Sum of infinite cosine series I'm sure this has been asked before, but I'm not sure how to search for this.


  Evaluate the sum $$\sum_{k=1}^{\infty} \frac{\cos(k)}{2^k}.$$


I thought about expanding $\cos(x)$ as a Taylor series, but that didn't go anywhere. Any hints?
HINT:

$$\cos x =\text{Re}\left(e^{ix}\right)$$

Then, the series of interest is 

$$\text{Re}\left(\sum_{k=1}^{\infty}\left(\frac{e^{i}}{2}\right)^k\right)$$
Amusement Time

Firstly I thought about a joke, namely: let's take the series, and white it like:

$$S = \frac{e^i}{2} + \frac{e^{2i}}{4} + \frac{e^{3i}}{8} + \frac{e^{4i}}{16} + \ldots $$

Now:

$$S = \frac{e^i}{2}\cdot\left[1 + \frac{e^i}{2} + \frac{e^{2i}}{4} + \frac{e^{3i}}{8} + \frac{e^{4i}}{16} + \ldots \right] = \frac{e^i}{2}\cdot (1 + S)$$

Then re-arranging the terms to get S:

$$S = \frac{e^i}{2-e^i}$$

Now we have to take the Real part:

$$\Re \frac{e^i}{2 - e^i} = \Re \frac{i\sin(1) + \cos(1)}{2 - i\sin(1) -\cos(1)} = \frac{\cos(1)}{2-\cos(1)} $$

Unfortunately this is not the sum of that series.

But I guess that a similar road shall be undertaken because the sum of that series is (thanks to Mathematica)

$$S = \frac{1 - 2\cos(1)}{4\cos(1) - 5}$$