How to prove $\sum_{n=-\infty}^ \infty {\rm sinc}\bigl( \pi(t-n)\bigr) = 1$?

Do you know how to prove the Fourier series of a function $f \in C^\infty[-1/2,1/2]$ converges for $x \in (-1/2,1/2)$ ?

Then look at the Fourier series of $f(x) = e^{2i \pi t x}$ at $0$.

$$c_n(f) = \int_{-1/2}^{1/2} f(x) e^{-2i \pi nx}dx = \int_{-1/2}^{1/2} e^{2i \pi (t-n)x}dx= \frac{\sin \pi (n-t)}{\pi (n-t)}$$

$$1=f(0) = \sum_n c_n(f) = \sum_n \frac{\sin \pi (n-t)}{\pi (n-t)}$$

The proof of the theorem in yellow is based on the Dirichlet kernel $$(\sum_{n=-N}^N c_n(f))-f(0) = \sum_{n=-N}^N c_n(f-f(0))=\int_{-1/2}^{1/2}(f(x)-f(0))\sum_{n=-N}^N e^{-2i \pi nx}dx$$ $$ =\int_{-1/2}^{1/2} (f(x)-f(0)) \frac{\sin(2 \pi (N+1/2) x)}{2\pi\sin(\pi x)}dx = \int_{-1/2}^{1/2} g(x) \sin(2 \pi (N+1/2) x)dx$$ $$= \int_{-1/2}^{1/2} g'(x) \frac{\cos(2 \pi (N+1/2) x)}{2 \pi (N+1/2)}dx= O(\frac1N)$$ Where $$g(x)=\frac{f(x)-f(0)}{2\pi\sin(\pi x)}=\frac{x}{2\pi\sin(\pi x)} \int_0^1 f'(xt)dt \in C^\infty[-1/2,1/2]$$