I'm learning about Fourier series and need some help with this following problem:
Consider the function $f(x) = \frac{\pi - x}{2}, \ x \in [0, 2\pi)$ extended periodically with period $2\pi$. Find the Fourier series of $f$. Examine what happens if we set $x = \frac{\pi}{2}$. What can you say about the pointwise and uniform convergence of the functional series $\sum_{k = 1}^{+\infty} \frac{\sin(kx)}{k}$?
My work and thoughts:
The function $f$ is piecewise smooth so the Fourier series converges pointwise to $f$ on $\mathbb{R}\setminus\{2k\pi, k \in \mathbb{Z}\}$ and for $\{2k\pi, k \in \mathbb{Z}\}$ the the Fourier series converges to $\frac{f(2\pi+) + f(2\pi-)}{2} = 0$.
The function $f$ is odd so the coefficients $a_k = 0, \, k \in \mathbb{N}$.
After some simple computations involving integration by parts I have found that the coefficient of $b_k = \frac{1}{k}$.
Therefore, the Fourier series of $f$ is $$ f(x) \sim \sum_{k = 1}^{+\infty}\frac{1}{k}\sin(kx). $$
For $x = \frac{\pi}{2}$ we have that $$ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} \ldots $$ so we conclude that the series $\sum_{k = 0}^{+\infty}\frac{(-1)^k}{(2k + 1)} = \frac{\pi}{4}$.
Is my work correct? Do I need to complete my answer with anything else regarding convergence? What do I need to say about the pointwise and uniform convergence of the functional series $\sum_{k = 1}^{+\infty} \frac{\sin(kx)}{k}$?