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The Fejer-Jackson-Gronwall inequality involving the sine function is as follows: $$\sum_{k=1}^n\frac{\sin kx}k>0\quad\text{for all}\ n=1,2,3,\ldots\ \text{and}\ 0<x<\pi.$$

Here I ask the following related question.

QUESTION: Do we have $$\sum_{k=1}^n(-1)^k\left(\frac{\sin kx}k\right)^m<0<\sum_{k=1}^n\left(\frac{\sin kx}k\right)^m$$ for all $m,n=1,2,3,\ldots$ and $0<x<\pi$ ?

Actually I formulated this question in 2013. My numerical computation suggests that the answer should be positive. How to prove this?

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    $\begingroup$ not sure if it helps but you can rewrite $sin(kx)=(e^{ikx}-e^{-ikx})/(2i)$. $\endgroup$ – user35593 Jun 21 '18 at 3:59
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    $\begingroup$ If $m$ is odd both inequalities are equivalent. To see this replace $x$ by $\pi-x$. $\endgroup$ – user35593 Jun 21 '18 at 4:10
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    $\begingroup$ Not sure how helpful this might be but, check out some of the proof techniques in "A New Bound for the Fejer Jackson Sum" -Brown : link.springer.com/article/10.1023/A:1006560422934 They might be adaptable to powers $m$. $\endgroup$ – Alex R. Jun 23 '18 at 21:53
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When $m$ is odd, since $\sin (k(\pi-x)) = (-1)^{k-1} \sin (\pi x)$, we only need to verify that the inequalities $$ \sum_{k=1}^{n} (-1)^{k} \Big( \frac{\sin (kx)}{k} \Big)^m < 0 < \sum_{k=1}^{n} \Big(\frac{\sin (kx)}{k} \Big)^{m} $$ hold for $0< x\le \pi/2$. If $m$ is even then $(\sin (k(\pi -x)))^2 = (\sin (kx))^2$, and once again it is enough to verify these inequalities for $0 < x\le \pi/2$.

We start by proving the inequality $$ F_m(x;n): = \sum_{k=1}^{n} \Big(\frac{\sin (kx)}{k} \Big)^{m} >0. $$ We may assume that $m\ge 3$, the cases for $m=1$ and $2$ already being known. Suppose that $K\ge 2$ is a natural number such that $\pi/(K+1) < x \le \pi/K$. Since $\sin(kx) \ge 0$ for all $1\le k\le K$ and $(\sin(kx))^m \ge -1$ for $k \ge K+1$, we find that $F_m(x;n) \ge 0$ if $n \le K$, and if $n \ge K+1$ then \begin{align*} F_m(x;n) &\ge \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k} \Big)^m - \sum_{k =K+1}^{\infty} \frac{1}{k^m} \\ &\ge \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k}\Big)^m - \int_{K}^{\infty} \frac{dt}{t^m} \\ &= \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k}\Big)^m -\frac{1}{(m-1)K^{m-1}}. \end{align*} Note that if $0\le t\le \pi/2$ then $\sin t \ge 2t/\pi$. Using this for $k\le K/2$ above, we conclude that $$ F_m(x;n) \ge \sum_{k\le K/2} \Big( \frac{2k x}{\pi k} \Big)^m - \frac{1}{(m-1)K^{m-1}} \ge \Big\lfloor \frac K2 \Big\rfloor \Big(\frac{2}{(K+1)}\Big)^m - \frac{1}{(m-1) K^{m-1}}. $$ Since $K\ge 2$ we have $2/(K+1) \ge 4/(3K)$ and $\lfloor \frac K2 \rfloor \ge K/3$, so that $$ F_m(x;n) \ge \frac{1}{K^{m-1}} \Big( \frac{1}{3} \Big(\frac 43\Big)^m - \frac{1}{m-1} \Big) > 0 $$ since $m\ge 3$. This completes the proof of one inequality.

We now turn to the other inequality, which is more involved. We rewrite this inequality as $$ G_m(x;n) := \sum_{k=1}^{n} (-1)^{k-1} \Big( \frac{\sin (kx)}{k} \Big)^m > 0, $$ which we want to establish for all $m\ge 2$ and all $n$ (the case $m=1$ following from the Jackson inequality).
Once again suppose that $K\ge 2$ is such that $\pi/(K+1) < x\le \pi/K$. Since $\frac{\sin y}{y}$ is decreasing and non-negative in $0 \le y\le \pi$, by pairing up two consecutive terms we see that $G_m(x;n)$ is positive if $n\le K$ (if there is an unpaired term at the end it is non-negative), and henceforth we assume that $n \ge K+1$. We now record a useful estimate for tails of our sum, which we shall prove later: for any $0 < x\le \pi/2$ and any integers $A <B$ we have $$ \Big| \sum_{k=A}^{B} (-1)^{k-1} \Big(\frac{\sin (kx)}{k}\Big)^m \Big| \le \frac{m}{2(m-1)} \frac{x}{A^{m-1}} + \frac{1}{A^m}. $$ With this estimate in place, we now finish the proof of the inequality. There are two cases $K \ge m$ and $2 \le K <m$ (which only arises for $m\ge 3$).

