A SHARP UPPER BOUND FOR THE SUM OF RECIPROCALS OF LEAST COMMON MULTIPLES II

Abstract

Let n and k be positive integers with $n\ge k+1$ and let $\{a_i\}_{i=1}^n$ be a strictly increasing sequence of positive integers. Let $S_{n, k}:=\sum _{i=1}^{n-k} {1}/{\mathrm {lcm}(a_{i},a_{i+k})}$ . In 1978, Borwein [‘A sum of reciprocals of least common multiples’, Canad. Math. Bull. 20 (1978), 117–118] confirmed a conjecture of Erdős by showing that $S_{n,1}\le 1-{1}/{2^{n-1}}$ . Hong [‘A sharp upper bound for the sum of reciprocals of least common multiples’, Acta Math. Hungar. 160 (2020), 360–375] improved Borwein’s upper bound to $S_{n,1}\le {a_{1}}^{-1}(1-{1}/{2^{n-1}})$ and derived optimal upper bounds for $S_{n,2}$ and $S_{n,3}$ . In this paper, we present a sharp upper bound for $S_{n,4}$ and characterise the sequences $\{a_i\}_{i=1}^n$ for which the upper bound is attained.

1 Introduction

Chebyshev [4] investigated the least common multiple of consecutive positive integers when he made the first important attempt to prove the prime number theorem stating that $\log \mathrm {lcm}(1, 2, \ldots , n)\sim n$ as n goes to infinity (see, for example, [13]). Hanson [8] and Nair [14] gave upper and lower bounds for $\mathrm {lcm}(1,2,\ldots ,n)$ and Nair’s lower bound was extended in [6, 11]. Goutziers [7] studied the asymptotic behaviour of the least common multiple of a set of integers not exceeding N. Bateman et al. [1] obtained an asymptotic estimate for the least common multiple of arithmetic progressions that is generalised in [12] to products of linear polynomials. In another direction, Behrend [2] strengthened an inequality of Heilbronn [9] and Rohrbach [15]. Erdős and Selfridge [5] proved a remarkable old conjecture that predicts that the product of any two or more consecutive positive integers is never a perfect power.

Erdős observed another interesting phenomena related to least common multiples. Let n and k be positive integers with $n\ge k+1$ and let $\{a_i\}_{i=1}^n$ be a strictly increasing sequence of positive integers. Let

$$ \begin{align*} S_{n, k}:=\sum_{i=1}^{n-k}\frac{1}{\mathrm{lcm}(a_{i},a_{i+k})}. \end{align*} $$

In 1978, Borwein [3] confirmed a conjecture of Erdős by showing that $S_{n, 1}\le 1-{1}/{2^{n-1}}$ with equality if and only if $a_{i}=2^{i-1}$ for $1\le i \le n$ . Recently, Hong [10] improved this upper bound and used the new result to get sharp upper bounds for $S_{n, 2}$ and $S_{n, 3}$ . He also characterised the sequences $\{a_i\}_{i=1}^\infty $ for which these upper bounds are attained. In this paper, we concentrate on $S_{n,4}$ . We will present an optimal upper bound for $S_{n, 4}$ and characterise the sequences $\{a_i\}_{i=1}^n$ for which this upper bound is attained.

As usual, for any real number x, we denote by $\lfloor x\rfloor $ and $\lceil x\rceil $ respectively the largest integer no more than x and the smallest integer no less than x. For brevity, we write $S_n :=S_{n, 4}$ .

The main result of this paper can be stated as follows.

Theorem 1.1. Let n be an integer with $n\ge 5$ and let $\{a_i\}_{i=1}^n$ be a strictly increasing sequence of positive integers. Then:

1. (i) $S_{5}\le 1/5$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5\}$ ;

2. (ii) $S_{6}\le {11}/{30}$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5, 6\}$ ;

3. (iii) $S_{7}\le {43}/{90}$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5, 6\}$ and $a_7=9$ ;

4. (vi) $S_{8}\le {101}/{180}$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5, 6\}$ , $a_7=9$ and $a_8=12$ ;

