Nice proofs of ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}?

Question 1

I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$ . For example, see Robin Chapman’s list or the answers to the question “Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$ ?”

Are there any nice ways to prove that $\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$

I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman’s list or Qiaochu Yuan’s answer in the linked question). I’m not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$ .

I would be particularly interested in a proof that isn’t an adaption of one that $\zeta(2) = \pi^2/6$ .

Question 2

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

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

Nice proofs of ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}?

Answer

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