Packing squares into a rectangle with a relatively small area

In this article we prove that all squares with sides of lengths 1m+1,1m+2,⋯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{1}{m+1}, \frac{1}{m+2}, \dots $$\end{document} can be packed in a rectangle with a one side fixed and a relatively small area. We give a similar result for all of three Moser’s Packing Problems.


Introduction
Amram Meir and Leo Moser [7] formulated the following problem. What is the smallest area of a rectangle into which we can pack squares with sides of lengths 1 2 , 1 3 , . . . ? Attempts to find an answer to the above question went in two directions.
One was checking for what values of t > 1 2 we can pack squares with sides 1 2 t , 1 3 t , . . . into a rectangle with area ∞ n=2 1 n 2t . The best answer was found by Terence Tao [9], who proved that for any real number t ∈ 1 2 , 1 and for any sufficiently large natural number m, squares with side lengths of 1 m t , 1 (m+1) t , . . . can be packed into a square of area ∞ n=m 1 n 2t . Unfortunately, his arguments did not seem to extend to the critical case t = 1.
On the other hand, considerations have been reduced to packing some finite number of initial squares into a rectangle with area ∞ n=2 1 n 2 = π 2 6 − 1 and proving that the remaining squares can be packed into a rectangle with a relatively small area. The best known estimate was found by Marc Paulhus [8], who showed that squares with side lengths of 1 2 , . . . , 1 m for m = 10 9 could be packed into a rectangle with one side equal to 1 2 and area equal to π 2 6 −1. Next, he gave (as it later turned out) an incorrect proof from which it was concluded that the remaining squares could be packed into a certain rectangle. Antal Joós [5] found a mistake in Paulhus' proof, and Paulina Grzegorek and Janusz Januszewski [2] gave the correct proof in the special case when m = 10 9 and one of the sides of the rectangle is equal to 1 2 . Many authors have considered packaging problems of a similar nature (see also [1,3,4,6]). The main goal of this article is to provide them with a mathematical basis for a such considerations. We formulate and prove a general version of the result proven by Paulina Grzegorek and Janusz Januszewski. The main result of the paper is Theorem 5.1 giving a formal proof that for each m we can pack all squares with sides of lengths 1 m+1 , 1 m+2 , . . . into a rectangle of a relatively small area.
Amram Meir and Leo Moser [7] also formulated two analogous and similar problems using the identities The main result of this paper also applies to the above two problems.
For the problem related with last identity ∞ n=1 1 n(n+1) = 1, the best known estimate was found Mingliang Zhu and Antal Joós [10]. They showed that the first 1.35 * 10 11 rectangles with dimensions 1 1 × 1 2 , 1 2 × 1 3 . . . can be packed into the unit square, and as a consequence that all of these rectangles can be packed into a square with side of length 1 + for < 1.49 * 10 −11 .

Simple inequalities
To begin with, we will give four simple inequalities, which we will use later in this article.
• The function ln(x) + 1 2x is increasing on the interval 1 2 , ∞ because its derivative for every real number x ∈ 1 2 , ∞ . In particular, if α ≥ 0 is a real number, then for each positive integer number n because e α n ≤ e α n. • For each real number y ∈ (0, 1), there is an inequality Vol. 97 (2023) Packing squares into a rectangle 807 because the expression x + ln(1 − x) + x 2 2 is equal to 0 for x = 0, and its derivative for any real number x ∈ (0, 1). • For each real number y ∈ 0, 1 3 , there is an inequality 3 2 is equal to 0 for x = 0 , and its derivative for any real number x ∈ 0, 1 3 . • Let α > 0 be a real number and n be a natural number such that n ≥ 4 e α −1 . Then e α n = n + (e α − 1)n ≥ n + 4.

Classical harmonic number inequalities
For any natural number n let H n = n j=1 1 j denote the n − th harmonic number, and γ = lim n→∞ (H n − ln (n)) be the Euler-Mascheroni constant gamma. As we see, harmonic numbers can be approximated by the natural logarithm. The following two facts give the precision of this approximation.
Proof. We define two sequences (a n ) ∞ n=1 , (b n ) ∞ n=1 as follows. Let for any natural number n. Obviously, by the definition of the constant γ, each of these two sequences tends to γ when n goes to infinity, i.e., lim n→∞ a n = γ, lim n→∞ b n = γ.
We will prove that the sequence (a n ) ∞ n=1 is decreasing, and the sequence (b n ) ∞ n=1 is increasing. Now we will prove that the difference a n+1 −a n is negative. By the definition of the sequence (a n ) ∞ n=1 in 2.5, we have Then, using 2.2, we obtain a n+1 − a n < − 1 2(n + 1) 2 + 1 2(n + 1)(n + 2) < 0 because n + 1 < n + 2. Therefore, since the sequence (b n ) ∞ n=1 is decreasing and goes to γ, each element of this sequence is larger than γ, i.e. a n > γ for any natural number n. Thus, we have proved the first inequality.
Similarly, we prove the second inequality. By the definition of the sequence In the special case when n = 1, we obtain Therefore, since the sequence (b n ) ∞ n=1 is increasing and goes to γ, each element of this sequence is less than γ, i.e. b n < γ for any natural number n. Thus, we have proved the second inequality.
Fact 2.1 can be weakened to a more popular and friendly form.

