For $x \in (0, \pi/2)$ find local maximum of $S_{N,f}(x) = \frac{4}{\pi} \sum_{k >0, \ k \ \text{odd}}^N \frac{\sin(kx)}{k}$
I have shown that $\frac{d}{dx} S_{N,f}(x) = \frac{2}{\pi} \frac{\sin((N+1)x)}{\sin(x)}$
I also know that $N$ is odd, hence $N+1$ is even. The zeros for $\frac{d}{dx} S_{N,f}(x)$ are $x= \frac{k\pi}{N+1}$ for $k \in \mathbb{Z}$.
Then $$\frac{d^2}{dx^2} S_{N,f}(x=\frac{k\pi}{N+1}) = \frac{(N+1)(-1)^k \sin(\frac{k\pi}{N+1})}{\sin^2(\frac{k\pi}{N+1})}.$$
How can I proceed?
I now saw that there was a hint on the back of my paper (I'm doing exercises), but I still need help.
Rewrite the partial sum as $S_{N,f}(x) = \int_{0}^{x} \frac{d}{dy} S_{N,f}(y) dy$ and use the fact that the denominator in the formula for $S_{N,f}(y)$ is increasing with $y$ for $y \in [0, \pi/2]$