# Prove $\sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} >0$

Prove that for $0<x< \pi$, $$\quad S_n(x) = \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} >0 \quad \forall n = 1,2,...$$

Having trouble with this problem. This is an olympiad-style question, so an answer that doesn't use calculus or analysis would be preferred. A possible approach is induction, but for this we need to find a function in terms of $n$ and $x$ so that we can actually use the inductive step. If anyone has any ideas they would be appreciated.

If you really want to go down the calculus route (at this point I don't mind), then $S_n' (x) = \cos x + \cos 3x + ... +\cos((2n-1)x)$ , which you can find a closed form for, but I don't know how useful that is.

• My first idea would be to try addition formula $\sin(k+2x)$ Jul 30, 2016 at 21:38
• @Carlos I don't understand your comment, at $x = \frac { \pi } {2}$ sequence is equal to $1 - \frac {1}{3} + \frac{1}{5} + ... + \frac { (-1)^{2n} }{2n-1}$, which approaches $\frac { \pi } {4}$ Jul 30, 2016 at 21:51
• One potentially useful avenue is to graph the first few terms. Looks suspiciously like a square wave. A quick check reveals where the 4/pi term comes from, the Fourier transform. No calculus, but at least you have a hint about what it might be related to. Jul 30, 2016 at 22:10
• Confirming Carlos suspicious: If $n\to\infty$, this is a fourier series approximating a square signal that is positive from $0$ to $\pi$ and negative from $\pi$ to $2\pi$, with amplitude of $\pi/4$. Not sure if this helps. Jul 30, 2016 at 22:20
• Is calculus not acceptable for olympiad? Many high school students learn calculus as official curriculum.. Jul 30, 2016 at 22:21

If we want to avoid integration, we can obtain the result via summation by parts. Let

$$F_m(x) := \sum_{k = 0}^{m-1} \sin \bigl((2k+1)x\bigr).\tag{1}$$

Using the addition theorem for the cosine, we obtain

$$\sin(rx) = \frac{2\sin x \sin(rx)}{2\sin x} = \frac{\cos\bigl((r-1)x\bigr) - \cos \bigl((r+1)x\bigr)}{2\sin x}$$

and hence by telescoping

$$F_m(x) = \frac{1 - \cos (2mx)}{2\sin x}\tag{2}$$

when $x$ is not an integer multiple of $\pi$. Since $1 - \cos (2mx) \geqslant 0$ for all $x \in \mathbb{R}$ and $m \in \mathbb{N}$, we have $F_m(x) \geqslant 0$ for $0 < x < \pi$ and $m \in \mathbb{N}$. Thus

\begin{align} S_n(x) &= \sum_{k = 0}^{n-1} \frac{\sin \bigl((2k+1)x\bigr)}{2k+1}\\ &= \sum_{k = 0}^{n-1} \frac{F_{k+1}(x) - F_k(x)}{2k+1}\\ &= \sum_{k = 0}^{n-1} \frac{F_{k+1}(x)}{2k+1} - \sum_{k = 0}^{n-1} \frac{F_k(x)}{2k+1}\\ &= \sum_{m = 1}^n \frac{F_m(x)}{2m-1} - \sum_{k = 1}^{n-1} \frac{F_k(x)}{2k+1}\tag{$F_0 \equiv 0$}\\ &= \frac{F_n(x)}{2n-1} + \sum_{k = 1}^{n-1} \biggl(\frac{1}{2k-1} - \frac{1}{2k+1}\biggr) F_k(x)\\ &\geqslant 0, \end{align}

since all terms on the right hand side are non-negative. The strict inequality $S_n(x) > 0$ for $x\in (0,\pi)$ and $n \in \mathbb{N}\setminus \{0\}$ follows because the term involving $F_1(x) = \sin x$ is strictly positive on $(0,\pi)$.

If we use integration, we first observe that since $\sin \bigl(m(\pi - x)\bigr) = (-1)^{m+1}\sin x$, we have the symmetry $S_n(\pi-x) = S_n(x)$, so it suffices to consider $0 < x \leqslant \pi/2$. Then we note that

$$S_n'(x) = \frac{\sin (2nx)}{2\sin x},\tag{3}$$

with more or less the same argument as above. Since clearly $S_n(0) = 0$, we have

$$S_n(x) = \int_0^x S_n'(t)\,dt = \int_0^x \frac{\sin (2nt)}{2\sin t}\,dt = \frac{1}{4n}\int_0^{2nx} \frac{\sin u}{\sin \frac{u}{2n}}\,du.$$

The denominator is strictly increasing on the interval of integration (since $0 < x \leqslant \pi/2$), and hence

$$\int_{2k\pi}^{2(k+1)\pi} \frac{\sin u}{\sin \frac{u}{2n}}\,du = \int_{2k\pi}^{(2k+1)\pi} \sin u\biggl( \frac{1}{\sin \frac{u}{2n}} - \frac{1}{\sin \frac{u+\pi}{2n}}\biggr)du > 0\tag{4}$$

for $0 \leqslant k < n$. Let $m = \bigl\lfloor \frac{2nx}{2\pi}\bigr\rfloor$, then

$$4nS_n(x) = \sum_{k = 0}^{m-1} \int_{2k\pi}^{2(k+1)\pi} \frac{\sin u}{\sin \frac{u}{2n}}\,du + \int_{2m\pi}^{2nx} \frac{\sin u}{\sin \frac{u}{2n}}\,du.$$

The last integral is strictly positive - if $2nx \leqslant (2m+1)\pi$, the integrand is strictly positive on the whole interval except the endpoints, if $(2m+1)\pi < 2nx < 2(m+1)\pi$, $(4)$ shows that $\int_{2m\pi}^{2nx}\dotsc\,du > \int_{2m\pi}^{2(m+1)\pi}\dotsc\,du > 0$ - unless $2nx = 2m\pi$. But in that case, we have $m \geqslant 1$, and the sum is strictly positive.

• Very elementary and beautiful argument as desired by OP.+1 Aug 1, 2016 at 7:49