Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$?

In the same spirit of the 1st proof of this answer. If we substitute $\pi $ for $ x $ in the Fourier trigonometric series expansion of $% f(x)=x^{4}$, with $-\pi \leq x\leq \pi $,

$$x^{4}=\frac{1}{5}\pi ^{4}+\sum_{n=1}^{\infty }\frac{8n^{2}\pi ^{2}-48}{n^{4}}\cos n\pi \cdot \cos nx,$$

we obtain

$$\begin{eqnarray*} \pi ^{4} &=&\frac{1}{5}\pi ^{4}+\sum_{n=1}^{\infty }\frac{8n^{2}\pi ^{2}-48}{n^{4}}\cos ^{2}n\pi \\ &=&\frac{1}{5}\pi ^{4}+8\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}} -48\sum_{n=1}^{\infty }\frac{1}{n^{4}}. \end{eqnarray*}$$

Hence

$$\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{\pi ^{4}}{48}\left( -1+\frac{1}{5}+ \frac{8}{6}\right) =\frac{\pi ^{4}}{48}\cdot \frac{8}{15}=\frac{1}{90}\pi ^{4}.$$

Consider the function $f(t):=t^2\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. Developping $f$ into a Fourier series we get $$t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).$$ If we put $t:=\pi$ here we easily find $\zeta(2)={\pi^2\over6}$. For $\zeta(4)$ we use Parseval's formula $$\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\ .$$ Here $$\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}$$ and the $c_k$ are the complex Fourier coefficients of $f$. Therefore $c_0={\pi^2\over3}$ and $|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}$ $\ (k\geq1)$. Putting it all together gives $\zeta(4)={\pi^4\over 90}$.

If you are specially interested only in $\zeta(4)$, the following proof would work but this is an adaptation Euler's idea. The idea is just to mimic Euler's proof for the Basel problem. Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$

To evaluate $\zeta(4)$, we can mimic Euler's idea and look at roots at $\pm \pi, \pm i \pi,\pm 2 \pi, \pm 2 i \pi,\pm 3 \pi, \pm 3 i \pi$.

Let $$p(z) = \left(1 - \left(\frac{z}{i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{2 i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{3 i \pi}\right)^4 \right) \times \cdots$$

It is not hard to guess that $p(z)$ is same as $$\frac{i \sin(z) \times \sin \left( \frac{z}{i} \right)}{z^2} = \left(1-\frac{z^2}{3!} + \frac{z^4}{5!} -\cdots \right) \times \left(1+\frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right)$$

Compare the coefficient of $z^4$ to get $$\zeta(4) = \frac{\pi^4}{90}$$

This proof could be extended for any even number to give that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $$

As expected for odd numbers, this doesn't work. For instance for $3$, if you try to work out by looking at $$p(z) = \left(1 - \left(\frac{z}{\omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{2 \omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{3 \omega \pi}\right)^3 \right) \times \cdots$$ where $\omega^3 = 1$ there is an asymmetry since $$\sin(z) \sin \left( \frac{z}{\omega}\right) \sin \left( \frac{z}{\omega^2}\right)$$ extends on both sides and the non-zero roots are at $$\pm \pi,\pm \omega \pi,\pm \omega^2 \pi,\pm 2 \pi,\pm 2 \omega \pi,\pm 2 \omega^2 \pi,\pm 3 \pi,\pm 3 \omega \pi,\pm 3 \omega^2 \pi,\ldots$$ and hence the $\zeta(3)$ terms nicely hides by canceling out and the resulting expression only gives $\zeta(6)$.