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. 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.
 A: 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,
$$\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}\frac{1}{1-\frac{1}{3}\big(\frac{\pi k}{n}\big)^2}=\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}\bigg(1+\frac{\frac{1}{3}\big(\frac{\pi k}{n}\big)^2}{1-\frac{1}{3}\big(\frac{\pi k}{n}\big)^2}\bigg)\geqslant S(n)\geqslant\frac{1}{2\pi^2}\sum_{k=1}^{(n-1)/2}\frac{1}{k^2}$$
$$\frac{1}{2\pi^2}\bigg(\sum_{k=1}^\infty\frac{1}{k^2}-\sum_{k=(n+1)/2}^\infty\frac{1}{k^2}+\sum_{k=1}^{(n-1)/2}\frac{\frac{1}{3n^2}\big(\frac{\pi }{2}\big)^2}{1-\frac{1}{3}\big(\frac{\pi }{2}\big)^2}\bigg)\geqslant S(n)\geqslant\frac{1}{2\pi^2}\bigg(\sum_{k=1}^\infty\frac{1}{k^2}-\sum_{k=(n+1)/2}^\infty\frac{1}{k^2}\bigg)$$
$$\frac{1}{2\pi^2}\zeta(2)+O\Big(\frac{1}{n}\Big)\geqslant S(n)\geqslant\frac{1}{2\pi^2}\zeta(2)+O\Big(\frac{1}{n}\Big)$$
$$S(n)=\frac{1}{12}+O\Big(\frac{1}{n}\Big)$$
A: 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.
