I know some nice ways to prove that ζ(2)=∑∞n=11n2=π2/6. For example, see Robin Chapman’s list or the answers to the question “Different methods to compute ∑∞n=11n2?”

Are there any nice ways to prove that ζ(4)=∞∑n=11n4=π490?

I already know some proofs that give all values of ζ(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 ζ(4).

I would be particularly interested in a proof that isn’t an adaption of one that ζ(2)=π2/6.

**Answer**

In the same spirit of the 1st proof of this answer. If we substitute π for x in the Fourier trigonometric series expansion of f(x)=x4, with −π≤x≤π,

x4=15π4+∞∑n=18n2π2−48n4cosnπ⋅cosnx,

we obtain

π4=15π4+∞∑n=18n2π2−48n4cos2nπ=15π4+8π2∞∑n=11n2−48∞∑n=11n4.

Hence

∞∑n=11n4=π448(−1+15+86)=π448⋅815=190π4.

**Attribution***Source : Link , Question Author : Mike Spivey , Answer Author : Community*