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