Estimation of sum of series with cosine

Question

Prove the following:

$$ \sum_{{\large j = 1} \atop {\large j \neq k}}^{n} {1 \over \left\vert\cos\left(k\pi/ n\right) - \cos\left(j\pi/n\right)\right\vert} \leq cn^{2} \qquad\mbox{where}\quad 1 \leq k \leq n\quad\mbox{is fixed.} $$ I was able to get the upper bound to be $cn^{3}$. Any method can be used for getting the desired bound. I also tried taking the upper bound in terms of an integral. I will appreciate any suggestions.

Answer 1

By sum-to-product formulae, the sum $S(n)$ equals $$S(n)=\frac12\sum^n_{j=1}\left|\csc\frac{(j+k)\pi}{2n}\csc\frac{(j-k)\pi}{2n}\right|$$

By Cauchy-Schwarz inequality, $$S(n)\le\frac12\sqrt{ \underbrace{\sum^n_{j=1}\csc^2\frac{(j+k)\pi}{2n}}_{S_1} \cdot\underbrace{\sum^n_{j=1}\csc^2\frac{(j-k)\pi}{2n}}_{S_2}}$$

Regarding $S_1$, $$ \begin{align} S_1 &=\sum^n_{j=1}\csc^2\frac{(j+k)\pi}{2n} \\ &=\sum^{n+k}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn \\ &=\sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn +\sum^{n+k}_{j=n+1}\csc^2\frac{\pi}{2}\frac jn\\ &=\sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn +\sum^{k}_{j=1}\csc^2\frac{\pi}{2}\left(1+\frac jn\right)\\ &=\sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn +\sum^{k}_{j=1}\sec^2\frac{\pi}{2}\frac jn\\ &\le \sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn+k \sec^2\frac{k\pi}{2n}\\ &\le \sum^{n}_{j=k+1}\frac{n^2}{j^2}+k \sec^2\frac{k\pi}{2n} \qquad\text{NB: }x\csc\frac{\pi x}{2}\le1 \text{ for }|x|\le 1\\ &\le\frac{\pi^2}{6}n^2+k \sec^2\frac{k\pi}{2n} \end{align} $$

Thus, $S_1=O(n^2)$.

Regarding $S_2$, $$ \begin{align} S_2 &=\sum^{k-1}_{j=1}\csc^2\frac{(k-j)\pi}{2n}+\sum^n_{j=k+1}\csc^2\frac{(j-k)\pi}{2n} \\ &=\sum^{k-1}_{j=1}\csc^2\frac{\pi j}{2n}+\sum^{n-k}_{j=1}\csc^2\frac{\pi j}{2n} \\ &\le\sum^{k-1}_{j=1}\frac{n^2}{j^2}+\sum^{n-k}_{j=1}\frac{n^2}{j^2} \\ &\le \frac{\pi^2}{3}n^2 \end{align} $$

Hence, $S_2=O(n^2)$. Therefore, $$S(n)=\sqrt{O(n^2)\cdot O(n^2)}=O(n^2)$$

Answer 2

Since on $(0,\pi/2)$ we have $\frac{2}{\pi}x<\sin x<x$ it follows that $$ S(n) = \sum_{1\leq j<k\leq n}\frac{1}{\left|\sin\frac{\pi(k-j)}{2n}\sin\frac{\pi(k+j)}{2n}\right|}\sim n\sum_{1\leq j<k\leq n}\frac{1}{(k-j)\sin\frac{\pi(k+j)}{2n}} $$ and by letting $k-j=d$ we get $$ S(n)\sim n\sum_{d=1}^{n-1}\frac{1}{d}\sum_{j=1}^{n-d}\frac{1}{\sin\frac{\pi(d+2j)}{2n}}$$ where the innermost sum can be regarded as a Riemann sum. Since $$ \int_{\frac{\pi d}{2n}}^{\pi-\frac{\pi d}{2n}}\frac{dx}{\sin x}=2\log\cot\frac{\pi d}{4n}\sim 2\log\frac{4n}{\pi d} $$ we have $$ S(n) \sim n^2\cdot \frac{1}{n}\sum_{d=1}^{n-1}\frac{-2\log\frac{\pi d}{4n}}{\pi \frac{d}{n}}\sim n^2\log^2(n) $$ always by Riemann sums. We cannot improve the bound beyond this since the existence of a constant $c$ such that $S(n)\leq cn^2$ would imply the integrability of $\frac{1}{\left|\cos(x)-\cos(y)\right|}$ over $(0,\pi)^2$, which does not hold.

Numerical experiments confirm that $S(n)$ is not $O(n^2)$. In the following graph we have $\frac{1}{n^2}S(n)$ in blue and $1+\frac{1}{4}\log^2(n)$ in yellow:

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If $k\in[1,n]$ is fixed, the same techniques can be applied to $$ S_k(n) = \sum_{j\neq k}\frac{1}{\left|\sin\frac{\pi(k-j)}{2n}\sin\frac{\pi(k+j)}{2n}\right|}\sim n\sum_{j\neq k}\frac{1}{|k-j|\sin\frac{\pi(k+j)}{2n}}.$$ It is reasonable to expect that the $S_k$s do not differ really much from each other, such that, optimistically, $S_k(n)\ll n\log^2(n)$. By letting $d=k-j$ as before we have $$S_k(n)\sim n\sum_{\substack{d=k-n\\d\neq 0}}^{k-1}\frac{1}{|d|\sin\frac{\pi(2k-d)}{2n}}$$ and by using $\sin(x)\sim\min(x,\pi-x)$ (for $x\in(0,\pi)$) we have $$S_k(n)\sim n^2\sum_{\substack{d=k-n\\d\neq 0}}^{k-1}\frac{1}{|d|\min\left(2k-d,2n-2k+d\right)}$$ which can be estimated through $$ \sum_{d=1}^{k-1}\frac{1}{d(2k-d)}=\frac{1}{2k}\sum_{d=1}^{k-1}\left(\frac{1}{d}+\frac{1}{2k-d}\right)=\frac{H_{2k-1}-\frac{1}{k}}{2k}\ll\frac{\log(k)}{k}\ll\frac{\log(n)}{n}, $$ $$ \sum_{d=1}^{k-1}\frac{1}{d(2n-2k+d)}=\frac{1}{2n-2k}\left[H_{k-1}-H_{2n-k-1}+H_{2n-2k+1}\right]\ll\frac{\log\left(\frac{k(2n-2k)}{2n-k}\right)}{n-k}$$ and the corresponding sums for negative values of $d$. All things considered, $S_k(n)$ really is $\ll n\log^2(n)$.
A simpler alternative is to exploit $\sin(x)\sim x(\pi-x)$ on $(0,\pi)$ and the fact that $$\tilde{S}_k(n) = \frac{16n^4}{\pi^4}\sum_{j\neq k}\frac{1}{\left|(k-j)(k+j)(2n-k+j)(2n-k-j)\right|}$$ can be computed in explicit terms through partial fraction decomposition and harmonic numbers.