In a problem they ask me to calculate
-
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$
-
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}$
-
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^6}$.
To do this I have previously calculated the Fourier Series of $f(x)=x$ in $[-\pi,\pi]$ and $g(x)=x^2$ in $[-\pi,\pi]$:
$f(x)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n}\sin(nx)$
$g(x)=\displaystyle\frac{\pi^2}{3}+\sum_{n=1}^{\infty}(-1)^{n}\frac{4}{n^2}\cos(nx)$
For (1) is easy to get to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ by substituting $\pi$ in $g$.
For (2) I used the Parseval Theorem in $g$ and I get to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}$
But I don't know how to do (3) using what was calculated above.
Best Answer
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}$$\sum_{n=1}^\infty\frac{144}{n^6}=\frac{16\pi^6}{105}$$ $$\sum_{n=1}^\infty\frac{1}{n^6}=\frac{\pi^6}{945}$$