I do not know where I am making a mistake... The problem says that obtaining the Fourier series for the $2\pi$ periodic odd function $f(x)=x(\pi-x)$ on $[0,\pi]$ and using Parseval's relation I should obtain $$\sum_{n=1}^\infty\frac{1}{(2n-1)^6}=\frac{\pi^6}{960}.$$
So, I already know that the odd extension for the function is $$f(x)=\frac{8}{\pi}\sum_{n=1}^\infty\frac{1}{(2n-1)^3}\sin(2n-1)x$$ where $$b_n=\frac{8}{\pi}\cdot\frac{1}{(2n-1)^3}\Longrightarrow b_n^2=\frac{64}{\pi^2}\cdot\frac{1}{(2n-1)^6}.$$ On the other hand $$\frac{1}{\pi}\int_{-\pi}^\pi f^2(x)dx=\frac{1}{\pi}\int_{-\pi}^\pi [x(\pi-x)]^2 dx=\frac{16 \pi^4}{15}.$$ Putting this together with Parseval's $$\frac{64}{\pi^2} \sum_{n=1}^\infty \frac{1}{(2n-1)^6}=\frac{16 \pi^4}{15}$$ which results in $$\sum_{n=1}^\infty \frac{1}{(2n-1)^6}=\frac{\pi^6}{60}$$ and not what the problem states... ?
Any help will be appreciated. Thank you.
You have the correct fourier series. But as Jean-Claude Arbaut says you're doing the integral in Parseval Formula wrong. For the sine series use instead:
$$ \frac{2}{\pi} \int_0^\pi |f(x)|^2 dx = \sum_{n>1} |b_n|^2 $$
This gives the desired result:
$$ \sum_{n>1} \frac{1}{(2n-1)^6} = \frac{\pi^6}{960}$$
However I don't know how to get from here to:
$$ ζ (6) = \sum_{n>1} \frac{1}{n^6} = \frac{\pi^6}{945}$$
Maybe I have to compute the cosine fourier series for the even terms?