I'm doing the exercise $11.19$ from Apostol Real Analysis:
Let $S_n(x)=\frac{4}{\pi} \sum_{k=1}^n \frac{\sin((2k-1)x)}{2k-1}$.
Prove that $S_n(\frac{\pi}{2n}) \geq S_n(\frac{m \pi}{2n})$ for $m=1,2, ... ,2n-1$
I've already proved that $S_n(x)= \frac{2}{\pi} \int_0^x \frac{\sin(2nt)}{\sin(t)} dt $ and in $x_m$ has his local minimun if $m$ is odd and local maximum if $m$ is even.
Thanks in advance
Seems like you have done all the hard parts and the easy part is to show that $$a_m=\int_{\frac{(m-1)\pi}{2n}}^{\frac{m\pi}{2n}}\frac{\sin(2nt)}{\sin(t)}dt$$ is an alternating sequence with terms decreasing in magnitude for $1\le m\le n$ and the first term positive. You know that because $\sin(2nt)$ changes sign exactly when $2nt=\pi m$ or $t=\frac{m\pi}{2n}$ and $\sin(t)$ is an increasing function of $t$ for $0<t<\frac{\pi}2$. Then $$S_n\left(\frac{m\pi}{2n}\right)=\sum_{k=1}^ma_k$$ The alternating part guarantees that $S_n\left(\frac{2m\pi}{2n}\right)=S_n\left(\frac{(2m-1)\pi}{2n}\right)+a_{2m}<S_n\left(\frac{(2m-1)\pi}{2n}\right)$ and the decreasing part guarantees that $S_n\left(\frac{(2m+1)\pi}{2n}\right)=S_n\left(\frac{(2m-1)\pi}{2n}\right)+a_{2m}+a_{2m+1}<S_n\left(\frac{(2m-1)\pi}{2n}\right)$. The symmetry of the square wave means that $S_n\left(\frac{(2n-m)\pi}{2n}\right)=S_n\left(\frac{m\pi}{2n}\right)$. Did you see the nice pictures on Wikipedia?