The following is a question from a past exam in my university in a course called "Mathematical Methods for Statistics". It consists of two subquestions that may or may not be related (there is a high chance they are related based on similar exams by the same professor).
(a) Write the Fourier series for the function $f(x) := x(\pi - |x|)$ in the interval $[-\pi, \pi]$.
(b) Calculate the sum of the following series: $$ S :=\sum_{n = 0}^\infty \frac{1}{(2n + 1)^6} $$
Attempted solution
(a) $$ f(x) \sim \frac{8}{\pi} \sum_{n = 0}^\infty \frac{\sin((2n + 1) x)}{(2n + 1)^3} $$
(b) Based on past exams, I expect the result to follow from part (a) by evaluating $f(x)$ at some appropriately chosen $x$, but I can't figure out what this $x$ might be. The best I got was by assigning $x := \frac{\pi}{2}$, which yields: $$ f(\frac{\pi}{2}) = \frac{8}{\pi}\sum_{n = 0}^\infty (-1)^n \frac{1}{(2n + 1)^3} $$
If we define $a_n := \frac{(-1)^n}{(2n + 1)^3}$, we have that $S = \sum_{n = 0}^\infty a_n^2$. But how do I proceed from here?
Another possibility is that my answer to part (a) is wrong.
Thanks to Daniel Fischer's help, and assuming correctness of my answer to part (a), here's my answer to part (b).
By Parseval's identity,
$$ \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 (\pi - |x|)^2 dx = \frac{64}{\pi^2}\sum_{n = 0}^\infty \frac{1}{(2n + 1)^6} $$
Whence
$$ S = \frac{\pi^6}{960} $$