Determine the function to which the Fourier series converges for $f(x)$ given the following
$$f(x)=x,~~~~~-\pi<x<\pi$$
Solution: \begin{align*} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi} x dx\\ &=\frac{1}{\pi}\left[\frac{x^2}{2}\right]_{-\pi}^{\pi}\\ &=0. \end{align*} Since $f(x)$ is odd so $a_n=0$.
\begin{align*} b_n&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(nx)dx\\ &=\frac{1}{\pi}\int_{-\pi}^\pi x\sin(nx)dx\\ &=\frac{1}{\pi}\left[\frac{x\cos (nx)}{n}+\dfrac{\sin(nx)}{n^{2}}\right]_{-\pi}^{\pi}\\ &=-\frac{1}{n\pi}[\pi\cos(nx)-(-\pi)\cos(-n\pi)]\\ &= -\frac{1}{n\pi}\cdot 2\pi\cos(n\pi)\\ & = \frac{2\cdot(-1)^{n+1}}{n} \end{align*}
Therefore, the required fourier series is $$f(x)=\sum_{n=1}^\infty \frac{2\cdot(-1)^{n+1}}{n} \sin nx.$$
Help with the second part! Thanks in advance!
Your question follows from the following general theorem:
If the function $F$ is differentiable at the point $x_0$ then its Fourier series at the point $x_0$ converges to $F(x_0)$.