Does $\sum_{k \ \text{odd}, \ k>0}^{N} \frac{\sin(kx)}{k}$ converge to $f(x)$ in $L^2$? So I looked up that for something to converge in $L^2$ we must have that 
$$
\int_{I} |f_n(x) - f(x)|^2 dx \to 0 \text{ as } n \to \infty
$$
With 
$$
  f(x) =
\begin{cases}
-1,  & x\in [-\pi, 0) \\
1, & x\in [0, \pi].
\end{cases}
$$
And $f_n(x) = \sum_{k \ \text{odd}, \ k>0}^{N} \frac{\sin(kx)}{k}$
But it seems difficult to use the definition here. I have proved that $f_n(x)$ does not converge uniformly and neither pointwise to $f(x)$ is that a result I can use here?

First, note that for $f_n(x)=\sum_{k=1}^n \frac{\sin((2k-1)x)}{2k-1}$, we have
$$\lim_{n\to \infty}f_n(x)=\frac\pi4\text{sgn}(x)$$
for $x\in (-\pi,\pi)$, with $f_n(\pm \pi)=0$.  Therefore, we have
$$\lim_{n\to \infty}f_n(x)=f(x)=\begin{cases}\frac\pi4&,0<x< \pi\\\\0&,x=0,\pm \pi\\\\-\frac\pi4&,-\pi<x<0\end{cases}$$

Second, Dirichlet's Test for Uniform Convergence guarantees that the convergence is uniform on every closed interval $[a,b]$ for which either $0<a<b<\pi$ or $-\pi <a<b<0$ since there exists a number $M$ such that $\left|\sum_{k=1}^n \sin((2k-1)x)\right|\le M$ and $\lim_{n\to \infty}\frac{1}{2n-1}\to 0$.

Finally, since $f_n(x)$ is the Fourier Series for the function $f(x)$ as given by
$$f(x)=\begin{cases}\frac\pi4&,0<x< \pi\\\\0&,x=0,\pm \pi\\\\-\frac\pi4&,-\pi<x<0\end{cases}$$
and $f(x)\in \mathscr{L}^2([-\pi,\pi])$, then $||f_n(x)- f(x)||_2^2\to 0$.
