Hi I want to prove that this summation is much smaller for $n\neq n'$ than for the case where $n=n'$. I have seen this fact with simulation results. But I don't know how to prove it in mathematics.
$$ \sum_{m=0}^{M-1} \sum_{l=l_1}^{l_2} \sum_{l'=l_1}^{l_2} \sum_{o=n_1}^{n_2} \sum_{o'=n_1}^{n_2} \exp\left( \frac{2i\pi(n-n')(k+μ_v+l-o)}{M} \right) \times \exp\left( \frac{2i\pi m(l-l'-o+o')}{M} \right) $$
I haven't worked out the complete answer, but this should be a good start.
The sum on $m$ can be pulled through the other sums and past the first exponential. Then $$\sum_{m=0}^{M-1}e^{2\pi im(\ell-\ell'-o+o')/M}$$ is zero unless $\ell+o'\equiv\ell'+o\pmod M$, in which case it is $M$. So the whole thing comes down to $$M\sum_{\ell,\ell',o,o',\ell+o'\equiv\ell'+o\pmod M}e^{2\pi i(n-n')(k+\mu_v+\ell-o)/M}$$ If $n=n'$ you get $MN$ where $N$ is the number of solutions of $\ell+o'\equiv\ell'+o\pmod M$ with $\ell,\ell',o,o'$ in the required range. If $n\ne n'$ then you expect to get some cancellation in the remaining sum, and so a smaller answer.