<s>[INST] Compute $\zeta(6)$ using Fourier series How can i prove that $\sum_{n=1}^\infty \frac{1}{n^6} = \frac{\pi^6}{945}$ knowing that $f(t)=t^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos(nt)$ on $(-\pi, \pi)$. I've integrated four times but i don't get anywhere. Thank's for helping.
 [/INST] Starting with
$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2} \, \cos(n t) = \frac{t^{2}}{4} - \frac{\zeta(2)}{2}$$
then integrate from $0$ to $t$ to obtain
$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} \, \sin(n t) = \frac{t^{3}}{2 \cdot 3!} - \frac{\zeta(2) \, t}{2} + c_{0}.$$
When $t=0$ it is determined that $c_{0} = 0$. Integrate several more times evaluating each series at $t=0$ to determine the constants of integration to obtain:
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} \, \sin(n t) &= \frac{t^{3}}{2 \cdot 3!} - \frac{\zeta(2) \, t}{2} \\
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} \, \cos(n t) &= \frac{t^{4}}{2 \cdot 4!} - \frac{\zeta(2) \, t^{2}}{4} + \frac{7}{8} \, \zeta(4) \\
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^5} \, \sin(n t) &= \frac{t^{5}}{2 \cdot 5!} - \frac{\zeta(2) \, t^{3}}{12} + \frac{7}{8} \, \zeta(4) \, t \\
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^6} \, \cos(n t) &= \frac{t^{6}}{2 \cdot 6!} - \frac{\zeta(2) \, t^{4}}{48} + \frac{7}{16} \, \zeta(4) \, t^{2} - \frac{31}{32} \, \zeta(6).
\end{align}
In the last series let $t = \pi$ to obtain
$$\zeta(6) = \frac{\pi^{6}}{945}.$$
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