Packing squares into a rectangle with a relatively small area

Abstract

In this article we prove that all squares with sides of lengths $\frac{1}{m+1}, \frac{1}{m+2}, \dots$ 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.

1 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 $\frac{1}{2}, \frac{1}{3}, \dots$? Attempts to find an answer to the above question went in two directions.

One was checking for what values of $t>\frac{1}{2}$ we can pack squares with sides $\frac{1}{2^t}, \frac{1}{3^t}, \dots$ into a rectangle with area $\sum _{n=2}^{\infty } \frac{1}{n^{2t}}$. The best answer was found by Terence Tao [9], who proved that for any real number and for any sufficiently large natural number m, squares with side lengths of $\frac{1}{m^t}, \frac{1}{(m+1)^t}, \dots$ can be packed into a square of area $\sum _{n=m}^{\infty } \frac{1}{n^{2t}}$. 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 ${\sum _{n=2}^{\infty } \frac{1}{n^2}=\frac{\pi ^2}{6}-1}$ 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 $\frac{1}{2}, \dots , \frac{1}{m}$ for $m=10^9$ could be packed into a rectangle with one side equal to $\frac{1}{2}$ and area equal to $\frac{\pi ^2}{6}-1$. 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 $\frac{1}{2}$.

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 $\frac{1}{m+1}, \frac{1}{m+2}, \dots$ into a rectangle of a relatively small area.

Amram Meir and Leo Moser [7] also formulated two analogous and similar problems using the identities ${\sum _{n=1}^{\infty } \frac{1}{(2n+1)^2}=\frac{\pi ^2}{8}-1}$ and ${\sum _{n=1}^{\infty } \frac{1}{n(n+1)} =1}$. The main result of this paper also applies to the above two problems.

For the problem related with last identity ${\sum _{n=1}^{\infty } \frac{1}{n(n+1)} =1}$, the best known estimate was found Mingliang Zhu and Antal Joós [10]. They showed that the first $1.35*10^{11}$ rectangles with dimensions $\frac{1}{1} \times \frac{1}{2}$, $\frac{1}{2} \times \frac{1}{3} \dots$ 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 $1+\epsilon$ for $\epsilon < 1.49 * 10^{-11}$.

2 Approximation of harmonic numbers

2.1 Simple inequalities

To begin with, we will give four simple inequalities, which we will use later in this article.

• The function $\ln (x)+\frac{1}{2x}$ is increasing on the interval because its derivative

  $$\begin{aligned} \left( \ln (x)+\frac{1}{2x}\right) '=\frac{x-\frac{1}{2}}{x^2}>0 \end{aligned}$$

  for every real number . In particular, if $\alpha \ge 0$ is a real number, then

  $$\begin{aligned} \ln ( \left\lfloor e^{\alpha } n \right\rfloor )+ \frac{1}{2 \left\lfloor e^{\alpha } n \right\rfloor } \le \ln ( e^{\alpha } n)+ \frac{1}{2 e^{\alpha } n} \end{aligned}$$

  (2.1)

  for each positive integer number n because $\left\lfloor e^{\alpha } n \right\rfloor \le e^{\alpha }n$.

• For each real number , there is an inequality

  $$\begin{aligned} y+\ln (1-y)<-\frac{y^2}{2} \end{aligned}$$

  (2.2)

  because the expression $x+\ln (1-x)+\frac{x^2}{2}$ is equal to 0 for $x=0$, and its derivative

  $$\begin{aligned} \left( x+\ln (1-x)+\frac{x^2}{2}\right) '=\frac{-x^2}{1-x} < 0 \end{aligned}$$

  for any real number .

• For each real number , there is an inequality

  $$\begin{aligned} y+\ln (1-y) > - \frac{y^2}{2} - \frac{y^3}{2} \end{aligned}$$

  (2.3)

  because the expression $x+\ln (1-x) + \frac{x^2}{2} + \frac{x^3}{2}$ is equal to 0 for $x=0$ , and its derivative

  $$\begin{aligned} \left( x+\ln (1-x) + \frac{x^2}{2} + \frac{x^3}{2}\right) ' = \frac{3x^2\left( \frac{1}{3}-x\right) }{2(1-x)} > 0 \end{aligned}$$

  for any real number .

