Proving $\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+1}{n+1})}k>\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+3}{n+1})}k$

You have already determined the local maxima of a partial sum $S_n(x)$ in $[0,\pi]$ are at $x_{n,m} = \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$.

We first show that the vertical distance between successive peaks is diminishing in that

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right) > \ldots $$

Note that $\frac{2m+3}{n+1}\pi = \frac{2m+1}{n+1}\pi + \frac{2\pi}{n+1}$ and

$$S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x+ \frac{n\pi}{n+1} \right) \cos \left(\frac{n+1}{2}x+\pi\right)}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ =\frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x- \frac{\pi}{n+1} \right) \cos \frac{n+1}{2}x}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ = \frac{\cos \frac{n+1}{2}x}{\sin \frac{x}{2}\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)}\left[\sin \frac{x}{2}\sin \left(\frac{n}{2}x+ \frac{\pi}{n+1}\right) -\sin \frac{x}{2}\sin \left(\frac{n}{2}x- \frac{\pi}{n+1}\right) \right]$$

Applying angle addition identities and simplifying we eventually get

$$\tag{1}S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{\pi}{n+1}\sin (n+1)x}{\cos \frac{\pi}{n+1} - \cos \left(\frac{\pi}{n+1} +x \right)}$$

Now consider the points

$$\frac{2m+1}{n+1}\pi \leqslant \frac{2m+2}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+2}{n+1} \leqslant \frac{2m+2}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+3}{n+1}\pi, \\ \frac{2m+3}{n+1}\pi \leqslant \frac{2m+4}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+4}{n+1} \leqslant \frac{2m+4}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+5}{n+1}\pi$$

Using (1) with $x_- = \frac{2m+2}{n+1}\pi-\frac{y}{n+1}$, $x'_-= x_- + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi-\frac{y}{n+1}$, $x_+ = \frac{2m+2}{n+1}\pi+\frac{y}{n+1}$, and$x'_+=x_+ + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi+\frac{y}{n+1}$we get

$$G'(y) = \frac{d}{dy} \left[S_n \left(x_-) - S_n \left(x'_-\right) \right] - S_n(x_+) - S_n(x'_+)\right] \\ = \frac{\sin \frac{\pi}{n+1} \sin y}{n+1}\left[\frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi-y}{n+1}\right)} - \frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi+y}{n+1}\right)} \right] $$

Notice that $G(0) = 0$ and $G'(y) > 0$ for $0 < y < \pi$, so $G(\pi) > 0$ which implies

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right)$$

When $m = \lfloor \frac{n-1}{2}\rfloor -1 $ we have $\frac{2m+3}{n+1} < \pi < \frac{2m+5}{n+1} = \pi + \delta$. Hence,

$$S_n\left(\frac{2m+5}{n+1}\pi \right) = S_n(\pi + \delta) = S_n(\pi + \delta - 2\pi) = S_n(\delta - \pi)= - S_n(\pi - \delta) < 0,$$

Since $\frac{2m+3}{n+1}$ is the last relative extremum point before $\pi$, $S_n(\pi) = 0$, and $\frac{2m+3}{n+1} < \pi - \delta < \pi$, it follows that

$$S_n\left(\frac{2m+3}{n+1}\pi \right) > 0 > S_n\left(\frac{2m+5}{n+1}\pi \right)$$

Thus, for all $m = 0,1,\ldots, \lfloor\frac{n-1}{2}\rfloor -1$ we have

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right) > 0$$

Remark: One year ago, I analyzed the monotonicity of the function values at all local minimizers, when I tried to answer Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson). Here, I give the analysis of the monotonicity of the function values at all local maximizers.

Problem: Given positive integer $n\ge 3$, all the local maximizer of $f(x) = \sum_{k=1}^n \frac{\sin kx}{k}$ on $(0, \pi)$ are given by $x_m = \frac{2m+1}{n+1}\pi, m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor$. Prove that $f(x_m) > f(x_{m+1})$ for all $m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor - 1$.

Proof: Since $f(x) = \int_0^x f'(t) \mathrm{d} t + f(0)$, we have \begin{align} f(x_{m+1}) - f(x_m) &= \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+3}{n+1}\pi} f'(t) \mathrm{d} t = \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+2}{n+1}\pi} \Big( f'(t) + f'(t + \tfrac{\pi}{n+1})\Big) \mathrm{d} t. \tag{1} \end{align} Since $f'(x) = \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2} \cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}$, we have, for all $t\in [\frac{2m+1}{n+1}\pi, \frac{2m+2}{n+1}\pi]$, \begin{align*} &f'(t) + f'(t + \tfrac{\pi}{n+1})\\ =\ & \frac{\sin \frac{nt}{2} \cos \frac{(n+1)t}{2}}{\sin \frac{t}{2}} + \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi) \cos (\frac{(n+1)t}{2} + \frac{\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\ =\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\ & - \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi)\sin (\frac{(n+1)t}{2} + \frac{\pi}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{2}\\ =\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\ & - \frac{\cos (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2})\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\ \le\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\ & - \frac{\cos (\frac{nt}{2} - m\pi )\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{3}\\ \le\ & - \frac{\cos \frac{\pi - t}{2}}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\ \le\ & 0 \end{align*} where in (2) we have used $\sin(\alpha - m\pi) \sin (\beta - \frac{(2m+1)\pi}{2}) = -\sin \alpha \cos \beta$, and in (3) we have used \begin{align} &0 < \frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2} < \frac{nt}{2} - m\pi < \pi, \\ &0 \le \frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2} \le \frac{\pi}{2},\\ &0 < \frac{t}{2} < \frac{t}{2} + \frac{\pi}{2n+2} < \frac{\pi}{2}. \end{align} Also, note that $f'(\frac{2m+1}{n+1}\pi) + f'(\frac{2m+1}{n+1}\pi + \tfrac{\pi}{n+1}) = -1 < 0$. From (1), we have $f(x_{m+1}) < f(x_m)$. We are done.