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.
 A: 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.
