DTFS spectrum of {1, −0.5, 0, 0, 0.5}

The signal has period $N = 5$, so $\omega_0 = 2\pi/5$ and

$$a_k = \frac{1}{5}\sum_{n=0}^{4}x[n]\,e^{-jk(2\pi/5)n}$$

One period: $x[0] = 1$, $x[1] = -0.5$, $x[2] = 0$, $x[3] = 0$, $x[4] = 0.5$. Because of periodicity $x[4] = x[-1]$, so it is easier to sum over $n = -1, 0, 1$ (note $x[-1] = 0.5$):

$$\begin{aligned} a_k &= \frac{1}{5}\left[x[-1]e^{jk2\pi/5} + x[0] + x[1]e^{-jk2\pi/5}\right] \\ &= \frac{1}{5}\left[1 + 0.5e^{jk2\pi/5} - 0.5e^{-jk2\pi/5}\right] \\ &= \frac{1}{5}\left[1 + 0.5\,(2j)\sin\frac{2\pi k}{5}\right] \\ &= \frac{1}{5}\left[1 + j\sin\frac{2\pi k}{5}\right] \end{aligned}$$

Magnitude and phase:

$$|a_k| = \frac{1}{5}\sqrt{1 + \sin^2\frac{2\pi k}{5}}, \qquad \angle a_k = \tan^{-1}\left(\sin\frac{2\pi k}{5}\right)$$

Values over one period

(Since $a_{k+5} = a_k$: $a_{-1} = a_4$, $a_{-2} = a_3$.)

Magnitude and phase plots (repeat every 5)

 |a_k|  (repeats every 5)
 0.276     |     |
 0.232  |  |     |  |
   0.2  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
 43.56           |
 30.45           |  |
       -+--+--o--+--+--> k
-30.45  |  |
-43.56     |
       -2 -1  0  1  2

The magnitude spectrum is even and the phase spectrum is odd in $k$, as expected for a real $x[n]$.

Check (synthesis at $n = 0$): $\sum_{k=0}^{4}a_k = 5\times 0.2 + j(0) = 1 = x[0]$. Correct.