• Let $\alpha >0$ be a real number and n be a natural number such that ${n \ge \frac{4}{e^{\alpha }-1}}$. Then

  $$\begin{aligned} \left\lfloor e^{\alpha } n \right\rfloor = \left\lfloor n+(e^{\alpha } -1)n \right\rfloor \ge n+4. \end{aligned}$$

  Furthermore, using the above inequality twice, we obtain

  $$\begin{aligned} \left\lfloor e^{\alpha } \left\lfloor e^{\alpha }n \right\rfloor \right\rfloor \ge n+8. \end{aligned}$$

  (2.4)

2.2 Classical harmonic number inequalities

For any natural number n let $H_{n}=\sum \limits _{j=1}^{n} \frac{1}{j}$ denote the $n-th$ harmonic number, and ${\gamma =\lim _{n\rightarrow \infty } \left( H_{n}-\ln \left( n\right) \right) }$ 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.

Fact 2.1

For each positive integer number $n$, there are inequalities

$$\begin{aligned} \frac{1}{2(n+1)}<H_{n}-\ln (n)-\gamma <\frac{1}{2n}. \end{aligned}$$

Proof

We define two sequences $( a_{n})_{n=1}^{\infty }$, $( b_{n} )_{n=1}^{\infty }$ as follows. Let

$$\begin{aligned} \left\{ \begin{array}{c} a_{n} = H_{n} -\ln (n) - \frac{1}{2(n+1)}, \\ b_{n} = H_{n} -\ln (n) - \frac{1}{2n} \end{array} \right. \end{aligned}$$

(2.5)

for any natural number $n$. Obviously, by the definition of the constant $\gamma$, each of these two sequences tends to $\gamma$ when $n$ goes to infinity, i.e.,

$$\begin{aligned} \left\{ \begin{array}{c} \lim _{n\rightarrow \infty } a_{n} = \gamma , \\ \lim _{n\rightarrow \infty } b_{n} = \gamma . \end{array} \right. \end{aligned}$$

We will prove that the sequence $( a_{n} )_{n=1}^{\infty }$ is decreasing, and the sequence $( b_{n} )_{n=1}^{\infty }$ is increasing.

Now we will prove that the difference $a_{n+1}-a_{n}$ is negative. By the definition of the sequence $( a_{n} )_{n=1}^{\infty }$ in 2.5, we have

$$\begin{aligned} a_{n+1}-a_{n}&= H_{n+1} -\ln (n+1) - \frac{1}{2(n+2)} - \left( H_{n} -\ln (n) - \frac{1}{2(n+1)} \right) \\&= \left( H_{n+1}-H_{n} \right) + \left( \ln (n)-\ln (n+1) \right) + \frac{1}{2}\left( \frac{1}{n+1}-\frac{1}{n+2}\right) \\&= \frac{1}{n+1}+\ln \left( 1-\frac{1}{n+1}\right) +\frac{1}{2(n+1)(n+2)}. \end{aligned}$$

Then, using 2.2, we obtain

$$\begin{aligned} a_{n+1}-a_{n}&< -\frac{1}{2(n+1)^2}+\frac{1}{2(n+1)(n+2)} \\&< 0 \end{aligned}$$

because $n+1 < n+2$. Therefore, since the sequence $(b_{n})_{n=1}^{\infty }$ is decreasing and goes to $\gamma$, each element of this sequence is larger than $\gamma$, i.e.

$$\begin{aligned} a_{n} > \gamma \end{aligned}$$

for any natural number $n$. Thus, we have proved the first inequality.

Similarly, we prove the second inequality. By the definition of the sequence $( b_{n} )_{n=1}^{\infty }$ in 2.5, we have

$$\begin{aligned} b_{n+1}-b_{n}&= H_{n+1}-\ln (n+1)-\frac{1}{2(n+1)}-\left( H_{n}-\ln (n)-\frac{1}{2n} \right) \\&= \left( H_{n+1}-H_{n} \right) + \left( \ln (n)-\ln (n+1) \right) + \frac{1}{2}\left( \frac{1}{n}-\frac{1}{n+1}\right) \\&= \frac{1}{n+1}+\ln \left( 1-\frac{1}{n+1}\right) +\frac{1}{2n(n+1)}. \end{aligned}$$

