  Let $d$ 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 upon $t$, 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 $1/3 < t \leq 4/11 $ and
$n_0 = 1$ and builds upon recent work of Terence Tao in the two-dimensional
case.
