# Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson)

Show the following inequality for any $x\in [0, \pi]$ and $n\in \mathbb{N}^*$, $$\sum_{1\le k\le n}\frac{\sin kx}{k}\ge 0.$$

I have this question a very long time ago from a book or magazine but I cannot solve it by myself and did not know how to solve it until today.

My try: for $n=1, 2, 3$, one can check this by hand.

• You can write $\sin(xk) = \mathcal{Im}(e^{ixk)}$, but the result will involve transcendent functions
– Alex
Apr 29, 2013 at 15:31
• @Alex That seems not helpful for the problem. Apr 29, 2013 at 20:56

In short: let $f_n(x)$ denote the function on the lhs of the inequality. Of course, $f_1(x)=\sin x\geq 0$ on $[0,\pi]$. We will prove that $f_n(x)\geq 0$ on $[0,\pi]$ by induction on $n$. It is not too hard to determine the local minima of $f_n$ on $[0,\pi]$ by investigating its derivative. Then Ma Ming observed that $f_n$ coincides with $f_{n-1}$ on these local minima. And the induction step follows easily. Of course, $f_n(0)=f_n(\pi)=0$. We will actually prove that

$$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k}>0\qquad\forall x\in(0,\pi).$$

Remark: it is worth noting that the $f_n$'s are the partial sums of the Fourier series of the same sawtooth function. Just look at the case $n=6$, for instance, to see how they tend to approximate it nicely. See here to get an idea how to estimate the error in such approximations. As pointed out by math110, there are many proofs of this so-called Fejer-Jackson inequality. It can even be shown that the $f_n$'s are bounded below by a certain nonnegative polynomial on $[0,\pi]$. The proof below is at the calculus I level. I'm not sure it can be made more elementary.

Proof: first, $f_1(x)=\sin x$ is positive on $(0,\pi)$. Assume this holds for $f_{n-1}$ for some $n\geq 2$. Then observe that $f_n$ is differenbtiable on $\mathbb{R}$ with $$f_n'(x)=\sum_{k=1}^n\cos kx=\mbox{Re} \sum_{k=1}^n (e^{ix})^k.$$ For $x\in 2\pi \mathbb{Z}$, we have $f_n'(x)=n$. So the zeros of $f_n'$ are the zeros of $$\mbox{Re}\;e^{ix}\frac{e^{inx}-1}{e^{ix}-1}=\mbox{Re}\;e^{i(n+1)x/2}\frac{\sin (nx/2)}{\sin(x/2)}=\frac{\cos((n+1)x/2)\sin (nx/2)}{\sin(x/2)}.$$ This yields $$\frac{nx}{2}\in \pi\mathbb{Z}\quad\mbox{or}\quad \frac{(n+1)x}{2}\in \frac{\pi}{2}+\pi\mathbb{Z}$$ i.e. $$x\in \frac{2\pi}{n}\mathbb{Z}\quad\mbox{or}\quad x\in \frac{\pi}{n+1}+\frac{2\pi}{n+1}\mathbb{Z}.$$ Between $0$ and $\pi$, these are ordered as follows: $$0<\frac{\pi}{n+1}<\frac{2\pi}{n}<\frac{3\pi}{n+1}<\frac{4\pi}{n}<\ldots < \frac{2\lfloor n/2\rfloor \pi}{n}\leq \pi.$$ The sign of $f_n'$ changes at each of these zeros, starting from a positive sign on $(0,\pi/(n+1))$. It follows that $f_n$ is positive on the latter, positive on the last interval (if nontrivial, i.e. in the odd case), with local minima at $$\frac{2j\pi}{n}\qquad\mbox{for}\qquad j=1,\ldots,\lfloor n/2\rfloor.$$

But now here is Ma Ming's key observation: for these values, we have $$f_n\left(\frac{2j\pi}{n}\right)=f_{n-1}\left(\frac{2j\pi}{n}\right)+\sin\left(n\cdot\frac{2j\pi}{n}\right)=f_{n-1}\left(\frac{2j\pi}{n}\right)>0$$ by induction step. It follows that $f_n(x)>0$ on $(0,\pi)$. QED.

• Great. I already proved it. The last inequality for specific $x$ can be done by a simple induction on $n$, the last term $\sin nx \ge 0$! Apr 29, 2013 at 21:51
• You mean, using what I did? Then just edit my answer. I'll make it CW. Apr 29, 2013 at 21:53
• @MaMing Good observation. There is a problem with your inequality for the local maxima, but we don't need them. We only need to check that the local minima ar $\geq 0$. I'll edit. Apr 30, 2013 at 12:30

This is Fejer-Jackson inequality: This problem has some nice solutions, you can see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114058

Some years ago, I saw a nice solution in a Chinese forum. I do not know the original source. I translated it and rewrote it compactly. I give it here.

