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Consider the following sum:

\begin{gather} S = \sum\limits_{n = 1}^{N} \sum\limits_{n' = 1}^{N} \dfrac{\omega^{k(n-1+n'-1)}}{z - \Omega_n - \Omega_{n'}}, \\ \Omega_n = -i \alpha \dfrac{1}{1 - \gamma \omega^{-(n-1)}}, \end{gather} where $\omega = e^{i2\pi/N}$.

Basically, this question is a follow-up to my previous question, and the answer from that question allows to immediately get rid of one of the sums (thx to @metamorphy!), and obtain $\left(\text{it boils down to using } \sum\limits_{n=1}^{N} \dfrac{\omega^{k (n-1)}}{1 - \beta \omega^{n-1}} = \dfrac{N \beta^{\rm{mod}(-k, N)}}{1 - \beta^N} \right)$: \begin{gather} S = \sum\limits_{n = 1}^{N} \dfrac{\omega^{k(n-1)}}{\gamma(-z + \Omega_n)} \dfrac{N}{1 - \beta(z, n)^N} \left( \beta^{\rm{mod}(-(k+1),N)} - \gamma \beta^{\rm{mod}(-k, N)} \right), \\ \beta(z, n) = \dfrac{-z + \Omega_n - i \alpha}{-z + \Omega_n}. \end{gather}

After this step the problem is in powers of $\beta$, which, probably, won't allow to obtain an explicit answer.

May there be any other way? Might the limit of $N \to \infty$ help? I also though about replacing the discrete sum with an integral, but it did not look for me straightforward as discreteness turns out to be really important, when it comes to functions with roots of unity (simply because the integral of the exponent over the period is always zero, while the discrete sum is not).

Does anybody have any suggestions?

UPDATE 19.05.2021: In the latter single sum one can naively go into the limit $N \to \infty$, and try to calculate the integral using the residue theorem. However, this expression contains $N$ simple poles, all of which are contained in $1-\beta(z,n)^N$. So, we will end up in a sum of $N$ contributions, and this is what we have started from, except that now this sum is not exact, but approximate :).

Sl0wp0k3
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    I would try small values of $N$ for some pattern to be seen (and drop the "$-i\alpha$" since it can be absorbed into $z$). But I have little to no hope. My attempt to build a power series in $z$ (and then a power series in $\gamma$ for the coefficients) drowned me too deep. – metamorphy Jul 23 '21 at 13:52
  • Hello @metamorphy! Yes, it looks really bad. Nevertheless, thanks for the previous answer on a single sum! – Sl0wp0k3 Jul 23 '21 at 16:49
  • @metamorphy, I also want to admit that the opposite would be preferable: the limit of large $N$. A naive approach is to convert sums into integrals, however, every time I make such a transition for the functions involving roots of unity - I fail, turns out that discreteness for them is crucial, or it must be done in some clever way. – Sl0wp0k3 Jul 25 '21 at 23:16

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