Positivness of the sum of $\frac{\sin(2k-1)x}{2k-1}$.

Question

For $n\in \mathbb{N}$, $x\in (0,\pi)$. Prove that : $$f_n(x)=\sum_{k=1}^n \frac{\sin [(2k-1)x]}{2k-1} \geq 0.$$

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I've tried to do it by differentiation : I Calculate $f_n'(x)$ (sum of $\cos(2k-1)x$) but It has many roots. So, I couldn't go further.

Answer 1

Claim. $$ S_N(x)=\sum_{k=1}^N \frac{\sin (kx)}{k} > 0 $$

I just give the main ideas, I hope you will succeed with it.

1. Arguing by contradiction, consider a point $x_0 \in (0,\pi)$ where the sum $S_N(x)=\sum_{k=1}^N \frac{\sin (kx)}{k}$ reaches a negative minimum
2. Using the necessary condition for the existence of a minimum ($S_n'(x_0)=0$) show that $\sin (Nx_0)\geq 0$ and consequently the sum $g_{N-1}$ also takes negative values
3. In pursuing this argument we come to a contradiction to the fact that $S_1(x)=\sin(x)>0$ on $(0,\pi)$

Related (if you are interested) :

• Gibbs phenomenon
• Sharp inequalities for trigonometric sums

Answer 2

Suppose that $x/2\pi$ is a rational number $p/q$ where $p$ and $q$ are coprime. We can write $$f_n(x)=\Im\left(\sum\limits_{k\leq 2n-1\text{ and is odd}} \frac{\left(e^{2p\pi i/q}\right)^k}{k} \right)$$.

Let be $$g_n(z)=\sum_{k=0}^{n-1} z^{2k}=\frac{1-z^{2n-1}}{1-z^2}$$ for $z\neq\pm1$

Note that $f_n(x)$ is the imaginary part of the primitive of $g_n$ evaluated at $z=e^{i\pi x}$. I'm not good at calculus and integrating $g_n$ requires hypergeometric functions. I'm sorry but I can't go further.

My idea is to demostrate the statement for a dense subset (namely, $\{x\in(0,\pi):x/2\pi\in\mathbb Q\}$). $f_n$ is clearly continuous, so we can extend for the whole interval.