Cosines Abound

Let us write out a few of the terms of $a_n$ :

$a_1 = \displaystyle\prod_{m=1}^{1} \cos{(\frac{m}{1}*\pi}) = \cos{\frac{1\pi}{1}} = -1.$ $a_2 = \displaystyle\prod_{m=1}^{2} \cos{(\frac{m}{2}*\pi}) = \cos{\frac{\pi}{2}} * \cos{\frac{2\pi}{2}} = 0 * -1 = 0.$ $a_3 = \displaystyle\prod_{m=1}^{3} \cos{(\frac{m}{3}*\pi}) = \cos{\frac{\pi}{3}} * \cos{\frac{2\pi}{3}} * \cos{\frac{3\pi}{3}} = \frac{1}{2} * -\frac{1}{2} * -1 = \frac{1}{4}.$ $a_3 = \displaystyle\prod_{m=1}^{4} \cos{(\frac{m}{4}*\pi}) = \cos{\frac{\pi}{4}} * \cos{\frac{2\pi}{4}} * \cos{\frac{3\pi}{4}} * \cos{\frac{4\pi}{4}} = \frac{1}{\sqrt{2}} * 0 * -\frac{1}{\sqrt{2}} * -1 = 0.$

It is clear that every even term $a_2, a_4, ... , a_{2n} = 0$ because they contain $\cos{\frac{\pi}{2}} = 0$ at some point in their expansion. Thus we are left with only the odd terms to consider: $a_1, a_3, ... , a_{2n+1}$ . Seeing the first two of these, $a_1 = -1$ and $a_3 = \frac{1}{4}$ , we suspect that they form a geometric series with common ratio $r = -\frac{1}{4}$ .

Looking at $a_5$ in particular, we seek to prove our suspicion that $\cos{\frac{\pi}{5}} * \cos{\frac{2\pi}{5}} * \cos{\frac{3\pi}{5}} * \cos{\frac{4\pi}{5}} * \cos{\frac{5\pi}{5}} = -\frac{1}{16}.$ To this end, we multiple top and bottom by $\sin{\frac{\pi}{5}}$ and seek to simplify:

$\frac{\sin{\frac{\pi}{5}} * \cos{\frac{\pi}{5}} * \cos{\frac{2\pi}{5}} * \cos{\frac{3\pi}{5}} * \cos{\frac{4\pi}{5}} * \cos{\frac{5\pi}{5}}}{\sin{\frac{\pi}{5}}} =$ $\frac{\sin{\frac{2\pi}{5}} * \cos{\frac{2\pi}{5}} * \cos{\frac{3\pi}{5}} * \cos{\frac{4\pi}{5}} * \cos{\frac{5\pi}{5}}}{2*\sin{\frac{\pi}{5}}} =$ $\frac{\sin{\frac{4\pi}{5}} * \cos{\frac{3\pi}{5}} * \cos{\frac{4\pi}{5}} * \cos{\frac{5\pi}{5}}}{4*\sin{\frac{\pi}{5}}} =$ $\frac{\sin{\frac{8\pi}{5}} * \cos{\frac{3\pi}{5}} * \cos{\frac{5\pi}{5}}}{8*\sin{\frac{\pi}{5}}} =$ $\frac{-\sin{\frac{3\pi}{5}} * \cos{\frac{3\pi}{5}} * \cos{\frac{5\pi}{5}}}{8*\sin{\frac{\pi}{5}}} =$ $\frac{-\sin{\frac{6\pi}{5}} * \cos{\frac{5\pi}{5}}}{16*\sin{\frac{\pi}{5}}} =$ $\frac{\sin{\frac{6\pi}{5}}}{16*\sin{\frac{\pi}{5}}} =$ $\frac{-\sin{\frac{\pi}{5}}}{16*\sin{\frac{\pi}{5}}} =$ $-\frac{1}{16}.$

Using this same method finds $a_7 = \frac{1}{64}, a_9 = -\frac{1}{256}, ...$ and so on and so forth.

Thus we have found $\displaystyle | \sum_{i=1}^{\infty} a_i | = | -1 + \frac{1}{4} - \frac{1}{16} + \frac{1}{64} - \frac{1}{256} - ... | = | \frac{-1}{1-(-\frac{1}{4})} | = | \frac{-1}{\frac{5}{4}} | = | -\frac{4}{5} | = \frac{4}{5}.$ And $4 + 5 = \boxed{9}$ .