Sum of $(1/n^2)\sum_{k=2}^{n/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}$ in the large $n$ limit.

Question

This summation appears in studying circulant matrices. Consider an $n-$dimensional circulant matrix with with entries $c_1=c_{n-1}=1$ and $c_0=-2$ (all other coefficients beign $0$). This represents, for example, a random walk on a closed ring of $n$ states. The following calculation can be viewed as the average decay time for all modes (notice that the denominator of the series' terms correspond to the negative eigenvalues of the circulant matrix, $-\lambda_{k}$. Prefactor $1/n^2$ is a normalization constant). From the symmetry of the eigenvalues set one can distinguish two cases: $n$ odd or even.

If $n$ is odd, $$\frac{1}{n^2}\sum_{k=2}^{(n+1)/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}\,.$$ If $n$ is even, $$\frac{1}{n^2}\sum_{k=2}^{n/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}\,.$$

I've attempted to approximate this summation by using the continuous limit but I get the wrong values when comparing to numerical evaluation.

The goal is to find an expression for the large $n$ behavior.

Answer 1

Let's denote $$S(n)=\frac{1}{n^2}\sum_{k=2}^{(n+1)/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}=\frac{1}{2n^2}\sum_{k=1}^{(n-1)/2}\frac{1}{\sin^2\left(\frac{\pi k}{n}\right)}$$ It is not difficult to check that $\displaystyle \phi\geqslant\sin\phi\geqslant\phi-\frac{\phi^3}{6}$ for any $\phi\in \big[0;\frac{\pi}{2}\big]$. Therefore, we have $$\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}\frac{1}{\Big(1-\frac{1}{6}\big(\frac{\pi k}{n}\big)^2\Big)^2}\geqslant S(n)\geqslant\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}$$ and, opening the brackets in the first term, $$\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}\frac{1}{1-\frac{1}{3}\big(\frac{\pi k}{n}\big)^2}=\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}\bigg(1+\frac{\frac{1}{3}\big(\frac{\pi k}{n}\big)^2}{1-\frac{1}{3}\big(\frac{\pi k}{n}\big)^2}\bigg)\geqslant S(n)\geqslant\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}$$ $$\frac{1}{2\pi^2}\bigg(\sum_{k=1}^\infty\frac{1}{k^2}-\sum_{k=(n+1)/2}^\infty\frac{1}{k^2}+\sum_{k=1}^{(n-1)/2}\frac{\frac{1}{3n^2}\big(\frac{\pi }{2}\big)^2}{1-\frac{1}{3}\big(\frac{\pi }{2}\big)^2}\bigg)\geqslant S(n)\geqslant\frac{1}{2\pi^2}\bigg(\sum_{k=1}^\infty\frac{1}{k^2}-\sum_{k=(n+1)/2}^\infty\frac{1}{k^2}\bigg)$$ $$\frac{1}{2\pi^2}\zeta(2)+O\Big(\frac{1}{n}\Big)\geqslant S(n)\geqslant\frac{1}{2\pi^2}\zeta(2)+O\Big(\frac{1}{n}\Big)$$ $$S(n)=\frac{1}{12}+O\Big(\frac{1}{n}\Big)$$

Answer 2

It is also interesting to get the second term of the asymptotic at $n\to\infty$. We could expect that it would be $\,O\big(\frac{1}{n}\big)$, because heuristically $$S(n)=\frac{1}{2n^2}\sum_{k=1}^\frac{n-1}{2}\frac{1}{\sin^2\frac{\pi k}{n}}=\frac{1}{2\pi^2}\sum_{k=1}^\frac{n-1}{2}\frac{1}{k^2\Big(1+\frac{1}{6}\big(\frac{\pi k}{n}\big)^2+...\Big)^2}$$ $$\sim\frac{1}{2\pi^2}\sum_{k=1}^\frac{n-1}{2}\Big(\frac{1}{k^2}-\frac{1}{3}\frac{\pi^2}{n^2}+A\frac{\pi^4k^2}{n^4}+B\frac{\pi^6k^4}{n^6}+...\Big)$$ All the terms, except for the first one, give $\sim\frac{1}{n}$ as main terms. But we will see that in fact $$\boxed{\,\,S(n)=\frac{1}{12}-\frac{1}{12\,n^2}+o\Big(\frac{1}{n^2}\Big)\,\,}\tag{1}$$ and where the third term is $\sim\frac{1}{n^3}$ or, probably, $\sim\frac{1}{n^4}$.