In the special case when $n=1$, we obtain ${b_2-b_1=\frac{1}{2}+\frac{1}{4}-\ln (2)>0}$ because $\ln (2) \approx 0.69$. Otherwise, when $n+1 \ge 3$, then using  (2.3), we obtain

$$\begin{aligned} b_{n+1}-b_{n}&> -\frac{1}{2(n+1)^2}-\frac{1}{2(n+1)^3} + \frac{1}{2n(n+1)} \\&= \frac{1}{2n(n+1)^3}\left( -n(n+1)-n+(n+1)^2 \right) \\&= \frac{1}{2n(n+1)^3} \\&> 0. \end{aligned}$$

Therefore, since the sequence $(b_{n})_{n=1}^{\infty }$ is increasing and goes to $\gamma$, each element of this sequence is less than $\gamma$, i.e.

$$\begin{aligned} b_{n} < \gamma \end{aligned}$$

for any natural number $n$. Thus, we have proved the second inequality. $\square$

Fact 2.1 can be weakened to a more popular and friendly form.

Fact 2.2

Let $n$ be a positive integer number. Then, there are inequalities

$$\begin{aligned} \log (n)<H_{n}-\gamma <\log (n+1). \end{aligned}$$

Proof

Let $n$ be a positive integer number. Using Fact 2.1, we obtain

$$\begin{aligned} H_{n}-\gamma&> \ln (n)+\frac{1}{2(n+1)} \\ {}&>\ln (n) \end{aligned}$$

and

$$\begin{aligned} H_{n}-\gamma&< \ln (n)+\frac{1}{2n} \\ {}&\le \ln (n) + \frac{1}{n+1} \\ {}&= \ln \left( n+1\right) + \ln \left( 1-\frac{1}{n+1}\right) +\frac{1}{n+1}. \end{aligned}$$

Thus, by using 2.2 we arrive at

$$\begin{aligned} H_{n}-\gamma&< \ln (n+1)-\frac{1}{2(n+1)^2} \\ {}&< \ln (n+1). \end{aligned}$$

$\square$

3 Some convex function

Let n, N be fixed positive integer numbers. We define the function $G: {\mathbb {R}}_{+} \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} G(x)=\frac{1}{x}+\frac{x(n+1)}{N(n^2+n-3)}. \end{aligned}$$

(3.1)

The following three lemmas will be useful in approximating the value of the function G on the interval . The first lemma gives the upper bound of $G(n+2)$ as follows.

Lemma 3.1

Let n, N be fixed natural numbers such that $n \ge 17$ and $N \ge n + 7$. Then, there is an inequality

$$\begin{aligned} G(n+2) \le \frac{1}{n+1}+\frac{n+1}{Nn}. \end{aligned}$$

Proof

Let n, N be fixed natural numbers such that $n \ge 17$ and $N \ge n + 7$. The following inequalities are satisfied.

$$\begin{aligned} \left\{ \begin{array}{rclcrcl} (n+1)(n+3)&{}{}\le &{}{}(n-2)(n+7) &{}{} \text{ because } &{}{} 17&{}{}\le &{}{}n, \\ n+7&{}{}\le &{}{}N, &{}{}&{}{}&{}{}&{}{}\\ (n^2-2^2)(n+1)&{}{}\le &{}{}(n^2+n-3)n &{}{} \text{ because } &{}{} -4(n+1)&{}{}\le &{}{}-3n. \end{array} \right. \end{aligned}$$

Multiplying the inequalities by sides and canceling the common factor ${(n-2)(n+7)}$, we obtain

$$\begin{aligned} (n+1)^2(n+2)(n+3) \le N n(n^2+n-3). \end{aligned}$$

We divide both sides by $N n (n^2+n-3) (n+1)(n+2)$. Hence

$$\begin{aligned} \frac{n+3}{ N n(n^2+n-3)} (n+1) \le \frac{1}{(n+1)(n+2)}, \end{aligned}$$

and thus

$$\begin{aligned} \frac{ n(n+2)-(n^2+n-3)}{ N n(n^2+n-3)} (n+1) \le \frac{(n+2)-(n+1)}{(n+1)(n+2)}. \end{aligned}$$

Next, we decompose each fraction into the difference of two fractions and then we move its subtrahend to the other side

$$\begin{aligned} \frac{1}{n+2} + \frac{ n+2}{ N (n^2+n-3)} (n+1)&\le \frac{1}{n+1} + \frac{ 1}{ N n} (n+1), \\ G(n+2)&\le \frac{1}{n+1} + \frac{n+1}{ N n}, \end{aligned}$$

which concludes the proof. $\square$

The second lemma shows that the largest value of the function G on the interval is $G(n+2)$.