Problem: Let $$0 < x < \pi$$ and $$n$$ be a positive integer. Prove that $$\sin x+\dfrac{\sin 2x}{2}+\dfrac{\sin 3x}{3}+\ldots+ \dfrac{\sin nx}{n}>0.$$

Proof: Assume, for the sake of contradiction, that $$m$$ is the smallest positive integer such that there exists $$y\in (0, \pi)$$ satisfying $$\sum_{k=1}^m \dfrac{\sin ky}{k} \le 0$$. Clearly $$m \ge 2$$.

Let $$f(x) := \sum_{k=1}^m \dfrac{\sin kx}{k}$$. Since $$f(y) \le 0 = f(0) = f(\pi)$$, there exists $$z\in (0, \pi)$$ which is a global minimizer of $$f(x)$$ on $$[0,\pi]$$ (see Remark 1 at the end). Thus, we have $$f'(z) = 0$$ and $$f(z) \le 0$$.

We claim that $$\sin m z < 0$$. Indeed, if $$\sin mz \ge 0$$, using $$0 \ge f(z) = \sum_{k=1}^{m-1} \dfrac{\sin k z}{k} + \frac{\sin m z}{m}$$, we have $$\sum_{k=1}^{m-1} \dfrac{\sin k z}{k} \le 0$$. This contradicts the smallestness of $$m$$.

It follows from $$\sin m z < 0$$ that $$\sin \frac{mz}{2} \ne 0$$ and $$\cos\frac{(m+1)z}{2} \ne 0$$ (see Remark 2 at the end) which contradicts $$0 = f'(z) = \sum_{k=1}^m \cos kz = \frac{\cos\frac{(m+1)z}{2}\sin \frac{mz}{2}}{\sin \frac{z}{2}}.$$ This completes the proof.

Remark 1: Let $$f^\ast$$ be the minimum of $$f(x)$$ on $$[0, \pi]$$. If $$f^\ast < 0$$, since $$f(0) = f(\pi) = 0$$, the minimum of $$f(x)$$ on $$[0, \pi]$$ occurs on $$(0,\pi)$$. If $$f^\ast = 0$$, since $$f(y) \le 0$$, $$y$$ is a global minimizer.

Remark 2:
$$\sin \frac{mz}{2} = 0 \quad \Longrightarrow\quad \sin mz = 0,$$ and \begin{align} \cos\frac{(m+1)z}{2} = 0\quad & \Longrightarrow \quad \exists N\in \mathbb{Z},\ (m+1)z = (2N+1)\pi \\ &\Longrightarrow \quad \sin mz = \sin ((2N+1)\pi - z) = \sin z \ge 0 . \end{align} Contradiction.

• Nice solution! Let me ask you a small question please: You wrote that $f(z)\leq 0$. I cannot get it.
– ZFR
Apr 28, 2021 at 4:53
• @ZFR Thanks. Since $f(y) \le 0$ and $f(z)$ is the global minimum of $f(x)$ on $[0, \pi]$, we have $f(z) \le f(y) \le 0$. Apr 28, 2021 at 5:37
• One moment is a bit confusing to me: Since $f(x)$ is continuous on $[0,\pi]$ then it has global minimum $z\in [0,\pi]$ by Weierstrass theorem. If $z\in (0,\pi)$ then we can conclude that $f'(z)=0$ by Fermat's theorem. How do you know that $z$ is not one of the endpoints?
– ZFR
Apr 28, 2021 at 13:27
• I still cannot get your point. You claim that global minimum of $f(x)$ lies in $(0,\pi)$, right? I don't think that your reasoning is clear enough.
– ZFR
Apr 28, 2021 at 14:38
• I'd prefer this reasoning: Let's prove that the global minimum(let's denote it by $z$) of $f(x)$ lies in $(0,\pi)$. We have two possible cases: 1) There is $w\in [0,\pi]$ such that $f(w)<0$. Since $f(0)=f(\pi)=0$ then $w\in (0,\pi)$. Then $z$ which is global minimum of $f$ lies in $(0,\pi)$ and $f(z)<0$. 2) For each $w\in [0,\pi]$ we have $f(w)\geq 0$. Then $f(y)=0$ and $z=y$, i.e. $y$ is global minimum.
– ZFR
Apr 28, 2021 at 15:01