Suppose I have $f(x) = e^{-\pi x^2}$. Then it can be worked out that $$\hat{f}(y) = \int_{-\infty}^{\infty}f(x)e^{-2\pi i x y}\mathrm{d}x=f(y).$$
I can represent $f(x)$ as $$f(x) = \lim_{T\to\infty}\sum_{n=-\infty}^{\infty}c_ne^{2\pi i\frac{n}{T}x},$$ where $$c_n = \frac{1}{T}\hat{f}\left(\frac{n}{T}\right).$$
Here's what I don't understand. Suppose $x$ is real. Then $f(x)$ is real. Then $\hat{f}(x)$ is real. Then $c_n$ is real.
But $e^{2\pi i\frac{n}{T}x}$ is complex. So every term in the summation is complex.
So does that sum only converge to a real number in the limit case, where (I'd imagine) the complex parts must all cancel out?
If you consider the Fourier transform of a real valued function $f(x)$ it is easy to see that the definition leads to the following property $$ \hat{f} (-y) = \overline{\hat{f} (y)} $$ Since here you also know that the Fourier transform is real, which is always true for a symmetric real function $f(x)$, this automatically implies that $\hat{f}(y)=\hat{f}(-y)$ and hence you get $c_n = c_{-n}$ for the coefficients in your series.
As a results you rewrite the series as $$ f(x) = \lim_{T \rightarrow \infty} \left[ c_0 + \sum_{n=0}^\infty c_n \left(e^{2 \pi i \frac{n}{T}x} + e^{-2 \pi i \frac{n}{T}x}\right) \right] = \lim_{T \rightarrow \infty} \left[ c_0 + 2 \sum_{n=0}^\infty c_n \cos\left(2 \pi \frac{n}{T}x\right) \right] $$ which is the sum of only real terms.
So just as you expected the complex terms cancel out, but they do so for every value of $T$.