This is a claim one of my students made without justification on his exam. It definitely wasn’t the right way to approach the problem, but now I’ve been nerdsniped into trying to figure out if it is true.

Let ai be a sequence of positive reals such that ∑a2i=∞. Then [0,1]2 can be covered by translates of the squares [0,ai]2.

It is definitely not enough that ∑a2i>1: You can’t cover a unit square with 5 axis-aligned squares of sidelength 1/√5.

**Answer**

We actually only need ∑a2i>4 for this to hold*. In particular, note that this condition implies that there is some *finite* set I of the indexing subset such that ∑i∈Ia2i>4.

Let bi be the largest power of two less than or equal to ai for each i∈I. Clearly, 2bi>ai. In particular, ∑i∈Ib2i>1. Now, we can inductively construct a by squares of side length bi for i∈I. To be specific, for each k, let ck be the number of i∈I such that bi=2−k. We are equivalently trying to cover [0,1]2 by a collection of squares of powers of two in side length, with ck copies of [0,2−k]2.

However, this is easy: Let Gk be the partition** of [0,1]2 into translates [0,2−k]2 in the obvious way (i.e. by slicing into 2k rows and columns). Note that Gk+1 is always a refinement** of Gk. Then, we can place the squares from biggest to smallest; we can greedily place as many of the biggest c1 squares into G1 as possible. Then, if the square is not full, we put as many c2 squares into the uncovered cells of G2 as possible – and so on. Since each is a refinement of the last, we can always place a new square in without overlap of a previous square – unless everything is already full. This process must terminate eventually, since we only have finitely many squares. However, since their areas sum to more than one, and they are being placed without overlap, they must actually cover the square.

(*This inequality actually doesn’t need to be strict; if you trace through the argument, the fact that the inequality 2bi>ai is strict lets us weaken the inequality for the sum)

(**This is all up to the boundaries of the cells, which might intersect. But since we’re dealing with a finite set of squares, the union is closed, so measure zero sets don’t matter anyways)

Here’s an illustration of the process, where the squares are assumed to have powers of two as side lengths, c1=2 and c2=5 and c3=9 and c4=12. Observe that there’s really nothing that could possibly go wrong:

**Attribution***Source : Link , Question Author : David E Speyer , Answer Author : Milo Brandt*