Sum related to discrete Fourier Transform

Question

I am interested in calculation of the following sum: \begin{equation} \sum_{n=0}^{N-1}\frac 1 {1-a\cos(2\pi n/N)} \end{equation} where $0<a<1$. I Tried to pass to the exponential notation for the cosine in order to get a geomtric sum, but I could not get to any useful result. Is it possible to solve it analytically? If so, how?

Answer 1

Denote the sum by $S_N(a)$, and let $\omega=\exp(2\pi i/N)$. Then, if $|z|\neq 1$, \begin{align} S_N\left(\frac{2z}{1+z^2}\right) &=\sum_{n=0}^{N-1}\frac{1+z^2}{1-2z\cos(2\pi n/N)+z^2}\\ &=(1+z^2)\sum_{n=0}^{N-1}\frac{1}{1-z\omega^n}\frac{1}{1-z\omega^{-n}}\\ &=(1+z^2)\sum_{n=0}^{N-1}\frac{\sum_{u=0}^{N-1}(z\omega^n)^u}{1-(z\omega^n)^N}\frac{\sum_{v=0}^{N-1}(z\omega^{-n})^v}{1-(z\omega^{-n})^N}\\ &=\frac{1+z^2}{(1-z^N)^2}\sum_{u,v=0}^{N-1}z^{u+v}\sum_{n=0}^{N-1}\omega^{n(u-v)}\tag{*}\\ &=N\frac{1+z^2}{(1-z^N)^2}\sum_{u=0}^{N-1}z^{2u}=\color{blue}{N\frac{1+z^2}{1-z^2}\frac{1+z^N}{1-z^N}} \end{align} (the sum over $n$ in $(\text{*})$ is $0$ if $u\neq v$; otherwise, it is $N$). To get $S_N(a)$ back, put $$z=\frac{1-\sqrt{1-a^2}}{a}.$$