5. (v) if $n\ge 9$ , then

   (1.1) $$ \begin{align} S_{n}\le \frac{493}{420}-\frac{533}{105}\cdot\frac{1} {2^{\lfloor{n}/{4}\rfloor+1}} +\frac{\epsilon_{n}}{2^{\lfloor{n}/{4}\rfloor}}, \end{align} $$
   where
   $$ \begin{align*} \epsilon_n:= {\left\{\begin{array}{rl} 0 & \text{if} \ n\equiv 0\pmod 4,\\[5pt] \frac{2}{5} & \text{if} \ n\equiv 1\pmod 4,\\[5pt] \frac{11}{15} & \text{if} \ n\equiv 2\pmod 4,\\[5pt] \frac{107}{105} & \text{if} \ n\equiv 3\pmod 4, \end{array} \right.} \end{align*} $$
   and equality in (1.1) occurs if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4\}$ and $a_{4i+1}=5\times 2^{i-1} \ (1\le i\le \lfloor {(n-1)}/{4}\rfloor ), a_{4i+2}= 3\times 2^{i} \ (1\le i\le \lfloor {(n-2)}/{4}\rfloor )$ , $a_{4i+3}=7\times 2^{i-1} \ (1\le i\le \lfloor {(n-3)}/{4}\rfloor )$ and $a_{4i+4}=2^{i+2} \ (1\le i\le \lfloor {n}/{4}\rfloor -1)$ .

The rest of the paper is organised as follows. In Section 2, we prove several preliminary lemmas. In Section 3, we provide a proof for our main result.

2 Auxiliary lemmas

In this section, we supply several auxiliary lemmas that are needed in the proof of Theorem 1.1. The first is Hong’s upper bound [10, Theorem 1.2] which improves Borwein’s upper bound [3].

Lemma 2.1 [10, Theorem 1.2]

Let n be an integer with $n\ge 2$ and let $\{a_i\}_{i=1}^n$ be a strictly increasing sequence of positive integers. Then

(2.1) $$ \begin{align} \sum_{i=1}^{n-1} \frac{1}{\mathrm{lcm}(a_{i},a_{i+1})} \le \frac{1}{a_{1}}\bigg(1-\frac{1}{2^{n-1}}\bigg) \end{align} $$

with equality in (2.1) if and only if $a_{i}=2^{i-1}a_{1}$ for all integers i with $1 \le i \le n$ .

Lemma 2.2. Let m be an integer with $m\ge 3$ . Then

$$ \begin{align*} \frac{1}{7}+\frac{1}{9}+\frac{1}{9}\bigg(1-\frac{1}{2^{m-2}}\bigg) <\frac{1}{5}+\frac{1}{21}+\frac{1}{5}\bigg(1-\frac{1}{2^{m-2}}\bigg) \end{align*} $$

and

$$ \begin{align*} \frac{1}{9}+\frac{1}{12}+\bigg(\frac{1}{9}+\frac{1}{10}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) <\frac{1}{8}+\frac{1}{21}+\bigg(\frac{1}{7}+\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg). \end{align*} $$

Proof. Since $m\ge 3$ , a direct computation gives the desired inequalities.

Lemma 2.3. Let $S_n$ be given as above. Then:

1. (i) $S_5\le 1/5$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5\}$ ;

2. (ii) $S_6\le {11}/{30}$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5, 6\}$ ;

3. (iii) $S_7\le {43}/{90}$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4, 5, 6\}$ and $a_7=9$ .

Proof. We first deal with $S_5$ . Since $\mathrm {lcm}(a_1, a_{4})\ge a_{5}\ge 5$ ,

(2.2) $$ \begin{align} S_{5}=\frac{1}{\mathrm{lcm}(a_1, a_{5})}\le \frac{1}{5}. \end{align} $$

The equality in (2.2) holds if and only if $\mathrm {lcm}(a_1, a_{5})=5$ , which is true if and only if $a_1=1$ and $a_5=5$ . However, $a_1<a_2<a_3<a_4<a_5$ . So the equality in (2.2) holds if and only if $a_i=i$ for all $i\in \{1,2,3,4,5\}$ .

