# Regular way to fill a $1\times1$ square with $\frac{1}{n}\times\frac{1}{n+1}$ rectangles?

The series $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1$$ suggests it might be possible to tile a $1\times1$ square with nonrepeated rectangles of the form $\frac{1}{n}\times\frac{1}{n+1}$. Is there a known regular way to do this? Just playing and not having any specific algorithm, I got as far as the picture below, which serves more to get a feel for what I am looking for.

I think some theory about Egyptian fractions would help. It's nice for instance in the center where $\frac13+\frac14+\frac16+\frac14=1$. And on the right edge where $\frac12+\frac13+\frac16=1$.

Side note: The series is $\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots$. The similar looking $\left(\frac11-\frac12\right)+\left(\frac13-\frac14\right)+\left(\frac15-\frac16\right)+\cdots$ sums to $\ln(2)$, and there is a nice picture for that, if you interpret $\ln(2)$ as an area under $y= \frac{1}{x}$:

• Wow! This is the most interesting question I've seen on this site in a while :) Feb 24, 2015 at 23:12
• This is a research level problem in "Concrete Mathematics" (2nd edition) by Grapham, Knuth, Patashnik (ISBN-10: 0201558025): see page 66, exercise 37. In the section with hints you will find that every one of the authors has a different opinion. Feb 24, 2015 at 23:15
• This was asked at MO: mathoverflow.net/questions/34145/… and it appears to still be an open problem. Feb 24, 2015 at 23:20
• I wrote a paper related to this (math.ubc.ca/~gerg/index.shtml?abstract=CTGP). Therein you can find a reference to this problem statement, as well as my reason for believing that such a packing is possible. Feb 24, 2015 at 23:40
• Since no one seems to know how to fit the rectangles into a square of area equal to the sum of their areas, how about this question: What is the smallest square that contains all the $\frac1{n}\times \frac1{n+1}$ rectangles with no overlaps? Feb 25, 2015 at 4:02