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.
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1Are you sure it's not $\frac{N^2}{3}$? – Clement C. May 01 '18 at 19:32
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@ClementC. I also find $N^2 / 3$. – John Barber May 01 '18 at 19:34
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2Which is consistent with the fact that for $j \ll N$, $j/N \to 0$, and $$\frac{1}{1-\cos\frac{\pi j}{N}} \sim_{N\to \infty} \frac{2N^2}{\pi^2 j^2}$$ and $$\frac{2N^2}{\pi^2} \sum_{j=1}^N \frac{1}{j^2} \sim_{N\to\infty} \frac{2N^2}{\pi^2} \sum_{j=1}^\infty \frac{1}{j^2} = \frac{N^2}{3}$$ (just a heuristic, not a full-fledged proof) – Clement C. May 01 '18 at 19:38
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Yeah, thanks, i although found it, but i need in strong proof and don't know how to do it. – Виталий Моисеев May 01 '18 at 19:46
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@OP I took the liberty to edit your question to add the specific question to the body, and correct the empirical asymptotics. Does it look alright? – Clement C. May 01 '18 at 20:46
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Yeah, thank you. – Виталий Моисеев May 01 '18 at 21:58
2 Answers
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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. $\square$
Clement C.
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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 $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}} = \frac{N^2}{3}+O(N)$
Gabriel Romon
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"Continuous over $(0,\pi]$, and therefore bounded over" > you may want to specify here that you are crucially using the first part before to get this. Otherwise, "continuous and therefore bounded [over a non compact set]" is problematic. – Clement C. May 01 '18 at 21:15
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@ClementC. yes, the "therefore" takes into account the very first estimate. – Gabriel Romon May 01 '18 at 21:18
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Also, correct me if I'm wrong, but this argument actually establishes a bit more than the first term - given the asymptotics of the remainder of the converging series (your $o(1)$), you show that the $o(N^2)$ is a $O(N)$. – Clement C. May 01 '18 at 21:23
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@ClementC. The $o(N^2)$ could actually be $O(N\log N)$ for all we know, so I don't think you're allowed to transform the $o(N^2)+O(N)$ into $O(N)$ without a different estimate. – Gabriel Romon May 01 '18 at 21:30
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Well, you have $\sum_{j=1}^N 1/j^2 = \pi^2/6 + O(1/N)$, which implies what I said. – Clement C. May 01 '18 at 21:32
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