The Fourier coefficient $C_n$ of a periodic function $s(t)$ with period $T$ is given by
$$C_n= \frac{1}{T} \int_0^T s(t) e^{-2\pi int/T} \,dt$$
Now consider the Fourier term
$$\hat{S}\Big(\frac{n}{T}\Big)=\int_{-\infty}^{\infty} s(t) e^{-2\pi int/T}\,dt$$
Is the assumption that $C_n=\hat{S}\big(\frac{n}{T}\big)$ correct?
Also, $\hat{S}\big(\frac{n}{T}\big)$ can be seen as the projection of $S(t)$ on $e^{-2\pi int/T}$, can the same be said about $C_n$? If yes, how can we show the equivalence of $C_n$ and $\hat{S}\big(\frac{n}{T}\big)$? Thanks for your help.
It would be correct if you used $\chi_{[0,T]}(t)s(t)$ (where $\chi_{[0,T]}(t)$ is the box or indicator function of $[0,T]$) in the integrand. $s(t)$ itself as a periodic function is not square integrable. So the Fourier transform in the strict sense does not exist, especially not as a Fourier integral.
However, you can compute the Fourier transform in the sense of tempered distributions.
If $s(t)=\sum_{n\in\mathbb Z}C_ne^{2\pi i\,kt/T}$, then the Fourier transform is of the shape $S(f)=\sum_{n\in\mathbb Z}C_n\delta(f-n/T)$.