How to prove $\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots$ is positive? Let $0<x<\pi$. $n$ be a natural number.
How to prove
$$\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots+ \frac{\sin nx}{n}>0$$

Let $f_n(x)=\sum\limits_{k=1}^n\frac{\sin{kx}}{k}$.
For $n=1$ it's obvious.
Let $f_n(x)>0$ for any $0<x<\pi.$
It's enough to prove that $f_{n+1}(x)>0.$
Indeed, let $f_{n+1}(x)\leq0$ for some value of $x\in(0,\pi)$.
Since $f_{n+1}$ is a continuous function on $[0,\pi],$ $f_{n+1}(0)=f_{n+1}(\pi)=0,$ 
there is $x_0\in(0,\pi)$ for which $f_{n+1}$ gets a minimal value and we know that $f_{n+1}(x_0)\leq0.$
Now, $$f'_{n+1}(x_0)=0$$ or
$$\sum_{k=1}^{n+1}\cos{kx_0}=0$$ or
$$\sum_{k=1}^{n+1}2\sin\frac{x_0}{2}\cos{kx_0}=0$$ or
$$\sum_{k=1}^{n+1}\left(\sin\left(kx_0+\frac{x_0}{2}\right)-\sin\left(kx_0-\frac{x_0}{2}\right)\right)=0$$ or
$$\sin\left(nx_0+\frac{3x_0}{2}\right)=\sin\frac{x_0}{2},$$ which gives:
$$0\geq f_{n+1}(x_0)=f_n(x_0)+\frac{1}{n+1}\sin(n+1)x_0=$$
$$=f_n(x_0)+\frac{1}{n+1}\sin\left(\left(n+\frac{3}{2}\right)x_0-\frac{x_0}{2}\right)=$$
$$=f_n(x_0)+\frac{1}{n+1}\left(\sin\left(n+\frac{3}{2}\right)x_0\cos\frac{x_0}{2}-\cos\left(n+\frac{3}{2}\right)x_0\sin\frac{x_0}{2}\right)=$$
$$=f_n(x_0)+\frac{\sin\frac{x_0}{2}}{n+1}\left(\cos\frac{x_0}{2}-\cos\left(n+\frac{3}{2}\right)x_0\right)=$$
$$=f_n(x_0)+\frac{\sin\frac{x_0}{2}}{n+1}\left(\left|\cos\left(n+\frac{3}{2}\right)x_0\right|-\cos\left(n+\frac{3}{2}\right)x_0\right)\geq f_n(x_0),$$ which is a contradiction and by induction we are done!
