By which scheme should I add the elements in series $(\sum n^{-2})^2$ and $\sum n^{-4}$ to show their rational equivalence?

Question

We know that $\sum n^{-2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots=\frac{\pi^2}{6}$ and $\sum n^{-4}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots=\frac{\pi^4}{90}$ from "high mathematics" research about Riemann's Zeta function. One could notice that $$\frac{(\sum n^{-2})^2}{\sum n^{-4}} = \frac{\zeta^2(2)}{\zeta(4)} = \frac{(\pi^2 / 6)^2}{\pi^4 / 90} = \frac52.$$ Let us suppose that we do not know anything about medium terms in this equation, so we end up with $$2\left(\sum n^{-2}\right)^2 = 5\sum n^{-4}.$$ Now if we move terms we find out some of them are cancelling, but we don't know by which scheme should be terms on left side be added to be equal to terms on the right side. Could this scheme be derived in some way?