Now consider $S_6$ . Since $a_2\ge 2, a_2\mid \mathrm {lcm}(a_2, a_{6})$ and $\mathrm {lcm}(a_2, a_{6})\ge a_6\ge 6$ , we deduce that $\mathrm {lcm}(a_2, a_{6})\ge 6$ with equality if and only if $a_2=2$ and $a_6=6$ . So

(2.3) $$ \begin{align} S_{6}=\frac{1}{\mathrm{lcm}(a_1, a_{5})} +\frac{1}{\mathrm{lcm}(a_2, a_{6})} \le\frac{1}{5}+\frac{1}{6}=\frac{11}{30}, \end{align} $$

with equality in (2.3) if and only if $\mathrm {lcm}(a_1, a_{5})=5$ and $\mathrm {lcm}(a_2, a_{6})=6$ , which is true if and only if $a_1=1, a_2=2, a_5=5$ and $a_6=6$ , which is true if and only if $a_i=i$ for all $i\in \{1,2,3,4, 5, 6\}$ .

Finally, we consider $S_7$ . Since $a_3\ge 3, a_3\mid \mathrm {lcm}(a_3, a_{7})$ and $\mathrm {lcm}(a_3, a_{7})\ge a_7\ge 7$ , we deduce that either $\mathrm {lcm}(a_3, a_{7})=8$ which is true if and only if $a_3=4$ and $a_7=8$ , or $\mathrm {lcm}(a_3, a_{7})=9$ which is true if and only if $a_3=3$ and $a_7=9$ , or $\mathrm {lcm}(a_3, a_{7})\ge 10$ . We divide the rest of the proof into three cases.

If $\mathrm {lcm}(a_3, a_{7})\ge 10$ , then

$$ \begin{align*} S_{7}=\frac{1}{\mathrm{lcm}(a_1, a_{5})} +\frac{1}{\mathrm{lcm}(a_2, a_{6})}+\frac{1}{\mathrm{lcm}(a_3, a_{7})} \le\frac{11}{30}+\frac{1}{10}<\frac{43}{90} \end{align*} $$

as desired.

If $\mathrm {lcm}(a_3, a_{7})=8$ , then $a_3=4$ and $a_7=8$ . This implies that $a_4=5, a_5=6$ and $a_6=7$ . Since $(a_1, a_2)\in \{(1,2), (1,3), (2,3)\}$ , we have $\mathrm {lcm}(a_1, a_{5})=6$ and $\mathrm {lcm}(a_2, a_{6})\in \{14, 21\}$ . It then follows that

$$ \begin{align*} S_{7}\le\frac{1}{6}+\frac{1}{14}+\frac{1}{8}<\frac{43}{90}. \end{align*} $$

If $\mathrm {lcm}(a_3, a_{7})=9$ , then we must have $a_3=3$ and $a_7=9$ . So $\mathrm {lcm}(a_3, a_{7})=9$ . It then follows that

(2.4) $$ \begin{align} S_{7}=S_6+\frac{1}{\mathrm{lcm}(a_3,a_7)} \le\frac{11}{30}+\frac{1}{9}=\frac{43}{90}, \end{align} $$

with equality in (2.4) if and only if $a_i=i$ for all $i\in \{1,2,3,4,5,6\}$ and $\mathrm {lcm}(a_3, a_{7})=9$ , if and only if $a_i=i$ for all $i\in \{1,2,3,4,5,6\}$ and $a_7=9$ as required.

This completes the proof of Lemma 2.3.

Lemma 2.4. Let m be a positive integer with $m\ge 2$ and $\mathcal {A}=\{a_i\}_{i=1}^8$ a strictly increasing sequence of eight positive integers. Let

(2.5) $$ \begin{align} \Box_m=\Box_m(\mathcal {A}):=\sum_{i=1}^4 \bigg(\frac{1}{\mathrm{lcm}(a_i, a_{i+4})} +\frac{1}{a_{i+4}}\bigg(1-\frac{1}{2^{m-2}}\bigg)\bigg). \end{align} $$

Then both of the following statements are true.

1. (i) Either $\Box _2= {101}/{180}$ which is true if and only if $a_i=i$ for all $i\in \{1,2,3,4,5,6\}$ , $a_7=9$ and $a_8=12$ , or $\Box _2= {389}/{720}$ which holds if and only if $a_i=i$ for all integers $i\in \{1,2,3,4,5,6\}$ , $a_7=9$ and $a_8=16$ , or $\Box _2= {453}/{840}$ which is true if and only if $a_i=i$ for all integers $i\in \{1,2,3,4,5,6,7,8\}$ , or $\Box _2 < {453}/{840}$ .

