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)?

 A: 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.
A: Using an ad-hoc Mathematica program, I was able to pack the first $10.000$ rectangles into the unit square. I use no backtracking, but the next square is always placed in a way that the remaining $2 - 3$ rectangles (with sidelength $a$, $b$) minimize the square of their difference $(a-b)^2$. During placement I consider all of the up to $6$ possible placements for all rectangles in my list, where the new rectangle may fit in. Mathematica program available upon request at mail@thomas-vogler.de
A: I think the following result of Greg Martin related to this question deserves to be mentioned here. (It is already referenced in Chapter 3 of the book "Research Problems in Discrete Geometry" by P. Brass, W. O. J. Moser and J. Pach, which is mentioned in the accepted answer.)
Theorem: Let $\mathcal{A}$ be the countable collection of rectangles with dimensions $\frac{1}{1} \times \frac{1}{2}$, $\frac{1}{2} \times \frac{1}{3}$, $\frac{1}{3} \times \frac{1}{4}$, $\dots$ . If $\mathcal{A}$ can be packed into a square of area $1+\epsilon$ for every $\epsilon > 0$, then $\mathcal{A}$ can be packed into a square of area $1$.
The proof is a compactness argument. Under a suitable topology, valid packings (of various packing problems) form compact subsets of the space of all possible positionings of the tiles. If the weakened problem can be solved for every $\frac{1}{k} >0$, then one can extract a subsequence of the sequence of these solutions which converges to a solution of the problem.
A: It's been a long time since I considered this problem, so prompted by seeing this question I was intrigued to discover more on V. Bálint's bound of $(501/500)^ 2.$
A quick search revealed Bálint's paper A Packing Problem and Geometrical Series.
In this article it is only stated that with some patience one can pack the first 499 rectangles into the unit square. However, the main difficulty of the problem is to pack the larger rectangles and so it would have been nice to see a demonstration.
Bálint addresses the question in Two Packing Problems but I do not have easy access to this and so now I'm concerned that a similar claim, without a demonstration, may have been made in this paper too.
Please could someone with access to the paper lay my concern to rest?
I would very much like to have confidence in the later bound as its validity makes the problem yet more interesting. Can we get arbitrarily close to 1? I still see no good reason why this should be the case but it's a fascinating possibility that hints at the prospect of something quite deep going on.
A: I have written the following notes as an effort to understand the problem from one perspective.
In short, a function called packing efficiency(aka. packing density) is defined, the problem characterized as preimage of $1$ and two continuity properties proved.
The next step (not written in the notes above, pardon my laziness) would be combining these two properties to prove an infinite version of the first property and also of the second property (since both can be viewed as special cases) with the help of a theorem on interchange of limits, or by directly error estimates. It should be in $l^2$ norm. (if dimension $p$, should be in $l^p$ norm.) (Note carefully that the first property alone would not suffice for this purpose, because the estimates in the proof of the first property is not strong enough to deal with, say, boxes with sidelength $1/n$, whose sum of sidelengths diverge but sum of areas converge.)
The conclusion is, the continuity results would mean that, in a sense, the infinite problem is not so much different from the associated finite problem, which is NP-hard.
Maybe the continuity results have had already been known or at least intuitive, but I didn't find any previous references mentioning such results, anyway here's my two cents.
By the way, we can view continued fractions as perfect packing of maximal squares into rectangles.
There is also the following theory and stories on the related problem of dissecting a given square into squares: squaring
A: 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
