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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.

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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.

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    $\begingroup$ Aha...... So I replace $e^{i\theta}$ with $\cos \theta + i\sin \theta$ and find the real part? $\endgroup$ – Gummy bears Nov 16 '15 at 7:38
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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.

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  • $\begingroup$ Crap.... How did I miss that..... -_- Thanks though. $\endgroup$ – Gummy bears Nov 16 '15 at 7:37
  • $\begingroup$ Anurag's solution is better, once you get used to thinking of $\cos(\theta)$ as first and foremost the real part of $e^{i\theta}$. This answer goes the opposite way, thinking of $\cos(\theta)$ as being built out of $e^{i\theta}$ instead of the other way round. $\endgroup$ – alex.jordan Nov 16 '15 at 7:45
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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!

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\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?

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