Can the squares with side $1/n$ be packed into a $1 \times \zeta(2)$ rectangle?

Solution 1:

This answer answers the question only approximately, but at least does so rigorously.

Theorem 4 from

Moon, J.W.; Moser, L., Some packing and covering theorems, Colloq. Math. 17, 103-110 (1967). ZBL0152.39502.

states: Let there be given a set of squares of total area $A$ the largest of which has side $D$. Such a set can be packed in any rectangle of area $2A$ and shorter side $B$, if $D\le B$.

Therefore if a packing exists of the largest $n-1$ squares into the rectangle of size $1 \times \zeta(2)$, and $A = \sum^\infty_n {1\over{n^2}}$ is the total area of the remaining squares, the remaining squares fit into a rectangle of size $1 \times \max(1/n, 2A)$ which can be adjoined to the larger one, giving a packing of the entire set of squares into a rectangle of size $1 \times (\zeta(2)+\max(1/n, 2A))$.

Since you state that a packing exists for the first $10^{12}$ squares, this means that a rectangle not much larger than $1 \times \zeta(2)$ certainly can be packed with all the squares (a naive bound on A yields a width of $\zeta(2) + 4\times 10^{-12}$, if I am not mistaken).