Prove that $$\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$$ by the Fourier series of $x^2$.

By Parseval's identity, I can only show $\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}$. Could you please give me some hints?

  • 2
    $\begingroup$ I think you need to use the fourier series of x^3 to directly use Parseval's identity. $\endgroup$ – CapitalPi Sep 30 '16 at 16:20
  • $\begingroup$ ... or integrate the Fourier series of $x^2$, then apply Parseval's theorem. However, in order to implement this idea it is better to take a different quadratic polynomial, namely $B_2\left(\frac{x}{2\pi}\right)$. $\endgroup$ – Jack D'Aurizio Sep 30 '16 at 16:57

We may start from $$ f_1(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n} \tag{1}$$ that is the Fourier series of a sawtooth wave, equal to $\frac{\pi-x}{2}$ on the interval $I=(0,2\pi)$.
By termwise integration, we get that $$ \forall x\in I,\quad \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2} = \frac{2\pi x-x^2}{4}$$ hence: $$ \forall x\in I,\quad f_2(x) = \sum_{n\geq 1}\frac{\cos(nx)}{n^2}=\frac{\pi^2}{6}-\frac{\pi x}{2}+\frac{x^2}{4}\tag{2} $$ $$ \forall x\in I,\quad f_3(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n^3}=\frac{\pi^2 x}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}\tag{3} $$ (the integration constants are computed from the fact that $f_j(x)$ has to have mean zero over $I$)
and by Parseval's theorem $$ \zeta(6) = \frac{1}{\pi}\int_{0}^{2\pi}f_3(x)^2\,dx = \frac{2\pi^6}{9}\int_{0}^{1}\left[x(x-1)(2x-1)\right]^2\,dx=\color{red}{\frac{\pi^6}{945}}.\tag{4}$$ With the same approach it is not difficult to prove that for any $n\geq 1$, $\zeta(2n)$ is a rational multiple of $\pi^{2n}\int_{0}^{1}B_n(x)^2\,dx$, where $B_n(x)$ is a Bernoulli polynomial.

| cite | improve this answer | |
  • $\begingroup$ it is not 100% clear that you substract the mean such that $f_j(x)$ has zero mean $\endgroup$ – reuns Sep 30 '16 at 16:40
  • $\begingroup$ @user1952009: the expression for $f_2(x)$ is derived from $f_2(x)=\zeta(2)-\sum_{n\geq 1}\frac{1-\cos(nx)}{n^2}$ and the expression for $f_3(x)$ is derived from $f_3(\pi)=0$. Anyway, answer improved. Thanks for the suggestion. $\endgroup$ – Jack D'Aurizio Sep 30 '16 at 16:43
  • $\begingroup$ But where is the Fourier series of $x^2$? $\endgroup$ – Longitude Sep 30 '16 at 16:56
  • $\begingroup$ @Longitude: it is part of $(2)$. See my comment under your question. $\endgroup$ – Jack D'Aurizio Sep 30 '16 at 16:57

Considering $f(x)=x^3$ By Parseval identity we can prove that $\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^6}{945}$. Then $$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=0,$$$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx=0$$ and \begin{align}b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\\&=(-1)^{n+1}\frac{2\pi^2}{n}+(-1)^{n}\frac{12\pi}{n^3}\end{align} From the relation $$\frac{1}{\pi}\int_{-\pi}^{\pi}|f|^2dx=\frac{a_0}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)$$we get \begin{align}\sum_{n=1}^\infty(\frac{144}{n^6}+\frac{4\pi^4}{n^2}-\frac{48\pi^2}{n^4})=\frac{2\pi^6}{7}\end{align}$$\sum_{n=1}^\infty\frac{144}{n^6}=\frac{16\pi^6}{105}$$ $$\sum_{n=1}^\infty\frac{1}{n^6}=\frac{\pi^6}{945}$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.