It is known that the sine function can be written as an infinite product: $$ \frac{\sin x}{x}=\prod_{k=1}^{\infty}(1-\frac{x^2}{k^2 \pi^2}) $$ For $x$ sufficiently small I collect equal powers of $x$: $$ \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7)}{x}=1-\sum_{k=1}^\infty\frac{x^2}{(k \pi)^2}+\sum_{k=1}^{\infty}\frac{x^2}{(k \pi)^2}\sum_{l=k+1}^{\infty}\frac{x^2}{(l \pi)^2}+O(x^6) $$ or
$$ {1-\frac{x^2}{6}+\frac{x^4}{120}+O(x^6)}=1-\sum_{k=1}^\infty\frac{x^2}{(k \pi)^2}+\sum_{k=1}^{\infty}\frac{x^2}{(k \pi)^2}\sum_{l=k+1}^{\infty}\frac{x^2}{(l \pi)^2}+O(x^6) $$
The constant term cancels out. The $x^2$ term drops out if $$ \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}=\zeta(2) $$ which is a known value of the zeta function. Now we have to balance the $x^4$ terms. We get
$$ \sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{l=k+1}^{\infty}\frac{1}{l^2}=\frac{\pi^4}{120} $$
Just what is this monstrocity? Is it in any way related to $\zeta(4)$ which is $\frac{\pi^4}{90}$ ? So close and yet so far away... How to continue? Please help!