Is the DFT really the DTFS? In the continuous domain, I know that the Fourier Series is given by:
$$x(t)=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi}{T}kt}$$
So in the discrete domain, it would make sense that a discrete N-periodic signal could be represented with the 'Discrete Time Fourier Series' (I assume) as:
$$x[n]=\sum_{k=-\infty}^{\infty}X_k e^{j\frac{2\pi}{N}kn}$$
However, my book says that a discrete periodic signal 'is completely described' as:
$$x[n]=\sum_{k=0}^{N-1}X_k e^{j\frac{2\pi}{N}kn}$$
because of the inherent periodicity of discrete complex exponentials. Is the DFT (and DFS) then technically just one period of the DTFS? And are the technical limits of summation of the DTFS $(-\infty,\infty)$?
And if so, doesn't dropping all those terms from the summation technically cause the equality not to be true anymore, i.e.
$$x[n]\neq\sum_{k=0}^{N-1}X_k e^{j\frac{2\pi}{N}kn}$$
since $x[n]$ needs the sum of all harmonic complex exponentials?
 A: First of all, we're talking discrete time, so I'll be using $N$ for the period of our function rather than $T$.
In fact, we do have exact equality:
$$
x[n] = \sum_{k=0}^{N-1}X_k e^{j\frac{2\pi kn}{N}}
$$
where $X_k = \frac 1{N}\sum_{n=0}^{N} x[n] e^{-j \frac{2 \pi}N kn}$.  The easiest way to see this intuitively is with a little bit of linear algebra.  
Let $\tilde x[n]$ denote the function $\tilde x[n] = \sum_{k=0}^{N-1}X_k e^{j\frac{2\pi kn}{N}}$, which I attempt to prove is equal to $x[n]$ (for all $n$).  Since both $x$ and $\tilde x$ are $N$-periodic, it is enough to know that $x[n] = \tilde x[n]$ for $n = 0,1,2,\dots, N-1$.  In other words, we're going to think of $x[n]$ as really being the vector $\mathbf x = (x[0],x[1],\dots,x[N-1]) \in \Bbb C^N$.
Let $b_k[n] = e^{j \frac{2 \pi k n}{N}}$.  The values of $b_k[n]$ for $0 \leq n \leq N-1$ form a vector of $N$ entries.  The question is this: can we find $X_k$ such that $X_0b_0[n] + \cdots + X_{N-1} b_{N-1}[n]$ is exactly the same as the function $x[n]$?   In terms of vectors: if we define $\mathbf b_k = (e^{j \frac{2 \pi k (0)}{N}}, \dots, e^{j \frac{2 \pi k (N-1)}{N}})$, can we find coefficients $X_k$ so that 
$$
X_0 \mathbf b_0 + \cdots + X_{N-1} \mathbf b_{N-1} = \mathbf x
$$
Or as a system of equations, we have
$$
X_0 b_0[0] + \cdots + X_{N-1}b_{N-1}[0] = x[0]\\
X_0 b_0[1] + \cdots + X_{N-1}b_{N-1}[1] = x[1]\\
\vdots \\
X_0 b_0[N-1] + \cdots + X_{N-1}b_{N-1}[N-1] = x[N-1]
$$
As it turns out, this is a linear system of $N$ equations on $N$ variables (the variables $X_0,X_1,\dots,X_{N-1}$) that we can completely solve!
To put it another way: because the vectors $\mathbf b_0,\dots,\mathbf b_{N-1}$ form a basis of the $N$-dimensional space $\Bbb C^N$, every vector $\mathbf x \in \Bbb C^N$ can be uniquely expressed as a linear combination of them.  Because the functions $b_0[n],\dots,b_{N-1}[n]$ form a basis of the $N$-dimensional space of $N$-periodic functions, every $N$-periodic function $x[n]$ can be uniquely expressed as a linear combination of them.
If we were to write out this system of equations with a matrix, we would have $W \mathbf X = \mathbf x$ where $\mathbf X = (X_0,\dots,X_{N-1})$, $\mathbf x$ is as before, and $W$ is (a multiple of) the DFT matrix, as described in the link.

Another perspective: it turns out that there is a certain "aliasing" among the other complex exponentials that makes them very obviously redundant.  For example, with $k = N$, we find
$$
b_N[n] = e^{j \frac{2 \pi N n}{N}} = e^{j 2 \pi n} = [e^{j 2 \pi}]^n = 1\\
b_{N+1}[n] = e^{j \frac{2 \pi (N+1)n}N} = e^{j \frac{2 \pi Nn}N} \cdot e^{j \frac{2 \pi 1n}{N} } = 
1 \cdot b_1[n] = b_1[n]
$$
and so on so that in general, we have $b_{k + mN}[n] = b_k[n]$ for any integer $m$.
