Tricky Trigonometric Summation

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {\cos (n \theta)}{2^n} & \small \color{#3D99F6} \text{Since }\cos x = \frac {e^{xi}+e^{-xi}}2 \\ & = \sum_{n=0}^\infty \frac {e^{n \theta i}+e^{-n \theta i}}{2^{n+1}} \\ & = \frac 12 \left(\sum_{n=0}^\infty \frac {e^{n \theta i}}{2^n} + \sum_{n=0}^\infty \frac {e^{- n \theta i}}{2^n} \right) \\ & = \frac 12 \left(\frac 1{1-\frac {e^{\theta i}}2} + \frac 1{1-\frac {e^{-\theta i}}2} \right) \\ & = \frac 1{2-e^{\theta i}} + \frac 1{2-e^{-\theta i}} \\ & = \frac {2-e^{\theta i} + 2-e^{-\theta i}}{(2-e^{\theta i})(2-e^{-\theta i})} \\ & = \frac {4-(e^{\theta i} + e^{-\theta i})}{5-2(e^{\theta i}+e^{-\theta i})} & \small \color{#3D99F6} \text{Since }\cos x = \frac {e^{xi}+e^{-xi}}2 \\ & = \frac {4-2\cos \theta}{5-4\cos \theta} & \small \color{#3D99F6} \text{Note that }\cos \theta = \frac 15 \\ & = \frac {18}{21} = \frac 67 \end{aligned}$

Therefore, $\alpha + \beta = 6+7 = \boxed{13}$ .

Note that $\cos{(n\theta)}$ is the real part of $\cos{(n\theta)} + i\sin{(n\theta)} = e^{in\theta}$ . Therefore, the sum can be rewritten as $\large{\sum_{n=0}^{\infty} \frac{\cos{(n\theta)}}{2^{n}} = \operatorname{Re} \bigg(\sum_{n=0}^{\infty} \frac{e^{in\theta}}{2^n}\bigg)}$ Rearranging exponents $= \large{\operatorname{Re} \bigg(\sum_{n=0}^{\infty} \bigg( \frac{e^{i\theta}}{2} \bigg)^{n}\bigg)}$ Notice the resemblance this previous sum has to a geometric series. With first term $1$ and common ratio $\Large{\frac{e^{i\theta}}{2}}$ , we can evoke the formula for an infinite geometric series. namely $\Large{\frac{a}{1-r}}$ . Thus, $= \large{\operatorname{Re} \bigg(\frac{1}{1-\frac{e^{i\theta}}{2}} \bigg)}$ Changing our expression from exponential from to trigonometric form $= \large{\operatorname{Re} \bigg(\frac{1}{1- \frac{1}{2}\cos\theta - \frac{i}{2} \sin\theta} \bigg)}$ To find the real part of this expression, we need to rationalize the denominator which involves multiplying by the top and bottom by $\large{1- \frac{1}{2}\cos\theta + \frac{i}{2} \sin\theta}$ which leads to $= \large{\operatorname{Re} \Bigg(\frac{1- \frac{1}{2}\cos\theta + \frac{i}{2} \sin\theta}{(1-\frac{1}{2}\cos\theta)^{2} + \frac{1}{4}\sin^{2}\theta} \Bigg)}$ $= \large{ \frac{1- \frac{1}{2}\cos\theta}{(1-\frac{1}{2}\cos\theta)^{2} + \frac{1}{4}\sin^{2}\theta}}$ Now, since $\cos\theta = \large{\frac{1}{5}}$ , then $\cos^{2} \theta = \large{\frac{1}{25}}$ . This also means $1- \sin^{2}\theta = \large{\frac{1}{25}} \Leftrightarrow \sin^{2}\theta = \large{\frac{24}{25}}$ . Plugging all of this into our expression, we get $= \large{ \frac{1- \frac{1}{10}}{(\frac{9}{10})^{2} + (\frac{1}{4})(\frac{24}{25})}} = \frac{6}{7}$ Thus, our $\alpha$ and $\beta$ are $6$ and $7$ , so our answer is $\alpha + \beta = \boxed{13}$

$\underline{\hspace{ 11 in}}$

Much credit and inspiration is due to the authors of the Art of Problem Solving Volume 2: and Beyond , Richard Rusczyk and Sandor Lehoczky, for providing a basis for the solution to this problem!