# Inequality $\sum_{1\le k\le n}\frac{\sin kx}{k}\ge 0$

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.

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You can write $\sin(xk) = \mathcal{Im}(e^{ixk)}$, but the result will involve transcendent functions – Alex Apr 29 '13 at 15:31
@Alex That seems not helpful for the problem. – Ma Ming Apr 29 '13 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.

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Great. I already proved it. The last inequality for specific $x$ can be done by a simple induction on $n$, the last term $\sin nx \ge 0$! – Ma Ming Apr 29 '13 at 21:51
You mean, using what I did? Then just edit my answer. I'll make it CW. – 1015 Apr 29 '13 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. – 1015 Apr 30 '13 at 12:30

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

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