I´m trying to find the absolute maximum of $(2N-1)$ partial sum of the Fourier´s series of signum function on $[0,\pi]$, I have:
$S_{2N-1}[f](x)=\frac{4}{\pi}\displaystyle\sum_{k=0}^{N-1}{\frac{sen((2k+1)x)}{2k+1}}$
I also have proved the following integral representation:
$S_{2N-1}[f](x)=\frac{2}{\pi}\displaystyle\int_{0}^{x}\frac{sen(2Nt}{sen(t)}dt$
I have showed that the maximums on $[0,\pi]$ of $S_{2N-1}[f](x)$ are $x=\frac{k \pi}{2N}$ with $k=1,3,5,7,...,(2N-1)$ (k odd) .
How could I conclude that the absolute maximum is $x=\frac{\pi}{2N}$?
Thank your four your time.