Perfectly packing a cube by cubes of nearly harmonic sidelength

Abstract

AbstractLet d be an integer greater than $1$ , and let t be fixed such that $\frac {1}{d} < t < \frac {1}{d-1}$ . We prove that for any $n_0$ chosen sufficiently large depending on t, the d-dimensional cubes of sidelength $n^{-t}$ for $n \geq n_0$ can perfectly pack a cube of volume $\sum _{n=n_0}^{\infty } \frac {1}{n^{dt}}$ . Our work improves upon a previously known result in the three-dimensional case for when $\frac {1}{3} < t \leq \frac {4}{11} $ and $n_0 = 1$ and builds upon recent work of Terence Tao in the two-dimensional case.