TITLE: Prove that $\zeta (4)\le 1.1$
QUESTION [7 upvotes]: Prove the following inequality
$$\zeta (4)\le 1.1$$
I saw on the site some proofs for $\zeta(4)$ that use Fourier or Euler's way for computing its precise value, and that's fine and I can use it. Still, I wonder if there is a simpler  way around for proving
this inequality.  Thanks!

REPLY [4 votes]: One has
$${1\over (n-{1\over2})^3}-{1\over (n+{1\over2})^3}={3n^2+{1\over4}\over (n^2-{1\over4})^3}>{3\over n^4}\ .$$
Therefore one obtains the telescopic sum
$$\eqalign{\sum_{n=1}^\infty{1\over n^4}&=1+\sum_{n=2}^\infty{1\over n^4}< 1+{1\over3}\sum_{n=2}^\infty\Bigl({1\over (n-{1\over2})^3}-{1\over (n+{1\over2})^3}\Bigr)\cr &=1+{1\over3}{1\over(2-{1\over2})^3}=1+{8\over81}<1.1\ .\cr}$$