Lemma 3.2

Let n, N be fixed natural numbers such that ${n \ge 14}$ and ${N \ge n + 6}$. For each , there is an inequality

$$\begin{aligned} G(x) \le G(n+2). \end{aligned}$$

Proof

Let n, N be fixed natural numbers such that ${n \ge 14}$ and ${N \ge n + 6}$. The following inequalities are satisfied:

$$\begin{aligned} \left\{ \begin{array}{rclcrcl} (n+1)(n+2)&{}\le &{}(n-2)(n+6) &{} \text {because} &{} 14&{}\le &{}n, \\ (N+n)(n+6)&{}\le &{}2 N (n+3) &{} \text {because} &{} n (n + 6)&{}\le &{}N n,\\ (n-2)(n+3)&{}<&{}n^2+n-3 &{} \text {because} &{} -6&{}<&{}-3. \end{array} \right. \end{aligned}$$

Multiplying the inequalities by sides and canceling the common factor ${(n-2)(n+3)(n+6)}$, we obtain

$$\begin{aligned} (N+n)(n+1)(n+2) < 2 N (n^2 +n -3). \end{aligned}$$

We divide both sides by $2N(n^2+n-3)(n+2)(n+N)$ and multiply by ${(N+n)-2(n+2)=N-(n+4) >0}$. Then

$$\begin{aligned} \frac{(N+n)-2(n+2)}{2 N (n^2 +n -3)} (n+1) < \frac{(N+n)-2(n+2)}{(n+2)(N+n)} . \end{aligned}$$

Next, we decompose each fraction into the difference of two fractions and then we move its subtrahend to the other side

$$\begin{aligned} \frac{2}{n+N} + \frac{(n+N)}{2N(n^2+n-3)} (n+1) < \frac{1}{n+2} + \frac{(n+2)}{N(n^2+n-3)} (n+1). \end{aligned}$$

Thus

$$\begin{aligned} G\left( \frac{n+N}{2} \right) < G(n+2). \end{aligned}$$

(3.2)

We put $C=\frac{(n+1)}{N(n^2+n-3)}$. The function G (see  (3.1) for definition) has a second derivative

$$\begin{aligned} \left( G(x)\right) ''&= \left( \frac{1}{x}+Cx\right) '' \\&=\frac{2}{x^3} \\&>0 \end{aligned}$$

for $x>0$. Therefore, the function G is convex on the set of positive real numbers, and then for , the largest value of the function G(x) is taken at either of the two ends of the interval. By 3.2, we obtain that

$$\begin{aligned} G(x) \le G(n+2) \end{aligned}$$

for each real number . $\square$

The third lemma, with some additional assumptions, shows that the upper bound of a sum $\frac{1}{A}+\frac{1}{B}$ is G(A).

Lemma 3.3

Let n, A, B, N be positive integer numbers such that ${n<A<B \le N}$ and $A+B \le N+n$ and $H_{n}+H_{N} < H_{A}+H_{B}$. Then, there is an inequality

$$\begin{aligned} \frac{1}{A}+\frac{1}{B} < G(A). \end{aligned}$$

Proof

Let n, A, B, N be positive integer numbers such that $n<A<B\le N$ and

$$\begin{aligned} A+B \le N+n \end{aligned}$$

(3.3)

and

$$\begin{aligned} H_{n}+H_{N} < H_{A}+H_{B}. \end{aligned}$$

(3.4)

By applying the above condition and both inequalities of Fact 2.2, we obtain

$$\begin{aligned} \ln (N)+ \ln (n)&< H_{N}+H_{n} - 2 \gamma \\&< H_{A}+H_{B} -2 \gamma \\&< \ln (A+1) + \ln (B+1). \end{aligned}$$

Thus

$$\begin{aligned} N n< (A+1)(B+1). \end{aligned}$$

Both sides of the inequality are natural numbers, so

$$\begin{aligned} N n- A B&\le (A+1)(B+1)-1-AB \\&= A+B. \end{aligned}$$

Therefore from 3.3 we obtain

$$\begin{aligned} N n- A B \le n+ N. \end{aligned}$$

(3.5)

Obviously there are three inequalities

• $1 > \frac{A+B}{2B}$ because $B>A$,

• $1 > \frac{n+N}{2N}$ because $N>n$,

• $\frac{2}{n(n+1)} \ge \frac{2}{n A}$ because $A \ge n+1$.

