By the sampling theorem, every continuous function $x(t)$, whose Fourier transform $X(\omega)$ is bandlimited to $\omega \in [-1/2,+1/2]$, can be represented by
$$x(t) = \sum_{n=-\infty}^{\infty} x(n) \ \mathrm{sinc} \!\left(t - n\right) \ .$$
When I use the constant function $x(t) = 1$ (doesn't get more bandlimited than this), this becomes
$$1 = \sum_{n=-\infty}^{\infty} \mathrm{sinc} \!\left(t - n \right)$$ $$1 = \sum_{n=-\infty}^{\infty} \frac{\mathrm{sin} \!\left( \pi(t - n) \right)}{\pi(t - n)} $$ or rather $$\forall t\in\mathbb{R} \ : \ \ \ \ \ \pi = \lim_{N \rightarrow\infty} \sum_{n=-N}^{N} \frac{\mathrm{sin} \!\left( \pi (t - n) \right)}{t - n} $$
This is trivial for $t \in \mathbb{Z}$, but for the general case I fail to see how one could verify that last expression with standard tools of calculus. Is there any "elementary" way of verifying this limit value without having to rely on the sampling theorem?