# Inequality involving partial sum of $\frac{|\sin{kx}|}{k}$

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.

-
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. – nayrb Dec 23 '12 at 13:51
emm, and when $x$ is large, $\sin{nx}\not\approx nx$ should also be taken into account, which confuse me alot.. – Golbez Dec 23 '12 at 16:05
I don't understand why this partial sum will "converge".Does it converge when $x=\pi/2$?Do you mistake "converge" for "diverge"? – y zhao Dec 23 '12 at 16:39
@yzhao sorry, I mistype "converge" , it should be replaced by "diverge", thanks for your correction – Golbez Dec 24 '12 at 1:44
By the way, where does the question come from ? – Amr Feb 22 at 15:09
show 1 more comment

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.

-
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. – Sanchez Jan 21 at 0:48
I plead guilty to handwaving. Still, I think what I've done must be a large part of the solution. – Gerry Myerson Jan 22 at 11:50
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 $[\pi/n, \pi]$. – Sanchez Jan 22 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 $n$.

-
 Then the problem becomes an estimation of $\sum\cos{2kx}$, what should I do next to get the $\log{n}$ lower bound? – Golbez Feb 22 at 15:49 $\sum\cos(2kx)/k$ converges by Dirichlet's test, and $\sum_{k=1}^n1/k=\log n+\gamma+O(1/n^2)$. – Julián Aguirre Feb 25 at 10:39 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! – Golbez Feb 25 at 12:31 For small $n$ you need a more detailed analysis. – Julián Aguirre Feb 25 at 14:59