Multiplying these three inequalities by sides, we obtain

$$\begin{aligned} \frac{2}{n(n+1)} > \frac{(A+B)(N+n)}{2nNAB}. \end{aligned}$$

Therefore, by 3.5 we have

$$\begin{aligned} \frac{2}{n(n+1)}&> \frac{(A+B)(N n - A B)}{2nNAB} \\&= \frac{(A+B)N n - (A+B)A B}{2nNAB}. \end{aligned}$$

Recall that $A+B \le n+N$ (see 3.3). Thereby

$$\begin{aligned} \frac{2}{n(n+1)}&> \frac{(A+B)N n - (n + N)A B}{2nNAB} \\&= \frac{1}{2} \left( \frac{1}{A} + \frac{1}{B} - \frac{1}{n} - \frac{1}{N} \right) . \end{aligned}$$

Since $n<N$, we obtain

$$\begin{aligned} \frac{3}{n(n+1)}&> \frac{1}{2} \left( \frac{1}{A} + \frac{1}{B} - \frac{1}{n} - \frac{1}{N} \right) +\frac{1}{2n(n+1)} + \frac{1}{2N(N+1)} \\&= \frac{1}{2} \left( \frac{1}{A} + \frac{1}{B} + \left( \frac{1}{n(n+1)}- \frac{1}{n} \right) + \left( \frac{1}{N(N+1)}- \frac{1}{N} \right) \right) \\&= \frac{1}{2} \left( \frac{1}{A} + \frac{1}{B} - \frac{1}{n+1} - \frac{1}{N+1} \right) . \end{aligned}$$

Obviously $\ln (1-y)<-y$ for any real number (see 2.2). Then

$$\begin{aligned} \ln \left( 1-\frac{3}{n(n+1)}\right) < - \frac{1}{2} \left( \frac{1}{A} + \frac{1}{B} - \frac{1}{n+1} - \frac{1}{N+1} \right) . \end{aligned}$$

(3.6)

Finally by applying both inequalities of Fact 2.1 and the initial condition $H_{n}+H_{N} < H_{A}+H_{B}$(see 3.4), we obtain

$$\begin{aligned} \ln (A)+ \ln (B)&> \left( H_{A}-\frac{1}{2A}-\gamma \right) + \left( H_{B}-\frac{1}{2B} - \gamma \right) \\&> H_{N}-\gamma +H_{n}- \gamma -\frac{1}{2}\left( \frac{1}{A}+\frac{1}{B} \right) \\&> \ln (n) + \ln (N) - \frac{1}{2} \left( \frac{1}{A} + \frac{1}{B} - \frac{1}{n+1} - \frac{1}{N+1} \right) . \end{aligned}$$

Hence, by 3.6 we obtain

$$\begin{aligned} \ln (AB)> \ln (nN) +\ln \left( 1-\frac{3}{n(n+1)}\right) , \end{aligned}$$

and thus

$$\begin{aligned} AB (n+1) > N\left( n^2 + n - 3\right) . \end{aligned}$$

(3.7)

In the end, we will give an upper estimate of the sum $\frac{1}{A}+\frac{1}{B}$. From 3.7 we obtain

$$\begin{aligned} \frac{1}{A}+\frac{1}{B}&< \frac{1}{A}+\frac{A (n+1)}{N \left( n^2+n-3 \right) }, \\ \frac{1}{A}+\frac{1}{B}&< G(A), \end{aligned}$$

what was there to prove. $\square$

4 Tools

The basic tool of this article will be the following two propositions.

Proposition 4.1

Let $\alpha >0$ be a real number. For any natural number ${n\ge \frac{1}{e^{\alpha }-1}}$, there is an inequality

$$\begin{aligned} H_{\left\lfloor e^{\alpha } n \right\rfloor }-H_{n} < \alpha . \end{aligned}$$

(4.1)

