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using sum of geometric progression and inverse DFT we can show: $${\sum_{j=0}^{k-1}\frac{\exp(2\pi ij\frac{u+1}{k})}{\exp(2\pi i\frac{j}{k})-a}=k\frac{a^{u(mod\ k)}}{1-a^{k}}}$$

My question is how to compute

$${\sum_{j=0}^{k-1}\frac{\exp(2\pi ij\frac{u+1}{k})}{(\exp(2\pi i\frac{j}{k})-a)^m}}$$ here $m,u,k$ are natural numbers

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1 Answer 1

up vote 2 down vote accepted

This is

$$\frac{1}{n!} \frac{\mathrm{d}^{n}}{\mathrm{d}a^{n}} S (a)\;,$$

where $n=m-1$ and $S(a)$ is the sum you computed.

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