DTFS of a periodic impulse train

The signal is a periodic impulse train: a unit impulse at every multiple of $N$.

$$x[n] = \sum_{\ell=-\infty}^{\infty}\delta[n - \ell N]$$

 x[n]
  1  *              *              *
     |              |              |
 ----+--*--*--...---+--*--*--...---+---- n
     0              N              2N

It is periodic with period $N$, so $\omega_0 = 2\pi/N$.

Coefficients

$$a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]\,e^{-jk(2\pi/N)n}$$

Choose the period $n = 0, 1, \dots, N-1$. Within it $x[n] = \delta[n]$, which is 1 at $n = 0$ and 0 elsewhere:

$$a_k = \frac{1}{N}\sum_{n=0}^{N-1}\delta[n]\,e^{-jk(2\pi/N)n} = \frac{1}{N}e^{0} = \frac{1}{N} \quad \text{for all } k$$

Fourier series representation

$$x[n] = \sum_{k=0}^{N-1}\frac{1}{N}e^{jk(2\pi/N)n} = \frac{1}{N}\sum_{k=\langle N\rangle} e^{jk(2\pi/N)n}$$

Check: for $n = 0, \pm N, \dots$ every exponential equals 1, so the sum is $N\cdot\frac{1}{N} = 1$. For other $n$ the sum is a full geometric series over the $N$th roots of unity, which is $\frac{1}{N}\cdot\frac{1 - e^{j2\pi n}}{1 - e^{j2\pi n/N}} = 0$. This matches $x[n]$.

Spectrum: all $N$ harmonics have the same amplitude $1/N$ and zero phase. The spectrum is flat. A train of narrow impulses contains all frequencies equally, just as the CT impulse train $\sum\delta(t - kT)$ has $a_k = 1/T$. This result is the basis of the sampling theorem.