Proof

Let $\alpha >0$ be a real number and $n$ be a natural number such that $n\left( e^{\alpha }-1 \right) \ge 1$. Using both inequalities of Fact 2.1 we obtain

$$\begin{aligned} H_{\left\lfloor e^{\alpha } n \right\rfloor }-H_{n}&< \frac{1}{2 \left\lfloor e^{\alpha } n \right\rfloor } + \ln \left( \left\lfloor e^{\alpha } n \right\rfloor \right) +\gamma - \left( \frac{1}{2 (n+1)} + \ln \left( n\right) +\gamma \right) \end{aligned}$$

and since the function $\ln (x)+\frac{1}{2x}$ is increasing for $x \ge \frac{1}{2}$ (see 2.1), then

$$\begin{aligned} H_{\left\lfloor e^{\alpha } n \right\rfloor }-H_{n}&< \frac{1}{2 e^{\alpha } n} + \ln \left( e^{\alpha } n\right) +\gamma - \left( \frac{1}{2 (n+1)} + \ln \left( n\right) +\gamma \right) \\&= \ln (e^{\alpha })+\frac{1}{2}\left( \frac{1}{e^{\alpha } n} - \frac{1}{n+1}\right) \\&= \alpha + \frac{1 - \left( e^{\alpha } -1 \right) n}{2 e^{\alpha }n(n+1) } \\&\le \alpha \end{aligned}$$

because $n\left( e^{\alpha } -1 \right) \ge 1$. $\square$

The second proposition is as follows.

Proposition 4.2

Let n, N be natural numbers such that ${n \ge 17}$ and ${N \ge n+7}$. For each natural numbers A, B such that $n< A < B \le N$ and

$$\begin{aligned} H_{A}-H_{n} > H_{N}-H_{B}, \end{aligned}$$

(4.2)

there is an inequality

$$\begin{aligned} \frac{1}{A}+\frac{1}{B} \le \frac{1}{n+1}+\frac{n+1}{N n}. \end{aligned}$$

Remark. For $N>2n$, the constant $\frac{n+1}{nN}$ can be improved to $\frac{1}{\left\lfloor N \left( 1-\frac{1}{n+1}\right) \right\rfloor +1}$, but it is not asymptotically more useful.

Proof

Since for any number A it holds that $\frac{1}{A}<\frac{1}{A-1}$, 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 $A < B \le N$ and the condition 4.2 holds. Using 4.2 we obtain

$$\begin{aligned} \frac{1}{n+1}&= H_{A} - H_{n} \\&> H_{N} - H_{B}. \\ \end{aligned}$$

Thus

$$\begin{aligned} \frac{1}{n+1}&>\sum _{j=B+1}^{N} \frac{1}{j} \\&\ge \left( N-B \right) \frac{1}{N}, \end{aligned}$$

because $j \le N$. Therefore,

$$\begin{aligned} N&> (n+1)(N-B), \\ (n+1)B&> Nn. \end{aligned}$$

Thereby

$$\begin{aligned} \frac{1}{A}+\frac{1}{B} \le \frac{1}{n+1}+\frac{n+1}{N n}. \end{aligned}$$

Now we will prove Proposition 4.2 in the second case. Let n, N be natural numbers such that ${n \ge 17}$ and ${N \ge n+7}$. Let A, B be natural numbers such that $n< A < B \le N$ and such that A is the smallest number satisfying the condition 4.2, i.e.,

$$\begin{aligned} H_{A-1}-H_{n} \le H_{N}-H_{B}. \end{aligned}$$

Using the above inequality we obtain

$$\begin{aligned} \left( A-1-n \right) \frac{1}{A}&< \sum _{j=n+1}^{A-1} \frac{1}{j} \\&\le \sum _{j=B+1}^{N} \frac{1}{j} \\&< \left( N-B \right) \frac{1}{B} \\&< \frac{N-B}{A} \end{aligned}$$

because $A < B$. Therefore

$$\begin{aligned}{} & {} A-1-n< N-B,\\{} & {} A+B < N+n+1. \end{aligned}$$

Both sides of the above equation are natural numbers. Thus

$$\begin{aligned} A+B \le N+n. \end{aligned}$$

(4.3)

In particular, since $A < B$, we obtain

$$\begin{aligned} A \le \frac{A+B}{2} \le \frac{N+n}{2}. \end{aligned}$$

Thus

Recall that $n<A<B \le N$ are natural numbers such that $n \ge 17$, $N\ge n+7$, and . Then using Lemmas 3.3, 3.2 and 3.1 ends the proof of Proposition 4.2. (See 4.3 and 4.2 to check the assumptions of Lemma 3.3.) Thus

$$\begin{aligned} \frac{1}{A} + \frac{1}{B}&\le G(A) \\&\le G(n+2) \\&\le \frac{1}{n+1}+\frac{n+1}{Nn}. \end{aligned}$$