2. (ii) If $m\ge 3$ , then

   (2.6) $$ \begin{align} \Box_m\le\frac{493}{420}-\frac{533}{105}\cdot\frac{1}{2^{m+1}}, \end{align} $$
   with equality in (2.6) if and only if $a_i=i$ for all integers i with $1\le i\le 8$ .

Proof (i). Evidently, $\Box _2=\sum _{i=1}^4 {1}/{\mathrm {lcm}(a_i, a_{i+4})}.$ We consider the following cases.

Case 1: $a_5\ge 6$ . Then $a_8\ge 9$ . If $a_8\ge 10$ , then by the fact $\mathrm {lcm}(a_i, a_{i+4})\ge a_{i+4}$ for all $i\in \{1,2,3,4\}$ , we derive

$$ \begin{align*} \Box_2\le \frac{1}{a_5}+\frac{1}{a_6}+\frac{1}{a_7}+\frac{1}{a_8} \le \frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10} <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21}=\frac{453}{840}. \end{align*} $$

If $a_8=9$ , then $a_5=6$ , $a_6=7$ and $a_7=8$ . This implies that $a_1\in \{1,2\}$ , $a_2\in \{2, 3\}, a_3\in \{3, 4\}$ and $a_4\in \{4,5\}$ . It follows that $\mathrm {lcm}(a_1, a_5)=6$ , $\mathrm {lcm}(a_2, a_6)\in \{14, 21\}, \mathrm {lcm}(a_3, a_7)=\mathrm {lcm}(a_3, 8)\in \{8, 24\}$ and $\mathrm {lcm}(a_4, a_8)=\mathrm {lcm}(a_4, 9)\in \{36, 45\}$ . So

$$ \begin{align*} \Box_2\le\frac{1}{6}+\frac{1}{14}+\frac{1}{8}+\frac{1}{36} <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21}=\frac{453}{840}. \end{align*} $$

Case 2: $a_5=5$ . Then $a_i=i$ for all integers i with $1\le i\le 4$ . If $a_6\ge 7$ , then $a_7\ge 8$ and $a_8\ge 9$ . So $\mathrm {lcm}(a_1, a_5)=5, \mathrm {lcm}(a_2, a_6)\ge 8, \mathrm {lcm}(a_3, a_7)\ge 9$ and $\mathrm {lcm}(a_4, a_8)\ge 12$ . However, $1/9+ {1}/{12} < 1/6 + {1}/{21}$ . Thus

$$ \begin{align*} \Box_2\le\frac{1}{5}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12} <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21}=\frac{453}{840}. \end{align*} $$

In what follows, we let $a_6=6$ . If $a_7\ge 10$ , then $a_8\ge 11$ . It follows that $\mathrm {lcm}(a_3, a_7)\ge 12$ with equality holding if and only if $a_7=12$ , and $\mathrm {lcm}(a_4, a_8)\ge 12$ with equality occurring if and only if $a_8=12$ . Since $a_7<a_8$ ,

$$ \begin{align*} \Box_2<\frac{1}{5}+\frac{1}{6}+\frac{1}{12}+\frac{1}{12} <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21}=\frac{453}{840}. \end{align*} $$

It remains to consider the case $a_7\in \{7, 8, 9\}$ . We consider three subcases.

Subcase 2.1: $a_7=7$ . Then $\mathrm {lcm}(a_3, a_7)=21$ and $\mathrm {lcm}(a_4, a_8)=\mathrm {lcm}(4, a_8)\ge 8$ with equality if and only if $a_8=8$ . So

$$ \begin{align*}\Box_2\le \frac{1}{5}+\frac{1}{6} +\frac{1}{21}+\frac{1}{8}=\frac{453}{840}\end{align*} $$

with equality if and only if $a_i=i$ for all integers i with $1\le i\le 8$ .

