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.
Asked
Active
Viewed 658 times
2
-
After integrating four times you can plug in $t=\pi$ (note that this equation is valid for this $t$ too by continuity). – Wojowu Nov 02 '17 at 22:19
2 Answers
3
$$f_1(t) = \sum_{n\geq 1}\frac{\sin(nt)}{n} $$ is the Fourier series of a sawtooth wave, an odd, piecewise-linear function with mean zero which equals $\frac{\pi-t}{2}$ over the interval $(0,\pi)$. By termwise integration we have that $$ f_2(t) = \sum_{n\geq 1}\frac{\cos(nt)}{n^2} = \text{Re}\,\text{Li}_2(e^{it}) $$ is the Fourier series of an even, piecewise-parabolic function with mean zero which equals $\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4}$ over the interval $(0,\pi)$. Performing the same trick once again, $$ f_3(t) = \sum_{n\geq 1}\frac{\sin(nt)}{n^3}=\text{Im}\,\text{Li}_3(e^{it}) $$ is the Fourier series of an odd, piecewise-cubic function mith mean zero which equals $\frac{\pi^2 t}{6}-\frac{\pi t^2}{4}+\frac{t^3}{12}$ over the interval $(0,\pi)$. By Parseval's theorem $$ \pi\,\zeta(6) = \int_{-\pi}^{\pi}f_3(t)^2\,dt = 2\int_{0}^{\pi}\left(\frac{\pi^2 t}{6}-\frac{\pi t^2}{4}+\frac{t^3}{12}\right)^2\,dt =\frac{\pi^7}{945}$$ and we are done.
Corollary: for any $m\in\mathbb{N}^+$, $\zeta(2m)\in \pi^{2m}\mathbb{Q}$.
Jack D'Aurizio
- 353,855
1
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 $$\zeta(6) = \frac{\pi^{6}}{945}.$$
Leucippus
- 26,329