The DFT is one and the same as the DFS. The DFT maps a discrete and periodic sequence of values with period N, that is
x[n]=x[n+N]∀n∈Z
to another discrete and periodic sequence
X[k]=X[k+N]∀k∈Z
and the iDFT maps it back. the discreteness in one domain causes the periodicity in the other domain, and since the mapping is 1-to-1, you can say that periodicity in one domain causes discreteness in the other. both periodic sequences are fully defined by N complex numbers and a practical iDFT and DFT maps those two finite-length sets of numbers back and forth to each other.
Although not commonly done, it's perfectly okay to generalize the DFT and iDFT to be
X[k]=n0+N−1∑n=n0x[n] e−j2πnk/N
x[n]=1Nk0+N−1∑k=k0X[k] ej2πnk/N
n,n0,k,k0 can be any integers.
so the choice of k0=−N2 comes from the convenience of pairing X[−k] to X[k]. since, in general
X[k]=ℜ{X[k]}+jℑ{X[k]}
and if ℑ{x[n]}=0 for all n, then
X[−k]=X∗[k]=ℜ{X[k]}−jℑ{X[k]}
then you can say, about reconstructing x[n], that if N is even
x[n]=1NN2−1∑k=−N2X[k] ej2πnk/N
if N is odd, then
x[n]=1NN−12∑k=−N−12X[k] ej2πnk/N
and you can sorta plug in n=fst to reconstruct bandlimited x(t) outa x[n].
in the N even case, then you have an extra term at Nyquist, the k=−N2 term, for that you cannot know both the amplitude and phase of that sinusoid. you must assume one or the other. often people only keep the cos(πfst) term at Nyquist.
x[n]=1NN2−1∑k=−N2X[k] ej2πnk/N=1N(X[0]+X[N/2]cos(πn)+N2−1∑k=1X[k]ej2πnk/N + X[−k]e−j2πnk/N)=1N(X[0]+X[N/2]cos(πn))+1NN2−1∑k=1(ℜ{X[k]}+jℑ{X[k]})ej2πnk/N + (ℜ{X[k]}−jℑ{X[k]})e−j2πnk/N=1N(X[0]+X[N/2]cos(πn))+1NN2−1∑k=12ℜ{X[k]}ej2πnk/N+e−j2πnk/N2+j(2j)ℑ{X[k]}ej2πnk/N−e−j2πnk/N2j=1N(X[0]+X[N/2]cos(πn))+2NN2−1∑k=1ℜ{X[k]}ej2πnk/N+e−j2πnk/N2−ℑ{X[k]}ej2πnk/N−e−j2πnk/N2j=1N(X[0]+X[N/2]cos(πn))+2NN2−1∑k=1ℜ{X[k]}cos(2πnk/N)−ℑ{X[k]}sin(2πnk/N)
for N odd, you can go through the same song and dance, but there is no Nyquist term and it comes out as
x[n]=1NX[0]+2NN−12∑k=1ℜ{X[k]}cos(2πnk/N)−ℑ{X[k]}sin(2πnk/N)
for odd N, all of these discrete-time sinusoids can be understood as a properly-sampled bandlimited periodic, continuous-time, real signal, x(t):
x[n]≜
or setting n \leftarrow f_\text{s} t
x(t) = \frac1N X[0] + \frac2N \sum\limits_{k=1}^{\frac{N-1}{2}} \Re\{X[k]\} \cos(2 \pi (k/N) f_\text{s} t) - \Im\{X[k]\} \sin(2 \pi (k/N) f_\text{s} t)
but when N is even, the Nyquist term is a little more ambiguous:
\begin{align}
x(t) & = \frac1N \left( X[0] + X[N/2] \cos(2 \pi (f_\text{s}/2) t) \right) \\ & \quad + \frac2N \sum\limits_{k=1}^{\frac{N}{2}-1} \Re\{X[k]\} \cos(2 \pi k f_\text{s} t) - \Im\{X[k]\} \sin(2 \pi k f_\text{s} t) \\
& = \frac1N \left( X[0] + X[N/2] \cos(2 \pi (f_\text{s}/2) t) \right) \\ & \quad + \frac2N \sum\limits_{k=1}^{\frac{N}{2}-1} \Re\{X[k]\} \cos(2 \pi (k/N) f_\text{s} t) - \Im\{X[k]\} \sin(2 \pi (k/N) f_\text{s} t) \\
& = \frac1N \left( X[0] + X[N/2] \cos(2 \pi (f_\text{s}/2) t) + A_x \sin(2 \pi (f_\text{s}/2) t) \right) \\ & \quad + \frac2N \sum\limits_{k=1}^{\frac{N}{2}-1} \Re\{X[k]\} \cos(2 \pi (k/N) f_\text{s} t) - \Im\{X[k]\} \sin(2 \pi (k/N) f_\text{s} t) \\
\end{align}
where A_x can be any number, because that term will be zero at the sampling instances, when t = nT = n/f_\text{s}.
looking at it from a magnitude/phase POV, for N odd
x(t) = \frac{X[0]}{N} + \frac2N \sum\limits_{k=1}^{\frac{N-1}{2}} |X[k]| \cos(2 \pi (k/N) f_\text{s} t + \arg\{X[k]\})
and for N even
\begin{align} x(t) & = \frac{X[0]}{N} + \frac{X[N/2]}{N \cos(\theta)} \cos(2 \pi (f_\text{s}/2) t + \theta) \\ & \quad + \frac2N \sum\limits_{k=1}^{\frac{N}{2}-1} |X[k]| \cos(2 \pi (k/N) f_\text{s} t + \arg\{X[k]\}) \end{align}
where
\Re\{X[k]\} = |X[k]| \cos(\arg\{X[k]\})
\Im\{X[k]\} = |X[k]| \sin(\arg\{X[k]\})
or
|X[k]| = \sqrt{\Re\{X[k]\}^2 + \Im\{X[k]\}^2}
\arg\{X[k]\} = \operatorname{atan2}(\Im\{X[k]\}, \Re\{X[k]\})
and \theta is the phase of the Nyquist component and can be any real value except \pm \frac{\pi}{2} or \frac{\pi}{2}+ m \pi for integer m (because you cannot divide by zero). the samples x[n] \triangleq x(n/f_\text{s}) will come out the same regardless of the phase of the Nyquist component as long as the amplitude of the component is adjusted accordingly by dividing by \cos(\theta).