<s>[INST] Can we cover the unit square by these rectangles? 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)?

 [/INST] it is hard to prove this problem directly, but it is not hard to prove (as someone mentioned in some comments) :
If (n-1) rectangles has been put into the 1x1 square, then all rectangles can be put into the square of length (1+1/n)
Proof:
we have put (n-1) rectangles into unit square
denote $P_n = \frac{1}{n} \times \frac{1}{n+1} \sim \frac{1}{n} \times \frac{1}{n}$ , put a rectangle into a square, it is ok
put rectangles from n to infinity in this way of rows and columns into a large rectangle
first row, $P_n  \cdots P_{2n-1}$
second row, $P_{2n} \cdots P_{4n-1}$
third row, $P_{4n} \cdots P_{8n-1}$
.... as the following picture showing

the left vertical is the familiar geometric sequence , the length is :  $\frac{1}{n} + \frac{1}{2n} + \frac{1}{4n} + \cdots  = \frac{2}{n} $ ,
the first horizontal row  is the sum of part of harmonic sequence ，$\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n-1} $
it is easy to estimate the upper bound as :
$$
\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n-1} + \frac{1}{2n} - \frac{1}{2n} \\
= \frac{1}{n} +( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n-1} + \frac{1}{2n}) - \frac{1}{2n} \\
< \frac{1}{n} +( \ln2) - \frac{1}{2n} = \ln 2 +  \frac{1}{2n}
$$
(the sum in bracket is easy to check, take $\frac{1}{x}$, the area of integral under the curve is larger than sum of rectanlges)
the  length of first row is less than $\ln 2 + \frac{1}{2n}$ ,
the  length of second row is less than $\ln 2 + \frac{1}{4n}$ , ...
the first row is longer than the above rows , when n takes big number, like n = 1000, this length $\ln 2 + \frac{1}{2n}$ is obviously less than 1,
for $\ln 2 < 0.7$ .
then rectangles from n to infinity are put into a rectangle of $\frac{2}{n} \times (\ln2 + \frac{1}{2n})$ , and divide this rectangle into two equal smaller rectangles $\frac{1}{n} \times (\ln2 + \frac{1}{2n})  $ along the unit square as following
then we prove all rectangles can be put into a square of length (1+1/n),
although in two stripes it is not efficient to pack them, and can be improved by some ways obviously.
Paulhus gave a graph of 1000 rectangles in unit sqaure in his paper (An Algorithm for Packing Squares) as

and I recovered a result of 1000 rectangle from an algorithm of Antal Joós in his paper (On packing of rectangles in a rectangle) similar to Paulhus' method, (took 2 seconds by a Mathematica program ) like

as Paulhus result, so at least the result (1+ 1/1000) is better than Balint's result in this tricky way , and it seems not hard to go further like 10000
and Paulhus even claimed that he had put 10^9 rectangles into unit sqaure by computer !!!???
if someone has proved the final problem or check results in computer in very large number, please tell me
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