The following question was a research exercise (i.e. an open problem) in R. Graham, D.E. Knuth, and O. Patashnik, "Concrete Mathematics", 1988, chapter 1.
It is easy to show that
$$\sum_{1 \leq k } \left(\frac{1}{k} \times \frac{1}{k+1}\right) = \sum_{1 \leq k } \left(\frac{1}{k} – \frac{1}{k+1}\right) = 1.$$
The product $\frac{1}{k} \times \frac{1}{k+1}$ is equal to the area of a $\frac{1}{k}$by$\frac{1}{k+1}$ rectangle. The sum of the areas of these rectangles is equal to 1, which is the area of a unit square. Can we use these rectangles to cover a unit square?
Is this problem still open?
What are the best results we know about this problem (or its relaxations)?
Best Answer
This problem actually goes back to Leo Moser.
The best result that I'm aware of is due to D. Jennings, who proved that all the rectangles of size $k^{-1} × (k + 1)^{-1}$, $k = 1, 2, 3 ...$, can be packed into a square of size $(133/132)^2$ (link).
Edit 1. A web search via Google Scholar gave a reference to this article by V. Bálint, which claims that the rectangles can be packed into a square of size $(501/500)^2$.
Edit 2. The state of art of this and related packing problems due to Leo Moser is discussed in Chapter 3 of "Research Problems in Discrete Geometry" by P.Brass, W. O. J. Moser and J. Pach. The problem was still unsettled as of 2005.