Let $N>0$ and $a_0,a_1,...,a_N$ be any positive numbers.
How to prove that $$\dfrac{1}{2}+\sum_{n = 1}^{\infty}\prod_{k = 0}^{N}\text{sinc}(a_kn) = \int_{0}^{\infty}\prod_{k = 0}^{N}\text{sinc}(a_kx)\,dx$$
Where $\displaystyle \text{sinc}(x)=\frac{\sin(x)}{x}$, and $\displaystyle\sum_{k = 0}^{N}a_k \le 2\pi$.
This is amazing...
If $a_k$ is a slowly divergent series, then the sum $\sum_{k=0}^{N} a_k$ will exceed $2\pi$ only when N exceed some $\textbf{very large} $ number $N_0$.
For instance, if $a_k=1/p_k$, sum of the reciprocals of the primes, than $N_0\approx10^{180}$