For $x\in[-\pi,\pi]$, $F(x)=\left\{ \begin{array}{cl} -1 & \text{for}~-\pi\leq x\leq 0\\ 1 & \text{for}~0\leq x\leq \pi \end{array}\right..$
To what value does the Fourier series converge at the point $x=\pi$?
I have found the fourier series as show below and I think it converges to 0 on the conditions give but I'm not entirely sure.
$f(-x)=-f(x)$ so we have and odd function and expect $a_n=0$.
\begin{align} b_n &= \frac1\pi \left[\int^\pi_{-\pi}f(x)\sin(nx)dx\right] \\ &= \frac1\pi \left[\int^0_{-\pi}-\sin(nx)dx\right] + \frac1\pi \left[\int^\pi_0\sin(nx)dx\right] \\ &= -\frac1\pi \left[-\frac{\cos(nx)}{n}\right]^0_\pi + \frac1\pi \left[-\frac{\cos(nx)}{n}\right]^\pi_0 \\ &= \frac{1-(-1)^n}{n\pi} - \frac{(-1)^n-1}{n\pi} \\ &= \frac{1-(-1)^n}{2n\pi}. \end{align}
$b_{2m}=0,\quad b_{2m+1}=\frac{4}{(2m+1)\pi}$
\begin{align} f(x) &=\frac4\pi\left(\sin x+\frac{1}{3}\sin(3x)+\frac{1}{5}\sin(5x)+\cdots\right)\\ &=\frac4\pi \sum^{\infty}_{m=0} \frac{1}{2m+1}\sin((2m+1)x) \end{align}