How to prove that $\forall x \in \mathbb{R}$, $n \in \mathbb{N}$, we have \begin{align} \sum_{k=1}^{n}\frac{|\sin{kx}|}{k}\ge |\sin{nx}| ? \end{align}

I know that this partial sum will diverge for $x\not = m\pi$, but I don't know how to prove this inequality. I have tried Abel summation, but it doesn't work because I can't give a lower bound for $\sum |\sin{kx}|$. Thanks for your attention.

  • $\begingroup$ It looks true at any rate. $|\sin kx|/k \approx |kx|/k = |x|$, which is summed over $n$ times giving $|nx|$. On the other hand, $|\sin nx| \approx |nx|$. Maybe try taylor series and be careful with the error terms? Another mild thought is to try complex numbers. $\endgroup$ – abnry Dec 23 '12 at 13:51
  • $\begingroup$ emm, and when $x$ is large, $\sin{nx}\not\approx nx$ should also be taken into account, which confuse me alot.. $\endgroup$ – Golbez Dec 23 '12 at 16:05
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    $\begingroup$ I don't understand why this partial sum will "converge".Does it converge when $x=\pi/2$?Do you mistake "converge" for "diverge"? $\endgroup$ – zy_ Dec 23 '12 at 16:39
  • $\begingroup$ @yzhao sorry, I mistype "converge" , it should be replaced by "diverge", thanks for your correction $\endgroup$ – Golbez Dec 24 '12 at 1:44
  • $\begingroup$ By the way, where does the question come from ? $\endgroup$ – Amr Feb 22 '13 at 15:09

Given $n\ge 1$ and $x\in \Bbb R$, denote $$f_n(x)=\sum_{k=1}^n\frac{|\sin kx|}{k} \quad\text{and}\quad g_n(x)=f_n(x)-|\sin nx|.$$

Lemma: For every $n\ge 1$,

(i) $f_n$ is increasing on $[0,\frac{\pi}{n+1}]$; (ii) $g_n$ is increasing on $[0,\frac{\pi}n]$; (iii) $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.

Proof of Lemma: Note that when $x\in [0,\frac{\pi}n]$, $$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k},\quad\text{and}\quad g_n(x)=f_n(x)-\sin nx,$$ so $$f_n'(x)=\sum_{k=1}^n \cos kx \quad\text{and}\quad g_n'(x)=f_n'(x)-n\cos nx.$$ (i) Given $x\in[0,\frac{\pi}{n+1}]$, noting that $\cos\frac{kx}{2}\ge 0$ for $k=0,\pm1,\dots, \pm (n+1)$, we have $$f_n'(x)=\sum_{k=1}^n \frac{\cos kx +\cos(n+1-k)x}{2}=\cos \frac{(n+1)x}{2} \cdot\sum_{k=1}^n \cos\frac{(n+1-2k)x}{2}\ge 0.$$

(ii) Since the cosine function is decreasing on $[0,\pi]$, when $x\in [0,\frac{\pi}{n}]$, $\cos k x\ge \cos nx$ for $k=1,\dots,n$, so $g'(x)\ge 0$.

(iii) When $n=1$, the statement is clearly true; when $n=2$, since $f_2$ is concave on $[\frac{\pi}{4},\frac{\pi}{2}]$, $f_2(\frac{\pi}{4})>1$ and $f_2(\frac{\pi}{2})=1$, the statement is also true. By induction, we may assume that $f_{n-1}\ge 1$ on $[\frac{\pi}{2(n-1)},\frac{\pi}{2}]$ for some $n \ge 3$, and the conclusion $f_n\ge 1$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$ follows from the facts below. Firstly, $f_n\ge f_{n-1}$; secondly, $f_n$ is increasing on $[0,\frac{\pi}{n+1}]\supset [\frac{\pi}{2n},\frac{\pi}{2(n-1)}]$; thirdly, $$\sin \frac{\pi t}{2}\ge t,\ \forall t\in[0,1]\Longrightarrow f_n(\frac{\pi}{2n})=\sum_{k=1}^n\frac{\sin \frac{k\pi}{2n}}{k}\ge 1.\qquad \square$$

Now we can prove that $g_n\ge 0$ by using the lemma. Since $g_n(\pi\pm x)=g_n(x)$, we may focus on $x\in[0,\frac{\pi}{2}]$. Since $g_n(0)=0$, by (ii), we know that $g_n(x)\ge 0$ on $[0,\frac{\pi}{n}]$. Since $g_n\ge f_n -1$, by (iii) we know that $g_n\ge 0$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$.


Let's assume $nx\le\pi$. A glance at the graph of $\sin x$ shows that for $0\le k\le n$ the line through the origin and the point $(kx,\sin kx)$ passes through or above the point $(nx,\sin nx)$, so $\sin kx\ge(k/n)\sin nx$. So $$\sum_{k=1}^n{\sin kx\over k}\ge\sum_{k=1}^n{\sin nx\over n}=\sin nx$$

I'm confident that the general case is just a matter of filling in a few details.

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    $\begingroup$ Can you say more about "filling in a few details"? It's very unclear to me how you may deal with the problem of different phase. $\endgroup$ – user27126 Jan 21 '13 at 0:48
  • $\begingroup$ I plead guilty to handwaving. Still, I think what I've done must be a large part of the solution. $\endgroup$ – Gerry Myerson Jan 22 '13 at 11:50
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    $\begingroup$ I'm not quite certain, since I guess that the phase problem is the main issue here. If you plot the graph of LHS - RHS, you can see that on $[0, \pi/n]$ it's quite positive, while there are troughs that are quite a bit smaller as we wade through $[\pi/n, \pi]$. $\endgroup$ – user27126 Jan 22 '13 at 19:48

You can get a lower bound as follows: $$ |\sin(k\,x)|\ge|\sin(k\,x)|^2=\frac{1-\cos(2\,k\,x)}{2}. $$ From this it is easy to get $$ \sum_{k=1}^n\frac{|\sin(k\,x)|}{k}\ge\sum_{k=1}^n\frac{1}{2\,k}-\sum_{k=1}^n\frac{\cos(2\,k\,x)}{2\,k}\ge\frac12\log n-C(x),\quad x\ne0,\pi, $$ for some $C(x)>0$. This will prove the inequality for large values of $n$.

  • $\begingroup$ Then the problem becomes an estimation of $\sum\cos{2kx}$, what should I do next to get the $\log{n}$ lower bound? $\endgroup$ – Golbez Feb 22 '13 at 15:49
  • $\begingroup$ $\sum\cos(2kx)/k$ converges by Dirichlet's test, and $\sum_{k=1}^n1/k=\log n+\gamma+O(1/n^2)$. $\endgroup$ – Julián Aguirre Feb 25 '13 at 10:39
  • $\begingroup$ I see. But this is only a rough estimation, a further discussion on $x$ should also be concerned, as $x\to 0,\pi$, the lefer side is approaching zero. So this inequality onlly holds for large $n$ and $x$ with some restrictions. What I need is only $|\sin{nx}|$. Anyway, thank you for your help! $\endgroup$ – Golbez Feb 25 '13 at 12:31
  • $\begingroup$ For small $n$ you need a more detailed analysis. $\endgroup$ – Julián Aguirre Feb 25 '13 at 14:59

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