Okay, I'm gonna expound a little.
Quoting (except for any typos that may result) from the 1989 text of O&S (Introduction to Chapter 8, The Discrete Fourier Transform, p 514):
Although several points of view can be taken toward the derivation and interpretation of the DFT representation of a finite-duration sequence, we have chosen to base our presentation on the relationship between periodic sequences and finite-length sequences. We will begin by considering the Fourier series representation of periodic sequences. While this representation is important in its own right, we are most often interested in the application of Fourier series results to the representation of finite-length sequences. We accomplish this by constructing a periodic sequence for which each period is identical to the finite-length sequence. As we will see, the Fourier series representation of the periodic sequence corresponds to the DFT of the finite-length sequence. Thus our approach is to define the Fourier series representation for periodic sequences and to study the properties of such representations. Then we repeat essentially the same derivations assuming that the sequence to be represented is a finite-length sequence. This approach to the DFT emphasizes the fundamental inherent periodicity of the DFT representation and ensures that this periodicity is not overlooked in applications of the DFT.
Section 8.1, p 516 on the DFS:
Eq. (8.11) $\quad \tilde{X}[k] = \sum\limits^{N-1}_{n=0} \tilde{x}[n] \ e^{-j2\pi n k/N} $
Eq. (8.12) $\quad \tilde{x}[n] = \frac{1}{N} \sum\limits^{N-1}_{k=0} \tilde{X}[k] \ e^{+j2\pi n k/N} $
Regarding the DFS, $\tilde{x}[n]$ (with the tilde) is defined to be periodic with period $N$ such that $$ \tilde{x}[n+N] = \tilde{x}[n] \quad \forall n $$ and $\tilde{X}[k]$ turns out to also be periodic with period $N$ (so $ \tilde{X}[k+N] = \tilde{X}[k] \quad \forall k $).
Later, in section 8.6, p 532 on the DFT:
Eq. (8.59) $\quad X[k] = \begin{cases} \sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N}, & 0 \le k \le N-1 \\ 0, & \text{otherwise} \end{cases} $
Eq. (8.60) $\quad x[n] = \begin{cases} \frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N}, & 0 \le n \le N-1 \\ 0, & \text{otherwise} \end{cases} $
Generally the DFT analysis and synthesis equations are written as
Eq. (8.61) $\quad X[k] = \sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N} $
Eq. (8.62) $\quad x[n] = \frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N} $
In recasting Eqs. (8.11) and (8.12) in the form of Eqs. (8.61) and (8.62) for the finite-duration sequences, we have not eliminated the inherent periodicity. As with the DFS, the DFT $X[k]$ is equal to samples of the periodic Fourier transform $X(e^{j\omega})$, and if Eq. (8.62) is evaluated for values of $n$ outside the interval $0 \le n \le N-1$, the result will not be zero but rather a periodic extension of $x[n]$. The inherent periodicity is always present. Sometimes it causes us difficulty and sometimes we can exploit it, but to totally ignore it is to invite trouble.
So the first obvious thing i would say is that the tildes used for the DFS (to explicitly depict a periodic sequence) are symbols and still do not change any mathematical fact. The direct relationship between the periodic $\tilde{x}[n]$ and the "finite-length" $x[n]$ is
$$ \tilde{x}[n] = x[n \bmod N] \qquad \forall n \in \mathbb{Z}, \ N \in \mathbb{Z}>0$$
where $ \qquad\qquad\qquad n \bmod N = n - N \left\lfloor \frac{n}{N} \right\rfloor $.
Now I know some folks will point to the Eqs. (8.59) and (8.60) definition of the DFT that has truncated (to $0$) values outside of the interval $0 \le n,k \le N-1$.
However, that definition is contrived. It could just as well be expressed as
$\quad X[k] = \begin{cases}
\sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N}, & 0 \le k \le N-1 \\
5, & \text{otherwise}
\end{cases} $
$\quad x[n] = \begin{cases}
\frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N}, & 0 \le n \le N-1 \\ 5, & \text{otherwise}
\end{cases} $
or
$\quad X[k] = \begin{cases}
\sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N}, & 0 \le k \le N-1 \\
5000, & \text{otherwise}
\end{cases} $
$\quad x[n] = \begin{cases}
\frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N}, & 0 \le n \le N-1 \\ 5000, & \text{otherwise}
\end{cases} $
or
$\quad X[k] = \begin{cases}
\sum\limits^{N-1}_{n=0} x[n] \ e^{-j2\pi n k/N}, & 0 \le k \le N-1 \\
\text{the man on the moon}, & \text{otherwise}
\end{cases} $
$\quad x[n] = \begin{cases}
\frac{1}{N} \sum\limits^{N-1}_{k=0} X[k] \ e^{+j2\pi n k/N}, & 0 \le n \le N-1 \\ \text{and his hot girlfriend}, & \text{otherwise}
\end{cases} $
Because that $0$ in that contrived DFT definition will never ever be used in any theorems regarding the DFT. When that contrived definition is used for the DFT, then when using any DFT theorems to do any real work (other than the linearity and scaling by constant theorems), then one must use modulo arithmetic in the arguments of $x[n]$ or $X[k]$. And using that modulo arithmetic is explicitly periodically extending the sequence.
So (sorta responding to hotpaw) there are two or three processes that you should think about when using the DFT on a real signal.
The sampling process: what happens to the spectrum of $x(t)$ when you sample it with a "dirac comb" or whatever you want to call the sampling function?
Windowing to finite length: what happens when you window either $x(t)$ or the sampled version, $x[n]$, with a rectangular window of length $N$?
Periodic extension: what happens when you periodically extend it by repeatedly shifting the windowed $x[n]$ by $N$ samples and overlap and add it?
Deal with each step by itself.
