Compute $\sum_{n=1}^\infty{\frac{1}{n^8}}$ using Parseval's Theorem I need to show that $$\sum_{n=1}^\infty{\frac{1}{n^8}} = \frac{\pi^8}{9450}$$ I have already shown that $$\sum_{n=1}^\infty{\frac{1}{n^4}} = \frac{\pi^4}{90}$$ by computing the Fourier series for the function $f(x)=x^2$ on the interval $(0,l)$ and then applying Parseval's Theorem. However, I don't know if $$\sum_{n=1}^\infty{\frac{1}{n^8}}$$ can be done in the same way because I am not given a function. My idea is to try and write $$\sum_{n=1}^\infty{\frac{1}{n^8}} = \sum_{n=1}^\infty({\frac{1}{n^4}})^2$$ so that I can use my previous calculations. Is this the right idea or is there an easier method? If this is the correct idea, then how does the square change the Fourier series?
 A: Over the interval $(0,2\pi)$ we have
$$ \frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n} \tag{1}$$
pointwise and uniformly over compact subsets. We may denote with $\text{Int}$ the operator which brings $f(x)$ into $C_f+\int_{0}^{x}f(t)\,dt$, where $C_f$ is chosen in such a way that $\int_{0}^{2\pi}(\text{Int}\,f)(x)\,dx = 0$. By applying such operator to both sides of $(1)$ we get
$$ \frac{1}{12}\left(-2\pi^2+6\pi x-3x^2\right) = \sum_{n\geq 1}\frac{\cos(nx)}{n^2}\tag{2} $$
uniformly over any compact subset of $[0,2\pi]$, then by applying $-\text{Int}^2$ to both sides of $(2)$ we get
$$ \sum_{n\geq 1}\frac{\cos(nx)}{n^4} = \text{Re}\,\text{Li}_4(e^{ix})=\frac{\pi^4}{90}-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac{1}{4}x^4.\tag{3} $$
Now we apply $\int_{0}^{2\pi}(\ldots)^2\,dx$ to both sides of $(3)$. We get:
$$ \pi \zeta(8) = \int_{0}^{2\pi}\left(\frac{\pi^4}{90}-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac{1}{4}x^4\right)^2\,dx=\begin{array}{c}\small\text{bunch of omitted}\\\small\text{computations}\end{array}=\frac{\pi^9}{9450}.\tag{4} $$
We are just integrating the square of a Bernoulli polynomial.

Anyway I believe that the most efficient approach for computing $\zeta(2m)$ for large values of $m\in\mathbb{N}^+$ is to recall Euler's generating function
$$ \frac{1-x\cot x}{2} = \sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}} x^{2m} \tag{4}$$
which comes from the application of $\frac{d}{dz}\log(\cdot)$ to both sides of $\frac{\sin z}{z}=\prod_{m\geq 1}\left(1-\frac{z^2}{\pi^2 m^2}\right)$.
Since
$$\left[1-2\sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}}z^{2m}\right]\cdot\left[1+\sum_{m\geq 1}\frac{(-1)^m}{(2m+1)!}z^{2m}\right]=1+\sum_{m\geq 1}\frac{(-1)^m}{(2m)!}z^{2m}\tag{5}$$
we get (once $\zeta(0)$ is defined as $-\frac{1}{2}$, consistent with the analytic continuation of the series $\sum_{n\geq 1}\frac{1}{n^s}$)
$$ \sum_{m=0}^{M}\frac{-2 \zeta(2m)(-1)^{M-m}}{\pi^{2m}(2M-2m+1)!}=\frac{(-1)^M}{(2M)!}\tag{6}$$
which allows to find $\zeta(2M)$ in terms of $\zeta(2M-2),\zeta(2M-4),\ldots,\zeta(2)$. This is equivalent to the well-known recursion for Bernoulli numbers.