$\square$

5 Main results

Theorem 5.1

Let $\alpha >0$ be a real number and $m \ge 17$ be a natural number such that $m \ge \frac{4}{(e^{\alpha }-1)}$. All squares with side lengths $\frac{1}{m+1}, \frac{1}{m+2}, \dots$ can be packed into the rectangle with dimensions $\alpha \times \epsilon$, where

$$\begin{aligned} \epsilon < \frac{1}{m-\frac{1}{e^{\alpha }-1}} \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) . \end{aligned}$$

Proof

Let $\alpha >0$ be a real number and $m \ge 17$ be a natural number such that $m \ge \frac{4}{(e^{\alpha }-1)}$. We define the sequence $(n_j)_{j=0}^{\infty }$ by the recursion

$$\begin{aligned} n_{j+1}=\left\lfloor e^{\alpha } \left\lfloor e^{\alpha }n_j \right\rfloor \right\rfloor \end{aligned}$$

with the initial condition $n_0=m$. The sequence $(n_j)_{j=0}^{\infty }$ is obviously non-decreasing. Moreover, from  (2.4) we obtain that for each non-negative integer number j it holds that $n_{j+1} \ge n_j + 8.$ In addition, by the definition of the floor function, for each index j we obtain

$$\begin{aligned} n_{j+1}&\ge e^{\alpha }\left\lfloor e^{\alpha } n_j \right\rfloor -1 \\&\ge e^{2 \alpha } n_j - e^{\alpha } -1 \\&= e^{2 \alpha }n_j -\frac{e^{2\alpha }-1}{e^{\alpha }-1}. \end{aligned}$$

Thus

$$\begin{aligned} n_{j+1}- \frac{1}{e^{\alpha }-1} \ge e^{2\alpha } \left( n_j - \frac{1}{e^{\alpha }-1} \right) . \end{aligned}$$

(5.1)

Let t be a natural number. We use  (5.1) t times

$$\begin{aligned} n_{t}- \frac{1}{e^{\alpha }-1}&\ge \cdots \\&\ge e^{ 2\alpha t} \left( n_0- \frac{1}{e^{\alpha }-1} \right) . \end{aligned}$$

Thus, since $n_0=m$, we obtain

$$\begin{aligned} n_t > e^{ 2\alpha t} \left( m - \frac{1}{e^{\alpha }-1}\right) . \end{aligned}$$

(5.2)

Let $\alpha >0$ be a real number and m be a natural number. Let $t=0$. We put $n=n_t$ and $N=n_{t+1}$, and $h_t= \frac{1}{n+1}+\frac{n+1}{n N}$. Take a rectangle of size $\alpha \times h_t$, i.e., with side length $\alpha$ and height length $h_t$. Using Proposition 4.1, we can inscribe squares with side lengths $\frac{1}{n+1}, \dots , \frac{1}{\left\lfloor e^{\alpha }n \right\rfloor }$ 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 $\alpha$. Similarly, starting from the upper left corner, we can inscribe squares with side lengths $\frac{1}{N}, \dots , \frac{1}{\left\lfloor e^{\alpha }n \right\rfloor +1}$ into the upper row as tangents to the upper side of length $\alpha$. An example of such packing for $\alpha =\frac{1}{5}$ and $n=23$ is shown in Fig. 1.

Fig. 1

An example of packing squares with sides $\frac{1}{24}, \dots , \frac{1}{28}$ from the left in the bottom row and squares with sides $\frac{1}{34}, \dots , \frac{1}{29}$ from the left in the top row

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 $\frac{1}{A}$ overlapped a certain square of the lower row with side length $\frac{1}{B}$, then the inequality $\frac{1}{A}+\frac{1}{B} > h_t$ would be satisfied. Of course, then $n<A<B \le N$ and the sum of the lengths of the sides of the squares from $\frac{1}{N}$ to $\frac{1}{B+1}$ would be less than that of those from $\frac{1}{n+1}$ to $\frac{1}{A}$, 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 $t=0, 1, \dots$ and then we merge all the rectangles into one large rectangle. Therefore, we can inscribe all squares with sides of length $\frac{1}{m+1}, \frac{1}{m+2}, \dots$ into a rectangle with dimensions $\alpha \times \epsilon$, where

$$\begin{aligned} \epsilon&= \sum _{t=0}^{\infty } h_j \\&=\sum _{t=0}^{\infty } \left( \frac{1}{n_j+1} + \frac{n_j+1}{n_j n_{j+1}} \right) \\&= \sum _{t=0}^{\infty } \left( \frac{1}{n_j+1} + \frac{1}{n_j n_{j+1}} \right) + \sum _{t=0}^{\infty } \frac{1}{n_{j+1}}. \end{aligned}$$

