DTFS coefficients are periodic

For a discrete-time periodic signal $x[n]$ with period $N$, the DTFS coefficients are

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

To show: $a_{k+N} = a_k$

Replace $k$ by $k + N$:

$$\begin{aligned} a_{k+N} &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-j(k+N)(2\pi/N)n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n}\,e^{-j2\pi n} \end{aligned}$$

Since $n$ is an integer, $e^{-j2\pi n} = \cos(2\pi n) - j\sin(2\pi n) = 1$. Hence

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

Repeating the argument, $a_{k+mN} = a_k$ for any integer $m$. So the sequence of DTFS coefficients is periodic in $k$ with period $N$.

Reason

The root cause is that discrete-time complex exponentials whose frequencies differ by $2\pi$ are identical:

$$\phi_{k+N}[n] = e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n} = \phi_k[n]$$

So there are only $N$ distinct harmonics, and the "$(k+N)$th harmonic" is the same signal as the $k$th; its coefficient must be the same. This is why the synthesis equation sums over only $N$ consecutive $k$:

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

This is different from the CT Fourier series, where every $e^{jk\omega_0 t}$ is distinct and the coefficients are not periodic.

Example

$x[n] = \cos(\pi n/2)$, $N = 4$: $x[n] = \frac{1}{2}e^{j(2\pi/4)n} + \frac{1}{2}e^{-j(2\pi/4)n}$, so $a_1 = a_{-1} = \frac12$. Computing from the analysis equation with samples $\{1, 0, -1, 0\}$:

The pattern $\{0, 0.5, 0, 0.5\}$ repeats every 4 values of $k$, e.g. $a_5 = a_1$ and $a_3 = a_{-1}$.

 a_k
   0.5  |     |     |     |     |
       -+--o--+--o--+--o--+--o--+--> k
       -3 -2 -1  0  1  2  3  4  5