Let us start with the first case $K \ge m$ (so that $x$ is small). Consider the function $f_1$ which is the characteristic function of the interval $(-x/(2\pi), x/(2\pi))$.
Its Fourier transform is $$ \hat{f_1}(\xi) = \int_{-x/(2\pi)}^{x/(2\pi)} e^{-2\pi i t \xi} dt = \frac{\sin (x\xi)}{\pi \xi}, $$ interpreted naturally as $x/\pi$ for $\xi =0$. Put $f_m$ to be the convolution of $f_1$ with itself $m$ times. Then $f_m$ is supported in $(-mx/(2\pi), mx/(2\pi))$ which is a subset of $(-1/2,1/2)$ (since $x \le \pi/K \le \pi/m$), and ${\hat f}_m(\xi) = {\hat f_1}(\xi)^m$. The Poisson summation formula now gives $$ 0 = \sum_{k \in {\Bbb Z}} f_m (k+1/2) = \sum_{k \in {\Bbb Z}} (-1)^k {\hat f_m}(k) = \sum_{k\in {\Bbb Z}} (-1)^{k} \Big( \frac{\sin(kx)}{\pi k} \Big)^m, $$ and rearranging we find that $$ \sum_{k=1}^{\infty} (-1)^{k-1} \Big( \frac{\sin (kx)}{k} \Big)^{m} = \frac 12 x^m. $$ Therefore for $n\ge K+1$ (which we may assume) we have $$ G_m(x;n) \ge \frac {x^m}{2} - \Big| \sum_{k=n+1}^{\infty} (-1)^{k-1}\Big(\frac{\sin (kx)}{k}\Big)^m\Big| \ge \frac{x^m}{2} - \Big(\frac{m}{2(m-1)}\frac{x}{(n+1)^{m-1}}+\frac{1}{(n+1)^m}\Big), $$ and since $n+1 \ge K+2 \ge \pi/x$, it follows that $$ G_m(x;n) \ge x^m \Big( \frac 12 - \frac{m}{2(m-1) \pi^{m-1}} - \frac{1}{\pi^m} \Big) > x^m \Big( \frac{1}{2} - \frac{1}{\pi} - \frac{1}{\pi^2} \Big) > 0. $$ This finishes the first case when $K \ge m$.

Now we turn to the second case $2 \le K < m$, which only happens for $m\ge 3$. As noted earlier, we may suppose that $n\ge K+1$, and write $$ G_m(x;n) = x^m \sum_{k=1}^{K} (-1)^{k-1} \Big( \frac{\sin (kx)}{k x}\Big)^{m} + \sum_{k=K+1}^{n} (-1)^{k-1} \Big( \frac{\sin (kx)}{k}\Big)^m. $$ As noted earlier $(\sin y)/y$ is non-negative and decreasing on $[0,\pi)$, and therefore by pairing consecutive terms in the first sum (and if there is a last term left unpaired it is non-negative) and discarding all but the first two terms, we see that the first sum above is $$ \ge (\sin x)^m - \Big(\frac{\sin 2x}{2} \Big)^m = (\sin x)^m (1- (\cos x)^m) \ge (\sin x)^{m} (\sin x)^2 \ge \Big( \frac{2}{\pi} x\Big)^m \Big( \frac{2}{K+1} \Big)^2, $$ where we used that $\sin t \ge 2t/\pi$ for $t\in [0, \pi/2]$. As for the second sum, using our inequality for tails this is bounded in size by $$ \frac{m}{2(m-1)} \frac{x}{(K+1)^{m-1} } + \frac{1}{(K+1)^{m-1}} \le x^m \Big( \frac{3}{4 \pi^{m-1}} + \frac{1}{\pi^m} \Big). $$ Since $K<m$, it follows that $$ G_m(x;n) > x^m \Big( \Big(\frac{2}{\pi}\Big)^m \frac{4}{m^2} - \frac{3}{4 \pi^{m-1}} - \frac{1}{\pi^m} \Big) >0, $$ since $m\ge 3$. This completes the proof of the second case.

Lastly it remains to verify the estimate for tails. This is proved by partial summation. The desired sum is $$ x^m \sum_{k=A}^{B} (-1)^{k-1} \Big(\frac{\sin (kx)}{kx} \Big)^m = x^m \int_{A^-}^{B^+} \Big( \frac{\sin (yx)}{yx} \Big)^{m} d\Big( \sum_{A \le k < y} (-1)^{k-1} \Big), $$ and integrating by parts (and note that $|\sum_{A \le k < y} (-1)^{k-1} | \le 1$ always) we can bound this in magnitude by $$ x^m \Big| \frac{\sin (Bx)}{Bx} \Big|^m + x^m \int_A^B \Big| \frac{d}{dy} \Big( \frac{\sin (yx)}{yx} \Big)^m \Big| dy. $$ Now \begin{align*} \Big| \frac{d}{dy} \Big( \frac{\sin (yx)}{yx}\Big)^m \Big| &= m \Big| \frac{\sin (yx)}{yx} \Big|^{m-1} \Big| \frac{\cos (yx)}{y} - \frac{sin (yx)}{ y^2x} \Big| \\ & \le m \Big| \frac{(\sin yx)^{m-1} (\cos yx)}{y^m x^{m-1}} \Big| + \frac{m}{y^{m+1} x^m} \le \frac{m}{2 y^mx^{m-1} } +\frac{m}{y^{m+1} x^m}, \end{align*} where in the last step we used that $|(\sin t)^{m-1} (\cos t)| \le |\sin t \cos t| \le 1/2$. Using this we conclude that our desired sum is $$ \le \frac{1}{B^m} + \int_A^B \Big( \frac{mx}{2y^m} + \frac{m}{y^{m+1}} \Big) dy < \frac{mx}{2(m-1) A^{m-1}} + \frac{1}{A^m}, $$ which completes the proof.

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