I have a periodic signal, with period $1$
$$x(t) = \begin{cases} 1 \qquad & 0 \le t - \lfloor t \rfloor < \tfrac12 \\ 0 \qquad & \tfrac12 \le t - \lfloor t \rfloor < 1 \\ \end{cases}$$
$\lfloor t \rfloor = \operatorname{floor}(t)$ is the floor()
function, returning the largest integer no greater than the argument $t$.
$$ x(t+1) = x(t) \qquad \forall t \in \mathbb{R} $$
The complex Fourier series for $x(t)$ is
$$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \ e^{i 2 \pi k t} $$
The complex Fourier coefficients are
$$\begin{align} c_k &= \int_{-1/2}^{1/2} x(t) \ e^{-i2\pi k t} \ \mathrm{d}t \qquad \qquad k \in \mathbb{Z} \\ &= \int_{0}^{1/2} 1 \ e^{-i2\pi k t} \ \mathrm{d}t \\ &= \tfrac{1}{-i2\pi k } \big( e^{-i\pi k} - 1 \big) \\ &= \tfrac{i}{2\pi k } \big( (-1)^k - 1 \big) \\ \end{align}$$
and are $0$ for even $k$.
Now, we sample $x(t)$ at $N\in 2\mathbb{N}$ time values,
$$\begin{align} x[n] &= x(t_n) \\ &= x\left(\tfrac{1}{N}n\right) \end{align}$$
where $\tfrac{1}{N}$ is the sampling period, $N$ is the sampling frequency, and $t_n = \frac{n}{N}$, with $n=0, N-1$.
Let $X[k]$ denote the DFT of this finite sequence $x[n]$.
$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-i2\pi nk/N} $$
One can show that $X[k] = 0$ for even $k$ as well, so $X[2k] = c_{2k}$, but $X[2k+1] \neq c_{2k+1}$.
I'm trying to come up with an intuitive explanation as to why $X[2k] = c_{2k}$, but $X[2k+1]\neq c_{2k+1}$. Obviously, we shouldn't really expect them to be equal in general since $X[k]$ is really just a Riemann sum approximation to $c_k$ with $N$ intervals, but in this case it seems there may be an explanation, since $X[2k] = c_{2k}$.
So far, all I can come up with is that since $x(t)$ isn't bandlimited, the DFT of the discrete sample of $x(t)$ is essentially trying to "fit" a bandlimited signal to the samples $x[n]$, and so for some reason this doesn't have any content at frequencies $2\pi (2k) = 4\pi k$ for any $k\in\mathbb{N}$.
Is there any specific reason for this? From the above argument, I have a feeling it has to do with aliasing, but I can't exactly make the connection.