from Wikipedia:
When the input data sequence $x[n]$ is $N$-periodic, DTFT can be computationally reduced to a discrete Fourier transform (DFT), because:
$ X_{1/T}(f)$ converges to zero everywhere except integer multiples of $\frac{1}{NT},$ known as harmonic frequencies.
Now is this just convergence, or is $ X_{1/T}(f)$ zero all the time if $x[n]$ is N-periodic? If it is just convergence, what convergence is it referring to? Maybe as $N \to \infty$?
The DTFT of a periodic function is a Dirac comb, so it is exactly zero everywhere except for the harmonics. Let $y[n]$, $n=0,1,\ldots,N-1$ be one period of the $N$-periodic signal $x[n]$. $x[n]$ can then be written as
$$x[n]=y[n]\star\sum_k\delta[n-kN]\tag{1}$$
where $\star$ denotes discrete-time convolution, and $\delta[n]$ is the unit impulse. The DTFT of (1) is
$$X(f)=Y(f)\cdot\frac{1}{N}\sum_k\delta\left(f-\frac{k}{NT}\right)= \frac{1}{N}\sum_kY\left(\frac{k}{NT}\right)\delta\left(f-\frac{k}{NT}\right)\tag{2}$$
where $\delta(f)$ is the Dirac impulse. The DTFT of $y[n]$ is
$$Y(f)=\sum_{n=0}^{N-1}y[n]e^{-jn2\pi fT}\tag{3}$$
So we have
$$Y\left(\frac{k}{NT}\right)=\sum_{n=0}^{N-1}y[n]e^{-jn2\pi k/N}\tag{4}$$
Note that (4) is the same as the DFT of $y[n]$. So from (2) we see that $X(f)$ is a weighted sum of shifted Dirac impulses, where the weights are given by the DFT coefficients of $y[n]$, i.e. of one period of the signal $x[n]$. Consequently, the DFT and the DTFT of $x[n]$ contain exactly the same information.