Find an asymptotic for $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$ I need to find the asymptotic behavior of $$\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$$
as $N\to\infty$.
I found (using a computer) that this asymptotically will be equivalent to $\frac{1}{3}N^2$, but don't know how to prove it mathematically.

Standard asymptotic calculus yields the estimate $$ \frac{1}{1-\cos x} = \frac{2}{x^2}+O(1)$$ hence $\displaystyle x\mapsto \frac{1}{1-\cos x} -\frac{2}{x^2}$ is bounded on a neighborhood of $0$, say $(0,\delta)$
Furthermore the function is continuous over $[\delta,\pi]$, thus bounded over $[\delta,\pi]$, hence bounded over $(0,\pi]$: write 
$$\displaystyle \forall x\in (0,\pi], \frac{1}{1-\cos x} = \frac{2}{x^2}+g(x)$$ where $g$ is bounded on $(0,\pi]$. 
With that estimate, $$\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}} = \sum_{j=1}^N \frac{2N^2}{\pi^2 j^2} + \sum_{j=1}^Ng(\frac{\pi j}n) = \frac{2N^2}{\pi^2}\left(\frac{\pi^2}6+o(1) \right) + O(N)=\frac{N^2}{3}+o(N^2)$$

As ClementC has noted, with the sharper estimate $\sum_{j=1}^N \frac{1}{j^2} = \frac{\pi^2}6 +O(\frac 1N)$ (which follows from integral methods applied to $\sum_{k=N+1}^\infty \frac{1}{k^2}$), one gets the better asymptotic $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}} = \frac{N^2}{3}+O(N)$
