Where did the length of time’s period disappear in Periodic Fourier Series Discrete Time
A DFT (Discrete Fourier Transform) is applied to an interval of a signal with uniformly sampled values on the underlying dimension, which is often time.  $N$ is a variable that is conventionally used to define the number of sample points.  The set of points in the interval is called the DFT frame.  Thus $N$ has units of $\frac{samples}{frame}$.

The DFT does not care about the sampling rate.  The sampling rate is used to convert frequencies in the DFT of cycles per frame to cycles per the underlying dimension's unit.

$2\pi$ has units of $\frac{radians}{cycle}$.  The $kth$ bin in a DFT represents a frequency of $k$ $\frac{cycles}{frame}$.

Therefore, the expression

$$ \frac{2\pi}{N}k $$ has units of

$$ \frac{\frac{radians}{cycle}}{\frac{samples}{frame}} \cdot \frac{cycles}{frame} = \frac{radians}{sample} $$

Now, look at the unnormalized definition of the DFT:

$$ X[k] = \sum_{n=0}^{N-1} x[n] e^{-i \left(\frac{2\pi}{N}k\right)n } $$

The units of $n$ is samples, so the units of the exponent is:

$$ \frac{radians}{sample} \cdot sample = radians $$

Which is what it should be.

$F_s$ is the usual symbol for the sampling rate, also called the sampling frequency and sometimes the sampling density.  With seconds as the unit of time, it has units of $\frac{samples}{second}$.  I do not like the use of $Hz$ as a unit here, it causes confusion with $Hz$ being a unit of the frequency domain.

$$ T = \frac{N}{F_s} $$

$$ \frac{seconds}{frame} = \frac{ \frac{samples}{frame} }{ \frac{samples}{second} } $$

The frequency of the $k$th bin.

$$ f_k = \frac{k}{T} = k \frac{F_s}{N} = \frac{k}{N} F_s $$

$$
\begin{aligned}
\frac{cycles}{second} &amp;=  \frac{ \frac{cycles}{frame} }{  \frac{seconds}{frame} }  \\
&amp;= \frac{cycles}{frame}   \frac{ \frac{samples}{second} }{ \frac{samples}{frame} }  \\
&amp;= \frac{ \frac{cycles}{frame} }{ \frac{samples}{frame} } \frac{samples}{second} \\
\end{aligned}
$$

Frequency spacing on the DFT is also known as the bin width.  The difference in frequency between any two adjacent bins (except DC and Nyquist) is going to be $\frac{F_s}{N}$ in units of $\frac{cycles}{second}$ also called $Hz$.