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

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.