Represent DFT coefficients with respect to Continuous time-Fourier series coefficients
No, it's not that simple. In general, sampling will introduce aliasing. The correct derivation is as follows. Let $T$ be the period of the continuous-time signal $x(t)$, and let $T_s$ be the sampling period satisfying $T=NT_s$ with integer $N$. It follows that the sampled sequence $x_d[n]=x(nT_s)$ is also periodic with period $N$.
We have
$$x(t)=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi k}{T} t}\tag{1}$$
For the sampled sequence we obtain
$$x_d[n]=x(nT_s)=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi k}{T} nT_s}=\sum_{k=-\infty}^{\infty}c_ke^{j\frac{2\pi k}{N} n}\tag{2}$$
where we've used $T_s=T/N$. Note that the term $e^{j\frac{2\pi k}{N} n}$ is $N$-periodic in $k$, so we can rewrite $(2)$ as
$$x_d[n]=\sum_{k=0}^{N-1}\sum_{l=-\infty}^{\infty}c_{k+lN}e^{j\frac{2\pi k}{N} n}\tag{3}$$
Comparing Eq. $(3)$ to the IDFT
$$x_d[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j\frac{2\pi kn}{N}}\tag{4}$$
yields
$$X[k]=N\sum_{l=-\infty}^{\infty}c_{k+lN}\tag{5}$$
Form Eq. $(5)$ it is obvious that the DFT coefficients $X[k]$ are a (scaled and) aliased version of the Fourier coefficients $c_k$ of the continuous-time function.