<s>[INST] Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? 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 $\zeta(2) = \pi^2/6$.
 [/INST] by using Fourier series of $f(x)=1, x\in[0,1]$
$$1=\sum_{n=1,3,5,..}^\infty\frac{4}{n\pi}\sin (n\pi x)$$
take triple integrals as follows
$$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}1.dxdxdx=\int_{0}^{1}\int_{0}^{x}\int_{0}^{x} \sum_{n=1,3,5,..}^\infty\frac{4}{n\pi}\sin (n\pi x) dxdxdx$$
$$\frac{1}{6}=\sum_{n=1,3,5,..}^\infty\frac{2}{(n\pi)^2}-\frac{8}{(n\pi)^4}$$
$$\frac{\pi^4}{6}=\sum_{n=1,3,5,..}^\infty\frac{2\pi^2}{n^2}-\frac{8}{n^4}$$
We know that
 $$\sum_{n=1,3,5,..}^\infty\frac{1}{n^2}=\frac{\pi^2}{8}$$
so
$$\frac{\pi^4}{96}=\sum_{n=1,3,5,..}^\infty\frac{1}{n^4}$$
then we use the inequality series
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{n=1}^\infty\frac{1}{(2n-1)^4}+\sum_{n=1}^\infty\frac{1}{(2n)^4}$$
simplify it to get
$$\sum_{n=1}^\infty\frac{1}{n^4}=\frac{16}{15}\sum_{n=1}^\infty\frac{1}{(2n-1)^4}$$
so,
$$\sum_{n=1}^\infty\frac{1}{n^4}=\frac{16}{15}\frac{\pi^4}{96}=\frac{\pi^4}{90}$$
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