# Let $n$ be any positive integer and let $x \in (0, \pi)$. Prove that $\sin x+ \frac{\sin 3x}{3} + \cdots +\frac{\sin(2n-1)x}{2n-1}$ is positive. [duplicate]

I came across the following problem in a high school calculus exam paper. I do have the solution, but it took me quite a while to work it out and I think it's very clumsy. I'm curious to know if there is a simpler solution.

Problem

Let $n$ be any positive integer and let $x \in (0, \pi)$. Prove that

$$\sin{x} + \frac{\sin{3x}}{3} + \frac{\sin{5x}}{5} + \cdots +\frac{\sin{(2n-1)x}}{2n-1}$$ is positive.

• Are you familiar with Lerch Transcendant $\phi$ functions? Because you can get this sum in closed form using them. (LOL -- if you think YOUR solution is clumsy, ...) Jun 14, 2017 at 23:06
• @MarkFischler Thinking that this is a highschool exam problem, I don't think it would be Lerch Transcendent functions. There's probably a trigonometric trick.
– user335907
Jun 14, 2017 at 23:11
• @james.nixon To me, it looks like that we can differentiate the sum w.r.t x, and get the sum of $\cos nx$, which then can be calculated by a geometric progress involved with $e^{i\theta}$. Jun 14, 2017 at 23:20
• Let $$f(x)=\sum_{k=1}^n\frac{\sin\big((2k-1)x\big)}{2k-1}$$ $f$ is continuous and periodic, so it has a maximum and a minimum. Moreover it is differentiable so its derivative must vanish at extrema. We have $$f'(x)=\sum_{k=1}^n\cos\big((2k-1)x\big)$$ Do you think you can look for solutions to $f'=0$? Jun 14, 2017 at 23:23
• @Fimpellizieri My first line of attack was the same, but it turned out to be very difficult to prove that $f(x) > 0$ at the zeros of $f'(x)$. Jun 15, 2017 at 0:05

Let the required sum be $S(x)$,

\begin{align*} S(x) &= \sum_{k=1}^{n} \frac{\sin(2k-1)x}{2k-1}\\ S'(x) &= \sum_{k=1}^{n} \cos(2k-1)x\\ &= \Re \left[\sum_{k=1}^ne^{i(2k-1)x}\right]\\ &= \Re \left[e^{ix}\cdot\frac{1-e^{i2nx}}{1-e^{i2x}}\right]\\ &= \Re \left[e^{inx} \cdot \frac{e^{-inx} -e^{inx}}{e^{-ix} - e^{ix}}\right]\\ &= \Re\left[(\cos nx +i\sin nx)\cdot\frac{\sin nx}{\sin x}\right]\\ &= \frac{\sin 2nx}{2\sin x} \end{align*}

Since $S(x) = S(\pi-x)$, it is sufficient to consider only $x \in (0,2\pi/4]$.

In the range $x\in(0,2\pi/4]$, $S'(x)$ has the same sign as $\sin 2nx$. Local minima of $S(x)$ (including the boundary case $x=0$) are at

\begin{align*} 2nx &= 2\pi k\\ x &= \frac{m}{2n}\cdot 2\pi, &&0\le m\le \frac n2,\quad m\in \mathbb N \end{align*}

The difference between two adjacent local minima is

\begin{align*} D &= S\left(\frac{m}{2n}\cdot 2\pi\right) - S\left(\frac{m-1}{2n}\cdot 2\pi\right)\\ &= \int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}S'(x)\ dx\\ &= \int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\frac{\sin 2nx}{2\sin x}dx\\ &= \int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m-1/2}{2n}\cdot 2\pi}\frac{\sin 2nx}{2\sin x}dx + \int_{\frac{m-1/2}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\frac{\sin 2nx}{2\sin x}dx\\ \end{align*}

The numerator $\sin 2nx$ is positive in the first integral, and negative in the second integral. Using the fact that $\sin x$ is increasing in $x\in(0,2\pi/4]$,

\begin{align*} D &> \frac{\int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m-1/2}{2n}\cdot 2\pi}\sin 2nx\ dx}{2\sin \frac{m-1/2}{2n}\cdot 2\pi} + \frac{\int_{\frac{m-1/2}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\sin 2nx\ dx}{2\sin\frac{m-1/2}{2n}\cdot 2\pi}\\ &= \frac{\int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m-1/2}{2n}\cdot 2\pi}\sin 2nx\ dx + \int_{\frac{m-1/2}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\sin 2nx\ dx}{2\sin\frac{m-1/2}{2n}\cdot 2\pi}\\ &= 0 \end{align*}

In other words, the values of $S$ at $x = 0, \frac{2\pi}{2n}, \frac{2\cdot 2\pi}{2n}, \cdots, \frac{\lfloor n/2\rfloor\cdot 2\pi}2$ are strictly increasing compared with the previous one, so they are all greater than $S(0) = 0$. Also, since these $x$'s are local minima, every other $x$ between them will also be greater than $S(0) = 0$.

• I came up with pretty much the same solution, with a few minor changes in the precise details. I still think that this is not a solution expected of a high school student. Jun 15, 2017 at 1:44