Summation of $\frac{\cos n \theta}{2^n}$ I would like to compute the following sum:
$$\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}$$
I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $\cos \theta = {e^{i\theta} + e^{-i\theta}\over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated. 
 A: Consider the series
$$S=\sum_{n=0}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^n.$$
This is a geometric series whose sum is 
$$S=\frac{2}{2-e^{i\theta}}.$$
Now the real part of $S$ is the sum you are looking for.
A: I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.
We can write the summation as the real part of
$$\sum_{n=0}^{\infty} \frac{\cos n\theta + i\sin m\theta}{2^n}$$
Using De Moivre's theorem:
$$\sum_{n=0}^{\infty} \left(\frac{\cos \theta + i\sin \theta}{2}\right)^n$$
We can then calculate the real portion of the infinite sum.
\begin{align*}
\operatorname{Re}\left(\frac{1}{1-r}\right)
&= \operatorname{Re}\left(\frac{1}{\frac{2-(\cos \theta + i\sin \theta)}{2}}\right) \\
&= \operatorname{Re}\left(\frac{2}{(2-\cos\theta) - i\sin\theta}\right) \\
&= \operatorname{Re}\left(\frac{2((2-\cos\theta) + i\sin\theta)}{4-4\cos\theta+1}\right) \\
&= \frac{4-2\cos\theta}{4-4\cos\theta+1}
\end{align*}
Hope this helps!
A: If you are only looking for a hint, write your sum as $$\sum_{n=0}^\infty\frac{e^{in\theta}+e^{-in\theta}}{2\cdot2^n}$$  Break it up as two sums, each of which are geometric, so you can use the geometric series formula.
A: \begin{align*}
\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}
&= \sum_{n=0}^\infty \frac{e^{i n \theta} + e^{- i n \theta}}{2^{n+1}} \\
&= \sum_{n=0}^\infty \frac{(e^{i \theta})^n + (e^{- i \theta})^n}{2^{n+1}} \\
&= \sum_{n=0}^\infty \frac{(e^{i \theta})^n}{2^{n+1}} + \sum_{n=0}^\infty  \frac{(e^{- i \theta})^n}{2^{n+1}} \\
&= \sum_{n=0}^\infty \frac12 \left(\frac{e^{i \theta}}{2}\right)^n 
+ \sum_{n=0}^\infty \frac12 \left(\frac{e^{-i \theta}}{2}\right)^n .
\end{align*}
These last two sums are geometric series. Can you finish it?
