# 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 :) – Zubin Mukerjee Feb 24 '15 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. – Moritz Feb 24 '15 at 23:15
• This was asked at MO: mathoverflow.net/questions/34145/… and it appears to still be an open problem. – Samuel Feb 24 '15 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. – Greg Martin Feb 24 '15 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? – marty cohen Feb 25 '15 at 4:02

Here is the desired packing:

Let $\mathcal{L}=\{L_1,L_2,\ldots\}$ denote the rectangles below the hyperbola $xy=1$, and let $\mathcal{U}$ denote those above the hyperbola. Index these sets in order of decreasing area. We reuse the construction of $\mathcal{L}$ from the original post.

We construct $\mathcal{U}$ inductively as follows. Suppose that $\mathcal{U}=\{U_1,U_2,\ldots,U_n\}$ have been constructed. Let $A_{n+1}$ be the top right corner of $L_{n+1}$, and note that $A_{n+1}$ lies on the hyperbola $xy=1$. Define $U_{n+1}$ to be the rectangle of largest area satisfying the following properties:

• $U_{n+1}$ is disjoint from $\{U_1,U_2,\ldots,U_n\}$

• $U_{n+1}$ has bottom left corner $A_n$

The tiling $\mathcal{L}$ represents the series $$\left(\frac11-\frac12\right)+\left(\frac13-\frac14\right)+\left(\frac15-\frac16\right)+\cdots,$$ and the tiling $\mathcal{U}$ represents the series $$\left(\frac12-\frac13\right)+\left(\frac14-\frac15\right)+\left(\frac16-\frac17\right)+\cdots$$ Their union $\mathcal{L}\cup \mathcal{U}$ represents the series $$\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots=1.$$

• These rectangles have the wrong sides. – mrf Jun 8 '15 at 5:18
• Sorry, I confused you with my second picture in the question. And in the comment thread, we kind of discussed this. But your rectangles are not the right dimensions. They have the right areas, but not the right dimensions. For instance, you have no rectangle that is literally $\frac{1}{3}\times\frac{1}{4}$. Instead, you have one that is $\frac{1}{2}\times\frac{1}{6}$. – alex.jordan Jun 8 '15 at 14:58