The general form of a DFS is:

which are harmonics of a fundamental frequency ${\displaystyle 1/N,}$  for some positive integer ${\displaystyle N.}$  The practical range of ${\displaystyle k,}$  is ${\displaystyle [0,\ N-1],}$  because periodicity causes larger values to be redundant. When the ${\displaystyle S[k]}$  coefficients are derived from an ${\displaystyle N}$ -length DFT, and a factor of ${\displaystyle 1/N}$  is inserted, this becomes an inverse DFT.^{[1]: p.542 (eq 8.4)  [2]: p.77 (eq 4.24)} And in that case, just the coefficients themselves are sometimes referred to as a discrete Fourier series.^{[3]: p.85 (eq 15a)}

Example edit

A common practice is to create a sequence of length ${\displaystyle N}$  from a longer ${\displaystyle s[n]}$  sequence by partitioning it into ${\displaystyle N}$ -length segments and adding them together, pointwise.(see DTFT § L=N×I) That produces one cycle of the periodic summation:

${\displaystyle s_{_{N}}[n]\ \triangleq \ \sum _{m=-\infty }^{\infty }s[n-mN],\quad n\in \mathbb {Z} .}$ 

Because of periodicity, ${\displaystyle s_{_{N}}}$ can be represented as a DFS with ${\displaystyle N}$  unique coefficients that can be extracted by an ${\displaystyle N}$ -length DFT.^{[1]: p 543 (eq 8.9) : pp 557-558    [2]: p 72 (eq 4.11)}

${\displaystyle {\begin{aligned}&s_{_{N}}[n]={\frac {1}{N}}\sum _{k=0}^{N-1}S_{N}[k]\cdot e^{i2\pi {\tfrac {k}{N}}n};\quad n\in \mathbb {Z} &&{\text{DFS}}\\&S_{N}[k]\triangleq \sum _{n=0}^{N-1}s_{_{N}}[n]\cdot e^{-i2\pi {\tfrac {k}{N}}n};\quad k\in [0,N-1]&&{\text{DFT}}\end{aligned}}}$ 

The coefficients are useful because they are also samples of the discrete-time Fourier transform (DTFT) of the ${\displaystyle s[n]}$  sequence:

${\displaystyle S_{1/T}(f)\triangleq \sum _{n=-\infty }^{\infty }e^{-i2\pi fnT}\ T\ s(nT)=\sum _{m=-\infty }^{\infty }S\left(f-{\tfrac {m}{T}}\right),\quad f\in \mathbb {R} }$ 

Here, ${\displaystyle s(nT)}$  represents a sample of a continuous function ${\displaystyle s(t),}$  with a sampling interval of ${\displaystyle T,}$  and ${\displaystyle S(f)}$  is the Fourier transform of ${\displaystyle s(t).}$  The equality is a result of the Poisson summation formula. With definitions ${\displaystyle s[n]\triangleq T\ s(nT)}$  and ${\displaystyle S[k]\triangleq S_{1/T}\left({\tfrac {k}{NT}}\right)}$ :

${\displaystyle S[k]=\sum _{n=-\infty }^{\infty }e^{-i2\pi {\tfrac {k}{N}}n}\ s[n];\quad k\in \mathbb {Z} .}$ 

Due to the ${\displaystyle N}$ -periodicity of the ${\displaystyle e^{-i2\pi {\tfrac {k}{N}}n}}$  kernel, the summation can be "folded" as follows:

${\displaystyle {\begin{aligned}S[k]&=\sum _{m=-\infty }^{\infty }\left(\sum _{n=0}^{N-1}e^{-i2\pi {\tfrac {k}{N}}n}\ s[n-mN]\right)\\&=\sum _{n=0}^{N-1}e^{-i2\pi {\tfrac {k}{N}}n}\underbrace {\left(\sum _{m=-\infty }^{\infty }s[n-mN]\right)} _{s_{_{N}}[n]}\\&=S_{N}[k].\end{aligned}}}$