From the complex Fourier series of general period $L$, $f(x+L)=f(x)$,
$$f(x)=\sum_{n=-\infty}^{\infty}c_n\exp\left(\frac{2\pi inx}{L}\right)\tag{1}$$
and the coefficients $c_n$ are given by
$$c_n=\frac{1}{L}\int_{-L/2}^{L/2}f(x)\exp\left(\frac{-2\pi inx}{L}\right)\tag{2}$$
combining the formula's $(2)$ and $(1)$ to give a single formula for $f(x)$ yields
$$\begin{align}f(x) &=\frac{1}{L}\sum_{n=-\infty}^{\infty}\left[\int_{-L/2}^{L/2}f(y)\exp\left(\frac{-2\pi iny}{L}\right)dy\right]\exp\left(\frac{2\pi inx}{L}\right)\\ &\color{red}{\approx}\frac{1}{L}\int_{n=-\infty}^{\infty}\left[\int_{-L/2}^{L/2}f(y)\exp\left(\frac{-2\pi iny}{L}\right)dy\right]\exp\left(\frac{2\pi inx}{L}\right)dn\end{align}$$
The lecturer noted that this is a is not a rigorous derivation, but even so, I would like to understand why the second equation is an approximation and not an equality.
The lecturer did mention that if $L$ is large the $\exp\left(\frac{2\pi inx}{L}\right)$ varies 'slowly' with $n$, so the discrete variable $n$ can be replaced with a continuous variable $n$ and then the sum becomes an integral. But I still fail to see why this justifies the approximation and not equality.
The reason why I object to the approximation is that we define integrals as limits to sums since we are going from a discrete quantity to a continuous one. Put simply, I have never seen a sum being approximated by an integral before, normally it's the other way round.
Even though this derivation is non-rigorous, is anyone able to explain why the approximation (marked red) is justified?
I have read this similar question, but it still does not address my question here.
Please note, I am not after a rigorous proof, I am just trying to understand the logic behind the approximation used here. Thanks.