Subcase 2.2: $a_7=8$ . Then $a_8\ge 9$ . Hence

$$ \begin{align*} \Box_2\le \frac{1}{5}+\frac{1}{6}+\frac{1}{24}+\frac{1}{12} <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21}=\frac{453}{840}. \end{align*} $$

Subcase 2.3: $a_7=9$ . Then $a_8\ge 10$ . It follows that either $\mathrm {lcm}(a_4, a_8)=12$ which is true if and only if $a_8=12$ , or $\mathrm {lcm}(a_4, a_8)=16$ which is true if and only if $a_8=16$ , or $\mathrm {lcm}(a_4, a_8)\ge 20$ . We then deduce that either

$$ \begin{align*} \Box_2=\frac{1}{5}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}=\frac{101}{180} \end{align*} $$

which is true if and only if $a_i=i$ for all $i\in \{1,2,3,4,5,6\}$ , $a_7=9$ and $a_8=12$ , or

$$ \begin{align*} \Box_2=\frac{1}{5}+\frac{1}{6}+\frac{1}{9}+\frac{1}{16}=\frac{389}{720} \end{align*} $$

which holds if and only if $a_i=i$ for all $i\in \{1,2,3,4,5,6\}$ , $a_7=9$ and $a_8=16$ , or

$$ \begin{align*} \Box_2\le \frac{1}{5}+\frac{1}{6}+\frac{1}{9}+\frac{1}{20} <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21}=\frac{453}{840} \end{align*} $$

as expected. This completes the proof of part (i).

(ii). Let $m\ge 3$ . Since $\mathrm {lcm}(a_i, a_{i+4})\ge a_{i+4}$ for all integers i with $1\le i\le 4$ ,

(2.7) $$ \begin{align} \Box_m &\le\sum_{i=1}^4\bigg(\frac{1}{a_{i+4}}+\frac{1}{a_{i+4}} \bigg(1-\frac{1}{2^{m-2}}\bigg)\bigg) =\bigg(2-\frac{1}{2^{m-2}}\bigg)\sum_{i=5}^8\frac{1}{a_i} \end{align} $$

with equality in (2.7) if and only if $a_i\mid a_{i+4}$ for all integers $i\in \{1, 2, 3, 4\}$ . Let $S_0:= {493}/{420} - {533}/{105} \cdot {1}/({2^{m+1}})$ . We divide the rest of the proof into two cases.

Case 1: $a_5\ge 6$ . Then $a_6\ge 7, a_7\ge 8$ and $a_8\ge 9$ . So by (2.7) and Lemma 2.2,

$$ \begin{align*} \Box_m & \le \bigg(2-\frac{1}{2^{m-2}}\bigg)\sum_{i=5}^8\frac{1}{a_i} \le \bigg(\frac{1}{6}+\frac{1}{7}+\frac{1}{8} +\frac{1}{9}\bigg)\bigg(2-\frac{1}{2^{m-2}}\bigg)\nonumber\\[4pt] & <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{7} +\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) =S_0 \end{align*} $$

since $m\ge 3$ . This gives the desired result for Case 1.

Case 2: $a_5=5$ . Then $a_i=i$ for all $i\in \{1, 2, 3, 4\}$ . We consider three subcases.

Subcase 2.1: $a_6=6$ . Then $a_7\ge 7$ and $\mathrm {lcm}(a_3, a_7)=\mathrm {lcm}(3, a_7)\ge 9$ . So

(2.8) $$ \begin{align} \Box_m & = \frac{1}{\mathrm{lcm}(1, 5)}+\frac{1}{\mathrm{lcm}(2, 6)} +\frac{1}{\mathrm{lcm}(3, a_{7})}+\frac{1}{\mathrm{lcm}(4, a_8)}\nonumber\\[6pt] &\quad +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{a_7}+\frac{1}{a_8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg). \end{align} $$

If $a_7=7$ , then it follows from $a_8\ge 8$ that $\mathrm {lcm}(4, a_8)\ge 8$ with equality if and only if $a_8=8$ . Therefore,

$$ \begin{align*} \Box_m & = \frac{1}{5}+\frac{1}{6}+\frac{1}{21}+\frac{1}{\mathrm{lcm}(4, a_8)}+\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{a_8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)\\[4pt] & \le \frac{1}{5}+\frac{1}{6}+\frac{1}{21}+\frac{1}{8}+\bigg(\frac{1}{5} +\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)=S_0 \end{align*} $$

with equality if and only if $a_i=i$ for all integers i with $1\le i\le 8$ .

