Proving $\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+1}{n+1})}k>\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+3}{n+1})}k$ A few days ago I asked a question about an interesting property of the partial sums of the series $\sum\sin(nx)/n$.
Here's the link:
Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$
The proof given there left some details for me, in particular I have to prove that
$$
\sum_{k = 1}^{n}{1 \over k}
\sin\left(k\pi\frac{2m+1}{n+1}\right) >
\sum_{k=1}^{n}{1 \over k}\sin\left(k\pi\frac{2m+3}{n+1}\right)
$$
for every natural $n$ and $m$ such that $m = 0, 1, \ldots, \left\lfloor\left(n-1\right)/2\right\rfloor$. I don't think this is relatively simple; anyway I tried a few expansions with sum-to-product and viceversa formulas, but everything I tried leads to $$\sum_{k=1}^{n}\frac{\cos(2(m+1)x_k)\sin(x_k)}{k} < 0$$
where $x_k = k\pi/(n+1)$. From here I don't see any further semplification. Also I noted that $\sin(x_k)$ is always positive since $k$ is at most $n$, but we can't say much about $\cos$.
Then I tried to pair terms since for $k' = n+1-k$ we have $\sin(x_k) = \sin(x_k')$, which gives $$\sum_{k=1}^{n/2}\left[\frac{\cos(2(m+1)x_k)}{k}+\frac{\cos(2(m+1)x_{k'})}{n+1-k}\right]\sin(x_k) < 0$$ for $n$ even. Then I hoped that the term between square brackets were always negative, but it's not always the case... How can I do?
 A: 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
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right) > 0$$
A: 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.
