$ x[n] = (-1)^n $
Computing the Fourier series coefficients...
$ a_k = \frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jK \frac{2\pi}{N}n} $
$ N = 2 $ due to the signal switching between 1 and -1.
$ a_k = \frac{1}{2} \sum_{n=0}^{1} x[n]e^{-jK \pi n} $
$ K = 0,1 $ due to the fundamental period of $ N = 2 $
$ a_0 = \frac{1}{2} \sum_{n=0}^{1} x[n] $
$ a_0 = (1 + -1) = 0 $
$ a_1 = \frac{1}{2} \sum_{n=0}^{1} x[n]e^{-j \pi n} $
$ a_1 = \frac{1}{2} (x[0] + x[1]e^{-j \pi}) $
$ a_1 = \frac{1}{2} (1 + (-1)(-1)) = 1 $
Now that I've solved an easy one (hopefully), I want to try a harder one.
