Separate and conquer

Relevant wiki: Taylor Series - Problem Solving

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {\cos \left(\frac {n \pi}3 \right)}{3^n} = \sum_{n=1}^\infty \frac {\Re \left( e^{\frac {n \pi}3 i} \right)}{3^n} = \Re \left[ \sum_{n=1}^\infty \left( \frac {e^{\frac \pi 3 i}}3 \right)^n \right] \\ & = \Re \left[ \frac {\frac {e^{\frac \pi 3 i}}3}{1 - \frac {e^{\frac \pi 3 i}}3} \right] = \Re \left[ \frac {e^{\frac \pi 3 i}}{3 - e^{\frac \pi 3 i}} \right] = \Re \left[ \frac {\frac 12 + \frac {\sqrt 3}2i}{3 - \frac 12 - \frac {\sqrt 3}2i} \right] \\ & = \Re \left[ \frac {1 + \sqrt 3 i}{5 - \sqrt 3 i} \right] = \Re \left[ \frac {(1 + \sqrt 3 i)(5 + \sqrt 3 i)}{(5 - \sqrt 3 i)(5 + \sqrt 3 i)} \right] \\ & = \Re \left[ \frac {2 + 6\sqrt 3 i}{28} \right] = \frac 1{14} \end{aligned}$

$\implies \alpha + \beta = 1 + 14 = \boxed{15}$

Relevant wiki: Taylor Series Manipulation

Call the expression $\mathcal G$ and write $\small{\color{teal}{\cos \left(\dfrac{n\pi}3\right)=\dfrac{e^{\large\frac{n\pi i}3}+e^{\large\frac{-n\pi i}3}}2}}$ .

$\mathcal G=\dfrac 12 \left[\displaystyle \sum_{n=1}^{\infty}\underbrace{\left(\dfrac{ e^{\frac{\pi i}3}}3\right)^n}_{\text{Infinite}\\\text{ GP}}+\displaystyle \sum_{n=1}^{\infty}\underbrace{\left(\dfrac{ e^{\frac{-\pi i}3}}3\right)^n}_{\text{Infinite}\\ \text{ GP}}\right]$

$=\dfrac 12\left[\dfrac{\dfrac{ e^{\frac{\pi i}3}}3 }{1-\dfrac{ e^{\frac{\pi i}3}}3}+\dfrac{\dfrac{ e^{\frac{-\pi i}3}}3 }{1- \dfrac{ e^{\frac{-\pi i}3}}3}\right]$

Taking LCM and simplifying gives $\mathcal G=\dfrac 1{14}$ . $\large 1+14=\boxed{15}$ .