Fourier sine series pointwise and uniform convergence

Consider the step function $f(x) = \begin{cases} -1 & -\pi < x < 0 \\ 1 & 0 \leq x < \pi \end{cases}$

1. Calculate the coefficients $a_n$ of the sin series $f(x) = \sum_{n=1}^\infty a_n\sin nx$.

2. What is the pointwise limit of this series, for each $x \in (-\pi,\pi)$?

3. Does this series converge uniformly? Prove that it does or does not.

I've managed to do the first part, $$a_n = \frac{2(1-(-1)^{n})}{\pi n}.$$ But I am getting confused with all the information available on the web regarding pointwise and uniform convergence of Fourier sine series. I believe I showed that for part 3 we have uniform convergence based on some estimates of the series but I would like to know the general result for these sorts of objects.

In general, if you have good references on calculus like Fourier Analysis for beginners please share them. Thank you!

• Are you sure with the expression of $a_n$ – hamam_Abdallah Aug 21 '17 at 20:44
• @Salahamam_ Fatima Well, I don't know much rigor behind the Fourier sine series. But if you just assume that $$\sin(nx)$$ are basis (after normalization) then $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x)\sin(nx)dx = \frac{1}{\pi} \left[ \int_{-\pi}^0 -\sin(nx) dx + \int_0^\pi \sin(nx)dx \right] = \frac{2}{\pi} \int_0^\pi \sin(nx)dx = \frac{2}{\pi} \left[ \frac{-\cos(nx)}{n} \right]_0^\pi = \frac{2(1-(-1)^{n})}{\pi n}.$$ – Vadim Aug 21 '17 at 20:46
• Dirichlet's test for series should work for pointwise convergence. – hamam_Abdallah Aug 21 '17 at 20:58
• @Salahamam_ Fatima I agree with this. Yet, do you know what is the pointwise limit? – Vadim Aug 21 '17 at 21:00
• I would use the $M$-test to see if there is uniform convergence. – Wolfy Aug 21 '17 at 21:10