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.
 A: Heuristic and discussion. First, as stated in the comments (this is not a proof) the fact that, when $j/N\to 0$, we have
$$
\frac{1}{1-\cos\frac{\pi j}{N}} \operatorname*{\sim}_{N\to \infty} \frac{2N^2}{\pi^2}\cdot\frac{1}{j^2}
$$
and that $\sum_{j=1}^\infty \frac{1}{j^2}$ gives a good heuristic/sanity check for the fact (which we seek to prove) that
$$
\lim_{N\to\infty}\frac{3}{N^2}\sum_{j=1}^N \frac{1}{1-\cos\frac{\pi j}{N}} =1\,. \tag{1}
$$
Moreover, using the inequality $1-\cos x \leq \frac{x^2}{2}$ allows one to easily prove that 
$$
\liminf_{N\to\infty}\frac{3}{N^2}\sum_{j=1}^N \frac{1}{1-\cos\frac{\pi j}{N}} \geq 1\,. \tag{2}
$$

Proof. 
Let us prove (1), though. Since
$1-\cos x  = 2\sin^2 \frac{x}{2}$ for all $x\in\mathbb{R}$, we can rewrite
$$
\begin{align}
\sum_{j=1}^N \frac{1}{1-\cos\frac{\pi j}{N}}
= \frac{1}{2} \sum_{j=1}^N \frac{1}{\sin^2\frac{\pi j}{2N}}
= \frac{2N^2}{\pi^2} \sum_{j=1}^N \frac{1}{j^2}\frac{1}{\left(\frac{2N}{\pi j}\sin\frac{\pi j}{2N}\right)^2}
\end{align}
$$
so to show the result it is sufficient to prove that
$$
\lim_{N\to\infty} \sum_{j=1}^N \frac{1}{j^2}\frac{1}{\left(\frac{2N}{\pi j}\sin\frac{\pi j}{2N}\right)^2} = \sum_{j=1}^\infty \frac{1}{j^2} = \frac{\pi^2}{6} \tag{3}
$$
I am not sure what is the most elegant way to prove (3), so I'll give it a shot with a big hammer: the Dominated Convergence Theorem. Namely, I'll show the following:

Lemma. Let $f\colon [0,1]\to\mathbb{R}$ be continuous, and $(a_n)_n$ be a non-negative sequence with $\sum_{n=1}^\infty a_n < \infty$. Then
  $$
\lim_{n\to\infty} \sum_{j=1}^n f\!\left(\frac{j}{n}\right) a_j = \sum_{j=1}^\infty a_j\,. \tag{4}
$$ 

Note that this implies (3), taking $f(x) = \frac{(2x/\pi)^2}{\sin^2 \frac{\pi x}{2}}$ extended by continuity at $0$.
Proof of the lemma.
Writing
$$
\sum_{j=1}^n f\!\left(\frac{j}{n}\right) a_j
= \sum_{j=1}^\infty g_n\!\left(j\right) a_j
$$
for $g_n(x) \stackrel{\rm def}{=} f\!\left(\frac{x}{n}\right) \mathbb{1}_{[0,n]}(x)$. Then $g_n$ converges pointwise to the constant function $f(0)$ and is dominated by $\lVert f\rVert_\infty$, so applying the DCT we get that 
$$
\lim_{n\to\infty} \sum_{j=1}^\infty g_n\!\left(j\right) a_j = \sum_{j=1}^\infty f(0) a_j
$$
as claimed. $\square$
A: 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 $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}} = \frac{N^2}{3}+O(N)$
