How to prove $\sum_{n=-\infty}^ \infty {\rm sinc}\bigl( \pi(t-n)\bigr) = 1$? Thank you by avance for your help.
So, I found on this website, that $\sum_{n=-\infty}^{\infty} {\rm sinc}( \pi n)= 1$.
But I could not find any way to prove it. I know it’s about fourrier, but I don’t know how to do so...
Does anyone know how to do prove $\sum_{n=-\infty}^{\infty}  {\rm sinc}\bigl( \pi(t-n)\bigr) = 1$ ?
Thank you !

After getting help of @reuns, here the full demonstration :
$f(x)=e^{2i\pi tx}$

$f(0)=e^0=1$


We take a period of 1.
So with $C_{n}(f)=\frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-i2\pi\frac{n}{T}t}dt$
We get : $C_{n}(f)=\int_{-1/2}^{1/2}f(t)e^{-i2n\pi t}dt$
We use $f(x)$ from above : 
$= C_{n}(f)=\int_{-1/2}^{1/2}f(t)e^{-i2n\pi x}dx$
$= C_{n}(f)=\int_{-1/2}^{1/2} e^{2i\pi tx} . e^{-2i\pi nx} dx$
$= C_{n}(f)=\int_{-1/2}^{1/2} e^{2i\pi (t-n)x}dx$
$=\left [\frac{e^{2i\pi (t-n)x}}{2i\pi (t-n)}  \right ]^{1/2}_{-½}$
$=\frac{e^{2i\pi (t-n)1/2} - e^{-2i\pi (t-n)1/2}}{2i\pi (t-n)}$
With Euler we know : $sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$
So $C_n(f)= \frac{sin (\pi(t-n))}{\pi (t-n)}$
We also know that $\frac{sin(x)}{x}=sinc(x)$
Based on Dirichlet kernel : $f(x)=\sum_{n=-\infty }^{\infty}C_n(f)e^{\frac{inx2\pi}{T}}$
And $f(0)=1$
So $f(0)=\sum_{n=-\infty }^{\infty}C_n(f)e^{0}=\sum_{n=-\infty }^{\infty}\frac{sin(\pi(t-n))}{\pi(t-n)}=1$
==> $\sum_{n=-\infty }^{\infty}\frac{sin(\pi(t-n))}{\pi(t-n)}=\sum_{n=-\infty }^{\infty}sinc(\pi(t-n))=1$
