<s>[INST] Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\zeta(2)$ and $\zeta(4)$.
How can I compute $ \zeta(6) $ now?
Fourier series of $ f $:
$$ S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$
$$ x=0,  \zeta(2)=\pi^2/6$$
 [/INST] Let $f(t):=t^3\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. The Fourier series of this function is
$$t^3=\sum_{k=1}^\infty {2(-1)^{k-1}(k^2\pi^2-6)\over k^3}\sin(kt)\qquad(-\pi\leq t\leq \pi).$$
We now use Parseval's formula
$$\|f\|^2=\sum_{k=0}^\infty |c_k|^2.$$
But 
$$\|f\|^2={1\over\pi}\int_{-\pi}^\pi x^6\>dt={2\pi^6\over7}$$
and
$$c_k^2={4(k^2\pi^2-6)^2\over k^6}=4\left(\frac{36}{k^6}-\frac{12\pi^2}{k^4}+\frac{\pi^4}{k^2}\right),k\geq1.$$ 
Noting that
$$ \sum_{k=1}^\infty \frac{12\pi^2}{k^4}=12\pi^2\frac{\pi^4}{90}=\frac{2\pi^6}{15}, \sum_{k=1}^\infty\frac{\pi^4}{k^2}=\pi^4\frac{\pi^2}{6}=\frac{\pi^6}{6}$$
we have
$$ \sum_{k=1}^\infty \frac{1}{k^6}=\left({\pi^6\over14}+\frac{2\pi^2}{15}-\frac{\pi^4}{6}\right)/36=\frac{\pi^6}{945}. $$
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