It's well know that : $$\frac{1}{e^{2\pi i x}-1}=\frac{1}{2\pi i x}+\frac{x}{\pi i }\sum_{k=1}^{\infty}\frac{1}{x^{2}-k^{2}}-\frac{1}{2}$$ And by the Mittag-Leffler theorem, the functions : $$\frac{1}{(e^{2\pi i x}-1)^{n}}$$ have partial fraction expansions in terms if their poles. What is the general form of these expansions ?
Mittag-Leffler expansion of $\frac{1}{(e^{2\pi i x}-1)^{n}}$
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sequences-and-series
complex-analysis
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2Try differentiating both sides. – 2017-02-15
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2You may also notice that $$\text{Res}\left(\frac{1}{(e^{2\pi i z}-1)^n},z=0\right)=(-1)^n\frac{i}{2\pi}$$ and we are dealing with $1$-periodic functions. – 2017-02-15
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0Recursion: $\frac{1}{(e^{i2\pi x}-1)^{n+1}}=-\frac{1}{(e^{i2\pi x}-1)^n}+\frac{i}{2\pi n}\frac{d}{dx}\frac{1}{(e^{i2\pi x}-1)^n}$ --- An explicit formula becomes complicate. :-) – 2017-02-21