{3})^2 H_{0} \implies \boxed{h_{3} = \dfrac{H_{2}}{r_{2}}(r_{2} - r_{3}) = \dfrac{2^2}{3^2} H_{0}}$

For $n \geq 1 \:\ \boxed{r_{n} = (\dfrac{2}{3})^n R_{0}}, \:\ H_{n - 1} = (\dfrac{2}{3})^{n - 1} H_{0}$ and $\boxed{h_{n} = \dfrac{1}{3}(\dfrac{2}{3})^{n - 1} H_{0}}$ and

$\sum_{n = 1}^{\infty} h_{n} = \dfrac{H_{0}}{3}\sum_{n = 1}^{\infty} (\dfrac{2}{3})^{n - 1} = H_{0}$

$\implies V_{n} = (\dfrac{3}{2})(\dfrac{8}{27})^{n} V_{cone} \implies V = \sum_{n = 1}^{\infty} V_{n} =$ $\dfrac{3}{2}V_{0}\sum_{n = 1}^{\infty} (\dfrac{8}{27})^n = \dfrac{12}{19}V_{cone}$

and,

$V^{*} = V_{cone} - V = V_{cone}(1 - \dfrac{12}{19}) = \dfrac{7}{19}V_{cone} \implies$ $\dfrac{V^{*}}{V_{cone}} = \dfrac{7}{19} = \dfrac{\alpha}{\beta} \implies \alpha + \beta\ = \boxed{26}$ .