I have a function $f:[0,1]\rightarrow \mathbb{R}$ given by $f(x)=1-x$. I have found its Fourier sine series to be $f(x)=\sum^{\infty}_{n=1} b_n \sin(n\pi x)$ with $b_n = \dfrac{2(1-x)}{n\pi}(1-\cos(n\pi))$. Now the question ask for what value $-1\leq x \leq 1$ does the Fourier sine series converge to $f(x)$.
I have tried to play around with $f(x)=\sum^{\infty}_{n=1} b_n \sin(n\pi x)$ and get: \begin{align} f(x)&=\sum^{\infty}_{n=1} \dfrac{2(1-x)}{n\pi}(1-\cos(n\pi)) \sin(n\pi x) = \sum^{\infty}_{\substack{n=1,\\ n\text{ odd}}} \dfrac{2(1-x)}{n\pi}\sin(n\pi x)\\&= f(x)\sum^{\infty}_{\substack{n=1,\\ n\text{ odd}}} \frac{2}{n\pi}\sin(n\pi x). \end{align} Now I'm trying to figure out the $x$ value that make $\sum^{\infty}_{n=1,\text{ n odd}} \dfrac{2}{n\pi}\sin(n\pi x)$ to be $1$. Am I doing using the right approach? If so, how should I go from here? If not, what are some suggestions? Thank you!