Given $f(x) = \frac{\pi ^2}{2}-x^2$, find the sine series in $[0,\pi]$.
I have computed $$b_k = \frac{2}{\pi}\int _0^\pi f(x)\sin kx dx = \begin{cases} \frac{8}{\pi k^3}, &k\mbox{ odd}\\ \frac{2\pi}{k}, &k\mbox{ even}\end{cases} $$ Which is a bit confusing since the end goal is to find $$f(x)\sim \sum_{k=1}^\infty b_k\sin kx $$
Prior to simplifying I arrived at $$b_k = \frac{2}{\pi}\left [ \frac{\pi ^2k^2((-1)^k+1)-4((-1)^k-1)}{2k^3}\right ] $$ What do I have to do to get the series written in a "sensible" fashion?
Or would I simply use this very form of $b_k$ [it looks ugly, I have this standard-textbook-exercise-syndrome, all the solutions are pretty] and prove via comparison, that the series converges absolutely? [actually no. It would be sufficient, but it's not going to work.]