Showing the Fourier sine series converges 
The Fourier sine series for $f(x) = x$, $-2 < x < 2$ is 
  $$f(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}$$
  For each $x$ in the interval to what does the Fourier since series for $f(x)$ converge, can we prove pointwise convergence, convergence in $L^2$, and/or uniform convergence?

Attempted solution - We have 
$$\int_{-2}^{2}x^2 dx = \frac{16}{3} < \infty$$ so the function $f(x) = x$ is in $L^2$ and the Fourier series converges in $L^2$. Now, 
$$\Bigg|\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}\Bigg| \leq \frac{(-1)^{n+1}}{n}$$ 
Take $M_n = \frac{(-1)^{n+1}}{n}$. Since $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ is convergent then by the Weirstrass M-test the original series is uniformly convergent and thus pointwise convergent. 
Assuming the latter above is correct, can I state that $S_n(f)\to f(x) = x$?

Dirichlet's Test proves the series converges. Take $a_n=1/n$ and $b_n=\sin(n\pi x/2)$. Notice that $\sum_{n=1}^N\sin(n\pi x/2)=\csc(\pi x/4)\sin(N\pi x/4)\sin((N+1)\pi x/4)$ (you can prove this by taking the imaginary part of the equivalent sum of $e^{in\pi x/2}$). Now show that this is bounded on $[-2,2]$.
