Relation between sawtooth Fourier coefficients and its DFT
Let me use the definitions of the Fourier series and the DFT as they are commonly used in signal processing. The Fourier series of a $T$-periodic function $f(t)$ is given by

$$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$

where $c_k$ are the complex Fourier coefficients:

$$c_k=\frac{1}{T}\int_0^Tf(t)e^{-j2\pi kt/T}dt\tag{2}$$

The length-$N$ DFT of a sequence $x[n]$ is defined by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{3}$$

and the inverse DFT is

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{4}$$

If you sample $f(t)$ at $t=n\tau$ with $\tau=T/N$ you can approximate the integral $(2)$ by the following sum:

$$c_k\approx\frac{1}{T}\sum_{n=0}^{N-1}f(n\tau)e^{-j2\pi kn\tau /T}\tau\tag{5}$$

With $x[n]=f(n\tau)$ and $\tau=T/N$, Eq. $(5)$ can be rewritten as

$$c_k\approx\frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}=\frac{1}{N}X[k],\qquad |k|\le\left\lfloor \frac{N}{2}\right\rfloor\tag{6}$$

Equation $(6)$ says that the Fourier series coefficients $c_k$ of the $T$-periodic function $f(t)$ are approximated by the (scaled) DFT coefficients of the sequence $x[n]=f(n\tau)$, with $\tau=T/N$.

Of course this is also the case for your function. With

$$f(t)=A\left(\frac{t}{T}-\frac12\right),\qquad 0\le t&lt; T\tag{7} $$

and $f(t)=f(t+T)$, we get

$$x[n]=f(nT/N)=A\left(\frac{n}{N}-\frac12\right),\qquad 0\le n&lt; N \tag{8}$$

The Fourier coefficients of $f(t)$ are 

$$c_k=\frac{jA}{2\pi k}\tag{9}$$

The DFT of $x[n]$ is given by

$$X[k]=\begin{cases}-\frac{A}{2},&amp;\quad k=0\\
\frac{A}{e^{-j2\pi k/N}-1} = e^{j\pi k/N} \frac{jA/2}{\sin(\pi k/N)}=\frac{A}{2}(-1+j\cot(\pi k/N)),&amp;\quad 0&lt;k&lt;N
\end{cases}\tag{10}$$

The first thing to notice is that $f(t)$ is an odd function (with a discontinuity at $t=0$), and, consequently, its Fourier coefficients are purely imaginary. On the other other hand, the sampled version of $f(t)$ is not odd because $x[0]\neq 0$; that's why we get a (relatively small) non-zero real part in the DFT coefficients. However, as predicted by Eq. $(6)$, the scaled DFT coefficients $X[k]/N$ still approximate the Fourier coefficients quite well. The zero real part of $c_k$ is "approximated" by the value $-A/2N$, which is small for large $N$. The figure below shows the imaginary part of $X[k]/N$ (red) and the (imaginary part of the) Fourier coefficients $c_k$ (blue):



The increasing difference between $c_k$ and $X[k]/N$ with increasing $k$ is due to aliasing caused by sampling $f(t)$. Note that $f(t)$ is not band-limited, so you will always have some amount of aliasing, no matter how fast you sample. The difference in the real part ($-A/2N$ vs $0$) is caused by the fact that $f(t)$ is odd, but $x[n]$ isn't because $x[0]=f(0^+)=A/2\neq 0$.


EDIT: Here comes the derivation of the DFT of $x[n]$ defined by $(8)$. From $(3)$ we have

$$X[k]=\frac{A}{N}\sum_{n=0}^{N-1}ne^{-j2\pi nk/N}-\frac{A}{2}\sum_{n=0}^{N-1}e^{-j2\pi nk/N}\tag{a}$$

Let's first consider the case $k=0$:

$$\begin{align}X[0]&amp;=\frac{A}{N}\sum_{n=0}^{N-1}n-\frac{A}{2}\sum_{n=0}^{N-1}1\\&amp;=
\frac{A}{N}\frac{(N-1)N}{2}-\frac{AN}{2}\\&amp;=-\frac{A}{2}\end{align}\tag{b}$$

For $k\neq 0$, the second term in Eq. (a) equals zero. From the formula

$$\sum_{n=0}^{N-1}nq^n=\frac{(N-1)q^{N+1}-Nq^N+q}{(1-q)^2},\qquad q\neq 1\tag{c}$$

(which can be obtained by taking the derivative of the standard formula for a geometric series), and by noting that $e^{-j2\pi k}=1$, we get after some manipulations

$$\begin{align}X[k]&amp;=\frac{A}{N}\sum_{n=0}^{N-1}ne^{-j2\pi nk/N}\\&amp;=\ldots\\&amp;=\frac{A}{e^{-j2\pi k/N}-1},\qquad k\neq 0\end{align}\tag{d}$$

from which all other representations shown in Eq. $(10)$ can be derived quite easily.