Finding the value of an infinite trigonometric series

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Find the sum $$\sum_{n=1}^\infty \cfrac{[[\cos(\frac {2\pi}{n})+1]]}{4^n}.$$

I honestly have no clue what to do except that I know that the sum for the $\cfrac {1}{4^n}$ component is $\cfrac 13$.

Can anyone help me in solving this problem? As much as possible, I need a solution that does not use calculus.

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Assuming $[[\cdot]]$ means the floor, consider what $\cos \frac{2\pi}{n}$ looks like - namely, for $n \ge 4$, that it'll always be the cosine of a positive angle less than or equal to $\frac{\pi}{2}$, which will always be a positive number strictly less than 1. Therefore, all terms after the third will be the same, so it's just a matter of replacing the first three terms. The first term will be doubled by adding one to the numerator, the second and third annihilated by the floor function after subtracting one and one-half respectively. That is to say, $\frac{1}{4} + \frac{1}{16} + \frac{1}{64} = \frac{21}{64}$ is replaced by one-half, so the final sum is

$$\frac{1}{3} - \frac{21}{64} + \frac{1}{2} = \frac{97}{192}$$