Thus since $n_{j+1} > n_{j}+1$, we obtain

$$\begin{aligned} \epsilon&< \sum _{t=0}^{\infty } \left( \frac{1}{n_j+1} + \frac{1}{n_j (n_{j}+1)}\right) + \sum _{t=1}^{\infty } \frac{1}{n_{j}} \\&= \sum _{t=0}^{\infty } \frac{1}{n_j} + \sum _{t=1}^{\infty } \frac{1}{n_j} \\&= \frac{1}{n_0}+ 2 \sum _{j=1}^{\infty } \frac{1}{n_j}. \end{aligned}$$

Thereby, we conclude from  (5.2) that

$$\begin{aligned} \epsilon&< \frac{1}{m-\frac{1}{e^{\alpha }-1}} \left( 1+ 2\sum _{t=1}^{\infty } \left( \frac{1}{e^{2 \alpha }} \right) ^t \right) \\&= \frac{1}{m-\frac{1}{e^{\alpha }-1}} \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) . \end{aligned}$$

$\square$

Let $\alpha >0$ be a real number. For a given natural number $m \ge 17$ such that $m \ge \frac{4}{(e^{\alpha }-1)}$, let S denote the sum of the areas of all squares with sides of lengths $\frac{1}{m+1}, \frac{1}{m+2}, \dots$, and let R denote the area of the rectangle with dimensions $\alpha \times \epsilon$ obtained from Theorem 5.1.

Corollary 5.2

The ratio $\frac{R}{S}$ is upper bounded by a constant that does not depend on m but depends solely on $\alpha$.

Proof

Since $j<j+1$ for any natural number j, we obtain

$$\begin{aligned} S&= \sum _{j=m+1}^{\infty } \frac{1}{j^2} \\&> \sum _{j=m+1}^{\infty } \frac{1}{j (j+1)} \\&= \sum _{j=m+1}^{\infty } \left( \frac{1}{j} - \frac{1}{j+1} \right) \\&= \frac{1}{m+1}. \end{aligned}$$

Using the above observation and Theorem 5.1, we conclude that

$$\begin{aligned} \frac{R}{S} < \frac{m+1}{m-\frac{1}{e^{\alpha }-1}} \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) \alpha . \end{aligned}$$

(5.3)

Obviously $2 \le m$ and $\frac{4}{e^{\alpha }-1} \le m$. Therefore, 5.3 can be weakened to

$$\begin{aligned} \frac{R}{S}&< \frac{m+\frac{m}{2}}{m-\frac{m}{4}} \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) \alpha \\&= 2 \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) \alpha . \end{aligned}$$

Remark. For large values of m, the fraction $\frac{m+1}{m-\frac{1}{e^{\alpha }-1}}$ in 5.3 is approximately equal to 1. $\square$

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 $\frac{1}{j} \times \frac{1}{j+1}$ is contained in a square with side $\frac{1}{j}$, and a square with side $\frac{1}{2j+1}$ is contained in a square with side $\frac{1}{2j}$, then the following two corollaries can be concluded from Theorem 5.1.

Corollary 5.3

Let $\alpha >0$ be a real number and $m \ge 17$ be a natural number such that $m \ge \frac{4}{(e^{\alpha }-1)}$. All rectangles with dimensions $\frac{1}{m+1} \times \frac{1}{m+2}, \frac{1}{m+2} \times \frac{1}{m+3} \dots$ can be packed into the rectangle with dimensions $\alpha \times \epsilon$, where

$$\begin{aligned} \epsilon < \frac{1}{m-\frac{1}{e^{\alpha }-1}} \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) . \end{aligned}$$

Corollary 5.4

Let $\alpha >0$ be a real number and $m-1 \ge 17$ be a natural number such that $m-1 \ge \frac{4}{(e^{\alpha }-1)}$. All squares with side lengths $\frac{1}{2m+1}, \frac{1}{2m+3}, \dots$ can be packed into the rectangle with dimensions $\frac{\alpha }{2} \times \frac{\epsilon }{2}$, where

$$\begin{aligned} \epsilon < \frac{1}{m-1-\frac{1}{e^{\alpha }-1}} \left( 1+ \frac{2 }{e^{2 \alpha } -1 } \right) . \end{aligned}$$