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)?


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
