# What is the Fourier series of $\frac1T\sum^{\infty}_{m=-\infty}\delta(f-\frac mT)$?

As the title mentioned, I've not known exactly about Fourier series and when I was reading an digital communication textbook, I wondered about below equation derivation of Fourier series like $$\alpha(f)=\frac1T\sum^{\infty}_{m=-\infty}\delta\left(f-\frac mT\right)$$ which is periodic with period $\frac1T$ and $\delta$ is the Dirac-Delta function. From Fourier series, we have $$\alpha(f)=\frac1{1/T}\sum^\infty_{m=-\infty}c_ne^{-i2\pi nf/(1/T)}=\sum^\infty_{m=-\infty}e^{-i2\pi nfT}$$ where $$c_n=\int^\frac1{2T}_{-\frac1{2T}}\alpha(f)e^{i2\pi nf/(1/T)}df=\int^\frac1{2T}_{-\frac1{2T}}\frac1T\delta(f)e^{i2\pi nf/(1/T)}df=\frac1T$$ As I read the Fourier series information on Wiki, but I don't know how that equation is changed simply although Fourier series is very complicated. Please help me understand detail procedure about that. Thank you.

Let $f$ be a function that periodic with period $1/T$, i.e. $f(t)=f(t+1/T)$, then this function can be expanded in the following form $$f(t)=\sum_{n=-\infty}^{\infty} c_n \mathrm{e}^{2 \pi \mathrm{i}n\omega t},$$ where $$c_n =\frac1T\int_{-1/(2T)}^{1/(2T)} f(t) \mathrm{e}^{-2 \pi \mathrm{i}n\omega t} dt.$$
I will not go into detail on the convergence results, i.e. under which circumstances does the series converge to the function $f$. However, it is instructive to note that this tells us that it is possible to expand a periodic function into building blocks of the form $\mathrm{e}^{2 \pi \mathrm{i}n\omega t}$.
The function that is of interest to you, i.e. $\frac{1}{T} \sum_{m=-\infty}^\infty \delta(f-m/T)$ indeed has the period $1/T$, which is easy to see. Therefore you can use the statement that I quoted above to compute the coefficients of the Fourier series to write the function as a superposition of complex exponential functions. This is what happens in the text you quoted.