If $a_7=8$ , then $a_8\ge 9$ and so $\mathrm {lcm}(4, a_8)\ge 12$ . Thus by (2.8),

$$ \begin{align*} \Box_m<\frac{1}{5}+\frac{1}{6}+\frac{1}{24}+\frac{1}{12} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}\bigg) \bigg(1-\frac{1}{2^{m-2}}\bigg)<S_0. \end{align*} $$

If $a_7=9$ , then $\mathrm {lcm}(3, a_7)=9$ , $a_8\ge 10$ and so $\mathrm {lcm}(4, a_8)\ge 12$ . Since $m\ge 3$ and by Lemma 2.2,

$$ \begin{align*} \Box_m & <\frac{1}{5}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{9}+\frac{1}{10}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) \\[4pt] & <\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) =S_0. \end{align*} $$

If $a_7\ge 10$ , then $a_8\ge 11$ . Hence $\mathrm {lcm}(3, a_7)\ge 12$ with equality holding if and only if $a_7=12$ , and $\mathrm {lcm}(4, a_8)\ge 12$ with equality occurring if and only if $a_8=12$ . Since $a_7<a_8$ and $m\ge 3$ ,

$$ \begin{align*} \Box_m & <\frac{1}{5}+\frac{1}{6}+\frac{1}{12}+\frac{1}{12} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{10}+\frac{1}{11}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) \\[4pt] & <\frac{1}{5}+\frac{1}{6}+\frac{1}{21}+\frac{1}{8}+\bigg(\frac{1}{5} +\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) =S_0. \end{align*} $$

Subcase 2.2: $a_6=7$ . Then $a_7\ge 8$ and $a_8\ge 9$ . So $\mathrm {lcm}(3, a_7)\ge 9$ with equality if and only if $a_7=9$ , and $\mathrm {lcm}(4, a_8)\ge 12$ with equality if and only if $a_8=12$ . Since ${1}/{14} + 1/9 + {1}/{12} < 1/6 + 1/8 + {1}/{21}$ , it then follows immediately that

$$ \begin{align*} \Box_m & =\frac{1}{\mathrm{lcm}(1, 5)}+\frac{1} {\mathrm{lcm}(2, 7)}+\frac{1}{\mathrm{lcm}(3, a_{7})} +\frac{1}{\mathrm{lcm}(4, a_{8})}\nonumber\\[4pt] &\quad +\bigg(\frac{1}{5}+\frac{1}{7}+\frac{1}{a_7}+ \frac{1}{a_8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)\nonumber\\[4pt] & <\frac{1}{5}+\frac{1}{14}+\frac{1}{9} +\frac{1}{12}+\bigg(\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+ \frac{1}{9}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)\nonumber\\[4pt] & <\frac{1}{5}+\frac{1}{6}+\frac{1}{21}+\frac{1}{8} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{7} +\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)=S_0. \end{align*} $$

Subcase 2.3: $a_6\ge 8$ . Then $a_7\ge 9$ and $a_8\ge 10$ . Thus $\mathrm {lcm}(a_2, a_6)=\mathrm {lcm}(2, a_6)\ge 8$ , $\mathrm {lcm}(a_3, a_7) =\mathrm {lcm}(3, a_7)\ge 9$ and $\mathrm {lcm}(a_4, a_8)=\mathrm {lcm}(4, a_8)\ge a_8\ge 10$ which implies that $\mathrm {lcm}(a_4, a_8)\ge 12$ since $4\mid \mathrm {lcm}(a_4, a_8)$ . It then follows from the inequality $ 1/9 + {1}/{12} < 1/6 + {1}/{21}$ that

$$ \begin{align*} \Box_m & =\frac{1}{\mathrm{lcm}(1, 5)}+\frac{1} {\mathrm{lcm}(2, a_6)}+\frac{1}{\mathrm{lcm}(3, a_{7})} +\frac{1}{\mathrm{lcm}(4, a_{8})} \\[4pt] &\quad +\bigg(\frac{1}{5}+\frac{1}{a_6}+\frac{1}{a_7}+ \frac{1}{a_8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg) \\& \le \frac{1}{5}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\bigg(\frac{1}{5}+\frac{1}{8} +\frac{1}{9}+\frac{1}{10}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)\nonumber\\[4pt] & <\frac{1}{5}+\frac{1}{6}+\frac{1}{21}+\frac{1}{8}+\bigg(\frac{1}{5}+\frac{1}{6} +\frac{1}{7}+\frac{1}{8}\bigg)\bigg(1-\frac{1}{2^{m-2}}\bigg)=S_0. \end{align*} $$

This completes the proof of part (ii).

