Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$

Solution 1:

Given $n\ge 1$ and $x\in \Bbb R$, denote $$f_n(x)=\sum_{k=1}^n\frac{|\sin kx|}{k} \quad\text{and}\quad g_n(x)=f_n(x)-|\sin nx|.$$

Lemma: For every $n\ge 1$,

(i) $f_n$ is increasing on $[0,\frac{\pi}{n+1}]$; (ii) $g_n$ is increasing on $[0,\frac{\pi}n]$; (iii) $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.

Proof of Lemma: Note that when $x\in [0,\frac{\pi}n]$, $$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k},\quad\text{and}\quad g_n(x)=f_n(x)-\sin nx,$$ so $$f_n'(x)=\sum_{k=1}^n \cos kx \quad\text{and}\quad g_n'(x)=f_n'(x)-n\cos nx.$$ (i) Given $x\in[0,\frac{\pi}{n+1}]$, noting that $\cos\frac{kx}{2}\ge 0$ for $k=0,\pm1,\dots, \pm (n+1)$, we have $$f_n'(x)=\sum_{k=1}^n \frac{\cos kx +\cos(n+1-k)x}{2}=\cos \frac{(n+1)x}{2} \cdot\sum_{k=1}^n \cos\frac{(n+1-2k)x}{2}\ge 0.$$

(ii) Since the cosine function is decreasing on $[0,\pi]$, when $x\in [0,\frac{\pi}{n}]$, $\cos k x\ge \cos nx$ for $k=1,\dots,n$, so $g'(x)\ge 0$.

(iii) When $n=1$, the statement is clearly true; when $n=2$, since $f_2$ is concave on $[\frac{\pi}{4},\frac{\pi}{2}]$, $f_2(\frac{\pi}{4})>1$ and $f_2(\frac{\pi}{2})=1$, the statement is also true. By induction, we may assume that $f_{n-1}\ge 1$ on $[\frac{\pi}{2(n-1)},\frac{\pi}{2}]$ for some $n \ge 3$, and the conclusion $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$ follows from the facts below. Firstly, $f_n\ge f_{n-1}$; secondly, $f_n$ is increasing on $[0,\frac{\pi}{n+1}]\supset [\frac{\pi}{2n},\frac{\pi}{2(n-1)}]$; thirdly, $$\sin \frac{\pi t}{2}\ge t,\ \forall t\in[0,1]\Longrightarrow f_n(\frac{\pi}{2n})=\sum_{k=1}^n\frac{\sin \frac{k\pi}{2n}}{k}\ge 1.\qquad \square$$

Now we can prove that $g_n\ge 0$ by using the lemma. Since $g_n(\pi\pm x)=g_n(x)$, we may focus on $x\in[0,\frac{\pi}{2}]$. Since $g_n(0)=0$, by (ii), we know that $g_n(x)\ge 0$ on $[0,\frac{\pi}{n}]$. Since $g_n\ge f_n -1$, by (iii) we know that $g_n\ge 0$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.