First, using $\displaystyle \frac{d}{dx}\cot x=-\frac{1}{\sin^2x}$ we can present the sum in the form $$S(n)=-\frac{1}{2\pi^2n^2}\frac{\partial}{\partial\frac{1}{n}}\sum_{k=1}^N\frac{n}{k^2}\frac{\pi k}{n}\cot\frac{\pi k}{n}$$ were we will put the value of $\displaystyle N=\frac{n-1}{2}$ after taking the derivative $\displaystyle\frac{\partial}{\partial\frac{1}{n}}$. Using also the presentation $\,\displaystyle x\cot x=\sum_{l=0}^\infty\frac{(-1)^lB_{2l}(2x)^{2l}}{(2l!)}$, where $B_{2l}$ is the corresponding Bernoulli number, we can rewrite the sum as $$S(n)=-\frac{1}{2\pi^2n^2}\frac{\partial}{\partial\frac{1}{n}}\sum_{l=0}^\infty\frac{(-1)^lB_{2l}(2\pi)^{2l}}{(2l!)}\sum_{k=1}^N\frac{k^{2l-2}}{n^{2l-1}}$$ Taking the derivative and rearranging a bit, $$=\frac{1}{2\pi^2}\sum_{k=1}^\frac{n-1}{2}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{l=1}^\infty\frac{(-1)^lB_{2l}(2\pi)^{2l}}{(2l!)}(2l-1)\sum_{k=1}^\frac{n-1}{2}\frac{k^{2l-2}}{n^{2l}}$$ Using the Euler-Maclaurin formula $$\sum_{k=1}^\frac{n-1}{2}k^{2l-2}=\int_1^\frac{n-1}{2}k^{2l-2}dk+\frac{1}{2}\Big(\big(\frac{n-1}{2}\big)^{2l-2}+1\Big)+...$$ Keeping only the terms up to $\displaystyle \sim\frac{1}{n^2}$ $$S(n)=\frac{1}{2\pi^2}\sum_{k=1}^\infty\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=\frac{n+1}{2}}^\infty\frac{1}{k^2}$$ $$-\,\frac{1}{2\pi^2}\sum_{l=1}^\infty\frac{(-1)^lB_{2l}(2\pi)^{2l}}{(2l!)}\frac{2l-1}{n^{2l}}\bigg(\frac{\big(\frac{n-1}{2}\big)^{2l-1}}{2l-1}-\frac{1}{2l-1}+\frac{1}{2}\Big(\big(\frac{n-1}{2}\big)^{2l-2}+1\Big)\bigg)+o\Big(\frac{1}{n^2}\Big)$$ Using the fact that $\displaystyle \sum_{l=0}^\infty\frac{(-1)^lB_{2l}(\pi)^{2l}}{(2l!)}=\frac{\pi}{2}\cot\frac{\pi}{2}=0$, and dropping all the terms smaller than $\displaystyle \sim\frac{1}{n^2}$, we get $$S(n)=\frac{1}{12}-\frac{1}{2\pi^2}\sum_{k=\frac{n+1}{2}}^\infty\frac{1}{k^2}+\frac{1}{\pi^2n}-\frac{1}{2\pi^2n^2}\frac{(-1)B_{2}}{2!}(2\pi)^2\Big(\frac{1}{2}-1\Big)+o\Big(\frac{1}{n^2}\Big)$$ Using $B_2=\frac{1}{6}$ and $\displaystyle\sum_{k=\frac{n+1}{2}}^\infty\frac{1}{k^2}=\int_\frac{n+1}{2}^\infty\frac{dk}{k^2}+\frac{1}{2}\Big(\frac{2}{n+1}\Big)^2+...\,$, we finally get the formula (1) : $$S(n)=\frac{1}{12}-\frac{B_2}{2n^2}+o\Big(\frac{1}{n^2}\Big)=\frac{1}{12}\Big(1-\frac{1}{n^2}\Big)+o\Big(\frac{1}{n^2}\Big)$$ Numeric check at WolframAlpha confirms the asymptotic.

Answer 3

Similar to @Svyatoslav's second answer $$\frac 1 {n^2}\sum_{k=2}^{\frac n 2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}=\frac 1 {2n^2}\sum_{k=2}^{{\frac n 2}}\frac{1}{\sin ^2\left(\frac{\pi (k-1)}{n}\right) }=\frac 1 {2n^2}\sum_{k=1}^{{\frac n 2}-1}\frac{1}{\sin ^2\left(\frac{\pi k}{n}\right) }$$

Using $$\frac{1}{\sin ^2(t)}=\sum_{m=-1}^\infty a_m\,t^{2m}$$ where the $a_m$ form tha apparently unknown sequence $$\left\{1,\frac{1}{3},\frac{1}{15},\frac{2}{189},\frac{1}{675}, \frac{2}{10395},\frac{1382}{58046625},\frac{4}{1403325},\frac{361 7}{10854718875},\cdots\right\}$$ $$\frac{1}{\sin ^2\left(\frac{\pi k}{n}\right) }=\sum_{m=-1}^\infty a_m\,\frac{\pi^{2m}}{n^{2m}} k^{2m}$$

$$\sum_{k=1}^{\frac n 2-1} k^{2m}=H_{\frac{n}{2}-1}^{(-2 m)}=\zeta (-2 m)-\zeta \left(-2 m,\frac{n}{2}\right)$$ where appear generalized harmonic numbers, the zeta function and the Hurwitz zeta function.

Using all the above gives (apparently at least) $$\frac 1 {n^2}\sum_{k=2}^{\frac n 2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}=\frac 1{12} \left(1-\frac 1{n^2}\right)+O\left(\frac{1}{n^{16}}\right)$$

Let $n=10^p$ and compute the absolute difference $$\left( \begin{array}{cc} p & \Delta_p \\ 1 & 2.5\times 10^{-3} \\ 2 & 2.5\times 10^{-5} \\ 3 & 2.5\times 10^{-7} \\ 4 & 2.5\times 10^{-9} \\ 5 & 2.5\times 10^{-11} \\ \end{array} \right)$$

I suppose that comments are not needed.