I am trying to determine the fourier series of $f(x)$ on $[-2,2]$ which I believe I have done correctly. I will show some details below of my calculation. However my question is how do I determine what the Fourier series of $f(x)$ on $[-2,2]$ converges to. My guess is that you must take the limit of $f(x)$, but do I take the limit as $x$ goes to $\infty$? However there is parts if the equation that are $(-1)^n$which confuses me if I was to take the limit.
$$f(x) = \begin{cases} 2,&-2\leq x \leq 0\\ x, &0 < x \leq 2\\ \end{cases}$$
So I know that,
$$f(x) = \frac{A_0}{2} + \sum _{n=1}^{\infty} A_n \cos \Big(\frac{n\pi x}{L}\Big) + \sum _{n=1}^{\infty} B_n \sin \Big(\frac{n\pi x}{L}\Big)$$
Then calculating $A_0,$
$$A_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx = \frac{1}{2} \int _{-2}^{2} f(x)dx = \frac{1}{2} \Bigg [\int_{-2}^02dx + \int_0^{2} x \Bigg]= \frac{1}{2}[4+2] = 3 $$
Then calculating $A_n$,
$$A_n = \frac{1}{L}\int_{-2}^{2} f(x) \cos\Big(\frac{n\pi x}{2} \Big) = \frac{1}{2}\Bigg[\int_{-2}^0 2\cos\Big(\frac{n\pi x}{2}\Big) + \int _0^2 x\cos \Big(\frac{n\pi x}{2}\Big) \Bigg]\\ = \frac{2}{\pi^2 n^2 }(-1)^{n} - \frac{2}{\pi^2 n^2 }$$
Then calculating $B_n, $
$$B_n = \frac{1}{L}\int_{-2}^{2} f(x) \sin\Big(\frac{n\pi x}{2} \Big) = \frac{1}{2}\Bigg[\int_{-2}^0 2\sin\Big(\frac{n\pi x}{2}\Big) + \int _0^2 x\sin \Big(\frac{n\pi x}{2}\Big) \Bigg]\\ = \frac{1}{2}\Bigg[\frac{-4}{n\pi}\cos \Big(\frac{n\pi x}{2}\Big)\Big|_{-2}^0 \Bigg] + \frac{1}{2}\Bigg[\frac{-2x}{\pi n}\cos \Big(\frac{n\pi x}{2 }\Big) + \frac{4}{n^2\pi ^2}\sin \Big(\frac{n\pi x}{2 }\Big)\Big|_0^2 \Bigg] \\ = \frac{1}{2}\Bigg[\frac{-4}{\pi n } + \frac{4}{\pi n } \cos(\pi n )\Bigg] + \frac{1}{2}\Bigg[\frac{-4}{\pi n }\cos(\pi n) + \frac{4}{\pi ^2 n ^2 } \sin (\pi n) \Bigg] \\ = \frac{1}{2} \Bigg[\frac{-4}{n\pi} + \frac{4}{n\pi}(-1)^n - \frac{4}{\pi n}(-1)^n \Bigg] = \frac{-2}{n\pi}$$
Thus,
$$f(x) = \frac{3}{2}+ \sum_{n=1}^{\infty}\Bigg[\frac{2}{\pi^2 n^2 }(-1)^{n} - \frac{2}{\pi^2 n^2 }\Bigg] \cos\frac{n\pi x}{2} + \sum_{n=1}^{\infty}\Bigg[\frac{-2}{n\pi }\Bigg] \sin \frac{n\pi x}{2}$$
So again my thought would be to take the limit of this function, and I have a feeling that it approaches $\frac{3}{2}$. But I am not sure if this is correct or how to show it, looking for some help with this, thanks!
Since $f (x) $ is piecewise smooth. But has a jump discontinuity at $x=0$ , the Fourier series will converge to $$1/2[ f (0-)+f (0+)]$$.
You don't need to calculate the series to find wheter it converges or not.
You can read more about it at http://tutorial.math.lamar.edu/Classes/DE/ConvergenceFourierSeries.aspx