# Evaluate the series $\sum_{n = 0}^\infty \frac{1}{(2n + 1)^6}$ by examining the real Fourier series of the function $f(x) := x(\pi - |x|)$

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.

• You have $(2n+1)^{-3}$. You want $(2n+1)^{-6}$. How do you get from the first to the second. Which theorem is related to that operation in the context of Fourier analysis? – Daniel Fischer Jun 23 '14 at 19:50
• @DanielFischer: the Parseval identity? – Evan Aad Jun 23 '14 at 19:56
• Exactly that one. – Daniel Fischer Jun 23 '14 at 19:57
• [Here] is Parseval identity. – Mhenni Benghorbal Jun 23 '14 at 20:03

$$\frac{1}{\pi} \int_{-\pi}^{\pi} x^2 (\pi - |x|)^2 dx = \frac{64}{\pi^2}\sum_{n = 0}^\infty \frac{1}{(2n + 1)^6}$$
$$S = \frac{\pi^6}{960}$$