Some convex function
Let n, N be fixed positive integer numbers. We define the function G : R + → R by .
The following three lemmas will be useful in approximating the value of the function G on the interval n + 2, n+N 2 . The first lemma gives the upper bound of G(n + 2) as follows.
Proof. Let n, N be fixed natural numbers such that n ≥ 17 and N ≥ n + 7.
The following inequalities are satisfied.
We divide both sides by Nn(n 2 + n − 3)(n + 1)(n + 2). Hence , Next, we decompose each fraction into the difference of two fractions and then we move its subtrahend to the other side 1 which concludes the proof.
The second lemma shows that the largest value of the function G on the interval n + 2, n+N 2 is G(n + 2). Lemma 3.2. Let n, N be fixed natural numbers such that n ≥ 14 and N ≥ n + 6.
We put C =
Both sides of the inequality are natural numbers, so Therefore from 3.3 we obtain Obviously there are three inequalities nA because A ≥ n + 1. Multiplying these three inequalities by sides, we obtain 2 n(n + 1) Therefore, by 3.5 we have 2 n(n + 1) Recall that A + B ≤ n + N (see 3.3). Thereby 2 n(n + 1) Since n < N, we obtain 3 n(n + 1) Vol. 97 (2023) Packing squares into a rectangle 813 Obviously ln(1 − y) < −y for any real number y ∈ (0, 1) (see 2.2). Then

Finally by applying both inequalities of Fact 2.1 and the initial condition
Hence, by 3.6 we obtain ln(AB) > ln(nN ) + ln 1 − 3 n(n + 1) , and thus In the end, we will give an upper estimate of the sum 1 what was there to prove.

Tools
The basic tool of this article will be the following two propositions.
The second proposition is as follows.

Proposition 4.2.
Let n, N be natural numbers such that n ≥ 17 and N ≥ n + 7.

For each natural numbers A, B such that n < A < B ≤ N and
there is an inequality Remark. For N > 2n, the constant n+1 nN can be improved to 1 N (1− 1 n+1 ) +1 , but it is not asymptotically more useful.
Proof. Since for any number A it holds that 1 A < 1 A−1 , it is sufficient to prove Proposition 4.2 for the minimum value of A, i.e. for A = n+1 and for A > n+1 such that A is the smallest number satisfying the condition 4.2.
First, we prove Proposition 4.2 in the special case when A = n + 1. Let n, N be natural numbers. Let A, B be natural numbers such that A = n + 1 and A < B ≤ N and the condition 4.2 holds. Using 4.2 we obtain Vol. 97 (2023) Packing squares into a rectangle 815 because j ≤ N . Therefore, (n + 1)B > Nn.
Now we will prove Proposition 4.2 in the second case. Let n, N be natural numbers such that n ≥ 17 and N ≥ n + 7. Let A, B be natural numbers such that n < A < B ≤ N and such that A is the smallest number satisfying the condition 4.2, i.e., Using the above inequality we obtain Both sides of the above equation are natural numbers. Thus In particular, since A < B, we obtain Thus A ∈ n + 2, N + n 2 .
Recall that n < A < B ≤ N are natural numbers such that n ≥ 17, N ≥ n +7, and A ∈ n + 2, N +n Nn .

Main results
Proof. Let α > 0 be a real number and m ≥ 17 be a natural number such that m ≥ 4 (e α −1) . We define the sequence (n j ) ∞ j=0 by the recursion n j+1 = e α e α n j with the initial condition n 0 = m. The sequence (n j ) ∞ j=0 is obviously nondecreasing. Moreover, from (2.4) we obtain that for each non-negative integer number j it holds that n j+1 ≥ n j + 8. In addition, by the definition of the floor function, for each index j we obtain Thus Let t be a natural number. We use (5.1) t times Vol. 97 (2023) Packing squares into a rectangle 817 Let α > 0 be a real number and m be a natural number. Let t = 0. We put n = n t and N = n t+1 , and h t = 1 n+1 + n+1 nN . Take a rectangle of size α × h t , i.e., with side length α and height length h t . Using Proposition 4.1, we can inscribe squares with side lengths 1 n+1 , . . . , 1 e α n into the bottom row, i.e., we inscribe the first square into the bottom left corner of the rectangle, and each successive square is inscribed as a tangent to the previous square and to the bottom side of length α. Similarly, starting from the upper left corner, we can inscribe squares with side lengths 1 N , . . . , 1 e α n +1 into the upper row as tangents to the upper side of length α. An example of such packing for α = 1 5 and n = 23 is shown in Fig. 1.
We claim that the squares of the bottom row do not overlap the squares of the top row. Otherwise, if a certain square of the lower row with side length 1 A overlapped a certain square of the lower row with side length 1 B , then the inequality 1 A + 1 B > h t would be satisfied. Of course, then n < A < B ≤ N and the sum of the lengths of the sides of the squares from 1 N to 1 B+1 would be less than that of those from 1 n+1 to 1 A , i.e., H A − H n > H N − H B would also be satisfied. Thus we would get a contradiction to Proposition 4.2.
We repeat the reasoning for t = 0, 1, . . . and then we merge all the rectangles into one large rectangle. Therefore, we can inscribe all squares with sides of length 1 m+1 , 1 m+2 , . . . into a rectangle with dimensions α × , where Thus since n j+1 > n j + 1, we obtain Thereby, we conclude from (5.2) that Let α > 0 be a real number. For a given natural number m ≥ 17 such that m ≥ 4 (e α −1) , let S denote the sum of the areas of all squares with sides of lengths 1 m+1 , 1 m+2 , . . . , and let R denote the area of the rectangle with dimensions α × obtained from Theorem 5.1. Proof. Since j < j + 1 for any natural number j, we obtain Using the above observation and The main implication of Corollary 5.2 is that if we want to improve some result on packaging in an asymptotically observable way, we need to pack more initial squares into a rectangle.
Since for any natural number j, a rectangle with dimensions 1 j × 1 j+1 is contained in a square with side 1 j , and a square with side 1 2j+1 is contained in a square with side 1 2j , then the following two corollaries can be concluded from Theorem 5.1.