Determine whether the following series is convergent or divergent, and find its sum if it is convergent. PROBLEM
Is the following complex series convergent, and if so, what is its sum?
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}}, \text{   } \theta \in \mathbb{R}$$
MY ATTEMPT #1
Let $$a_n = \frac{\cos(n\theta)}{3^n}.$$
By the Ratio Test
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\cos((n+1)\theta)}{3^{n+1}}\cdot\frac{3^n}{\cos(n\theta)}\right| = \frac{1}{3}\cdot\left|\frac{\cos((n+1)\theta)}{\cos(n\theta)}\right|.$$
Let $x = (n+1)\theta, y = n\theta$.  Then we have
$$\frac{x}{y} = \frac{n+1}{n} \to 1 \text{ as } n \to \infty$$
$$L = \lim_{n \to \infty}{\frac{1}{3}\cdot\left|\frac{\cos((n+1)\theta)}{\cos(n\theta)}\right|} = \frac{1}{3}\cdot\lim_{n \to \infty}{\left|\frac{\frac{\cos(x)}{x}}{\frac{\cos(y)}{y}}\right|}\cdot\lim_{n \to \infty}{\left|\frac{x}{y}\right|} = \frac{1}{3}\cdot{\frac{\lim_{x \to \infty}{\left|\frac{\cos(x)}{x}\right|}}{\lim_{y \to \infty}{\left|\frac{\cos(y)}{y}\right|}}} \to \frac{1}{3}\cdot\frac{0}{0}.$$
Applying L'Hopital's Rule, we get
$$\frac{1}{3}\cdot{\frac{\lim_{x \to \infty}{\left|\frac{\cos(x)}{x}\right|}}{\lim_{y \to \infty}{\left|\frac{\cos(y)}{y}\right|}}} = \frac{1}{3}\cdot{\frac{\lim_{x \to \infty}{\left|\frac{-x\sin(x) - \cos(x)}{x^2}\right|}}{\lim_{y \to \infty}{\left|\frac{-y\sin(y) - \cos(y))}{y^2}\right|}}}$$
But
$$\lim_{z \to \infty}{\left|\frac{-z\sin(z) - \cos(z)}{z^2}\right|} = \lim_{z \to \infty}{\left|\frac{z\sin(z) + \cos(z)}{z^2}\right|} = \left|\lim_{z \to \infty}{\frac{\sin(z)}{z}}\right| + \left|\lim_{z \to \infty}{\frac{\cos(z)}{z^2}}\right| = 0 + 0 = 0.$$
HENCE THIS IS NOT THE CORRECT APPROACH.
MY ATTEMPT #2
Still by the Ratio Test,
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\cos((n+1)\theta)}{3^{n+1}}\cdot\frac{3^n}{\cos(n\theta)}\right| = \frac{1}{3}\cdot\left|\frac{\cos((n+1)\theta)}{\cos(n\theta)}\right|.$$
Applying the trigonometric identity
$$\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$
to $\cos((n+1)\theta)$ does not appear to help either.
QUESTION

Does this mean that the Ratio Test is not an appropriate test of convergence for this series?  If so, what test should I use, and how?

 A: As hinted in the comments, the series
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}}$$
is convergent (by the Comparison Test) since $\theta \in \mathbb{R}$ implies that
$$\left|a_n\right| \leq \frac{1}{3^n},$$
where $\left(\frac{1}{3}\right)^n$ is a convergent geometric series.
Now, to compute the sum, we consider the geometric series
$$\sum_{n=0}^{\infty}{\bigg(\frac{e^{i\theta}}{3}\bigg)^n}$$
which (since $\left|e^{i\theta}/3\right| = 1/3 < 1$) has sum
$$\frac{1}{1 - \frac{e^{i\theta}}{3}} = \frac{3}{3 - e^{i\theta}} = \frac{\bigg(9 - 3\cos\theta\bigg)+i\bigg(3\sin\theta\bigg)}{\sin^2\theta + \bigg(3 - \cos\theta\bigg)^2}.$$
But by De Moivre's Theorem,
$$(e^{i\theta})^n = e^{i(n\theta)} = \cos(n\theta)+i\sin(n\theta).$$
Hence,
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}} = \Re\Bigg(\sum_{n=0}^{\infty}{\bigg(\frac{e^{i\theta}}{3}\bigg)^n}\Bigg) = \frac{9 - 3\cos\theta}{\sin^2\theta + \bigg(3 - \cos\theta\bigg)^2} = \frac{9 - 3\cos\theta}{10 - 6\cos\theta},$$
where $\Re(z)$ denotes the real part of $z \in \mathbb{C}$.
