Prove that $$\sum_{k=n}^{2n-2} \frac{|\sin k|}{k} < 0.7 \ln 2, \qquad (n\ge2)$$ and $$\cot\left(\frac{\pi}{2n}\right) \le \sum_{k=1}^n \left|\sin\left(x+\frac{k\pi}{n}\right)\right| \le \frac{n}{\sqrt{2}}, \qquad (n\ge 2)$$
I have posted this same question here, thank you.
A proof by brute force:
$\square$ Denote $a_k=\frac{|\sin k|}{k}$, $\ell=[\frac{n-2}{100}]$ and decompose the sum as $$S_n=\sum_{k=n}^{2n-2}a_k=\sum_{m=0}^{\ell-1}\underbrace{\sum_{j=0}^{99}a_{n+100m+j}}_{G_{nm}}+\underbrace{\sum_{k=n+100\ell}^{2n-2}a_k}_{R_n}$$ Also, denote $M_{100}=\max_{x\in[0,2\pi]}\sum_{k=0}^{99}|\sin(x+k)|$. It is easy to check (e.g. with Maple or Mathematica) that $M_{100}<64.27$ (approximately, $M_{100}\approx 64.2618$). This implies that $G_{nm}< \frac{M_{100}}{n+100m}$. Therefore, $$\sum_{m=0}^{\ell-1}G_{nm}<\frac{M_{100}}{100}\sum_{m=0}^{\ell-1}\frac{1}{m+\frac{n}{100}}$$ But since $\ell<\frac{n}{100}$, the sum on the right can be bounded from above similarly to harmonic series. For $\ell\geq1$, one has \begin{align}\sum_{m=0}^{\ell-1}\frac{1}{m+\frac{n}{100}}\leq\frac{100}{n}+\int_0^{\ell-1}\frac{dx}{x+\frac{n}{100}}\leq\frac{100}{n}+\left[\ln\left(x+\frac{n}{100}\right)\right]_{0}^{\frac{n}{100}-1}=\frac{100}{n}+\ln\left(2-\frac{100}{n}\right)\end{align} For example, for $n>3000$ this gives an upper bound $$\sum_{m=0}^{\ell-1}G_{nm}<\frac{M_{100}}{100}\left(\frac{1}{30}+\ln\left(2-\frac{1}{30}\right)\right)<0.4562$$
It remains to estimate the remainder $R_n$. For $\ell\geq 1$, one has \begin{align} R_n\leq\sum_{k=2n-101}^{2n-2}\frac{1}{k}<\int_{2n-102}^{2n-2}\frac{dx}{x}=\ln\frac{2n-2}{2n-102}, \end{align} so that for $n>3000$ one has $R_{n}<0.017$. Therefore, we have proved that for any $n>3000$ one has $$S_n<0.4562+0.017<0.7\ln2\approx 0.4852$$ Now it suffices to check the values $n=2\ldots 3000$ one by one. $\blacksquare$
P.S. The numbers 100 and 3000 in the above are somewhat arbitrary. They both can be increased to obtain a better bound on $S_n$.
