Don't understand a line for the proof of $\displaystyle \sum_{1≤k≤n}\frac{\sin{kx}}{k}≥0$ In this math exchange post Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson), in the answer posted by River Li (the third one), they write "We have $\sin{mz}<0$ due to the smallestness of $m$". Since $z\in{(0,\pi)}$, I don't understand how $\sin{mz}<0$ holds due to $m$ being small since $m$ is defined to be an integer anyway, i.e this line wouldn't hold if $m=10$ and $z=\frac{\pi}{5}$. Could someone explain the author's reasoning here and clarify what this line is trying to say? (I assume the proof given does ultimately make sense as the answer has two upvotes).
 A: Define $$f_m(x) = \sum_{n=1}^m \frac{1}{n} \sin nx. $$  Suppose there exists some $m$ such that $f_m(x)$ is negative in $(0,\pi)$.  Clearly $m > 1$ since $\sin x$ is strictly positive in $(0,\pi)$.  Then there must be a smallest $m > 1$ with this property, and we can find $y \in (0,\pi)$ such that $f_m(y) < 0$.
Now $f_m$ is continuous and differentiable on the closed interval $[0,\pi]$ and must therefore attain its minimum and the minimum is strictly negative since $f_m(y) <0$.  Neither end-point qualifies since $f_m(0)=f_m(\pi) = 0$ so the minimum must be at an interior point.  Let the minimum be at $z \in (0,\pi)$ so that $f_m(z) < 0$ and $f'_m(z) < 0$.
Now for your specific question.  If $\sin(mz) \geqslant 0$ then that term could be dropped from the sum and $f_{m-1}(z)$ would also be strictly negative.  But in that case $m$ would not be the smallest integer for which $f_m(x)$ has a negative value, contradicting the specific definition of $m$.  Thus $\sin mz < 0$.
The rest continues as before.
A: The integer $m$ was defined to be the least $n\in\Bbb N$ such that $$\sum_{k=1}^n\frac{\sin kx}k>0\quad\text{while}\quad\sum_{k=1}^{n-1}\frac{\sin kx}k\leqslant0.$$ Thus $m$ does not have to be "small" in any other or general sense.