3 Proof of Theorem 1.1

Let $m\ge 2$ be an integer and let $\Box _m$ be defined as in (2.5). Then $\Box _2=S_8$ , so the results for parts (i) to (iv) follow from Lemmas 2.3 and 2.4. It remains to prove (v).

We first deal with the upper bounds for $S_9, S_{10}$ and $S_{11}$ . For $r\in \{1,2,3\}$ ,

$$ \begin{align*} S_{8+r}=\Box_2+\sum_{i=1}^r\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})}. \end{align*} $$

By Lemma 2.4, either $\Box _2= {101}/{180}$ which is true if and only if $a_i=i$ for all $i\in \{1,2,3,4,5,6\}$ , $a_7=9$ and $a_8=12$ , or $\Box _2= {389}/{720}$ which holds if and only if $a_i=i$ for all integers $i\in \{1,2,3,4,5,6\}$ , $a_7=9$ and $a_8=16$ , or $\Box _2= {453}/{840}$ which is true if and only if $a_i=i$ for all integers $i\in \{1,2,3,4,5,6,7,8\}$ , or $\Box _2 < {453}/{840}$ .

If $\Box _2 < {453}/{840}$ , then it follows from $\mathrm {lcm}(a_5, a_9)\ge 10, \mathrm {lcm}(a_6, a_{10})\ge 12$ and $\mathrm {lcm}(a_7, a_{11})\ge 14$ that

$$ \begin{align*} S_9 & <\frac{453}{840}+\frac{1}{\mathrm{lcm}(a_5, a_9)} \le \frac{453}{840}+\frac{1}{10}=\frac{537}{840}, \\[3pt] S_{10} & <\frac{453}{840}+\sum_{i=1}^2\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} \le \frac{453}{840}+\frac{1}{10}+\frac{1}{12}=\frac{607}{840}, \\[3pt] S_{11} & <\frac{453}{840}+\sum_{i=1}^3\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} \le \frac{453}{840}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}=\frac{667}{840}. \end{align*} $$

If $\Box _2= {101}/{180}$ , then by Lemma 2.4, we must have $a_i=i$ for all integers i with $1\le i\le 6$ , $a_7=9$ and $a_8=12$ . So $a_9\ge 13$ , $a_{10}\ge 14$ and $a_{11}\ge 15$ . This implies that $\mathrm {lcm}(a_5, a_9)=\mathrm {lcm}(5, a_9)\ge 15$ with equality if and only if $a_9=15$ , $\mathrm {lcm}(a_6, a_{10})=\mathrm {lcm}(6, a_{10})\ge 18$ with equality if and only if $a_{10}=18$ , and $\mathrm {lcm}(a_7, a_{11})=\mathrm {lcm}(9, a_{11})\ge 18$ with equality if and only if $a_{11}=18$ . Hence

$$ \begin{align*} S_9 & =\frac{101}{180}+\frac{1}{\mathrm{lcm}(a_5, a_9)} \le\frac{101}{180}+\frac{1}{15}=\frac{113}{180}<\frac{537}{840}, \\[3pt] S_{10} & =\frac{101}{180}+\sum_{i=1}^2\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} \le\frac{101}{180}+\frac{1}{15}+\frac{1}{18}=\frac{123}{180}<\frac{607}{840}, \\S_{11} & =\frac{101}{180}+\sum_{i=1}^3\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} <\frac{101}{180}+\frac{1}{15}+\frac{1}{18} +\frac{1}{18}=\frac{133}{180}<\frac{667}{840} \end{align*} $$

as desired.

If $\Box _2= {389}/{720}$ , then by Lemma 2.4, we must have $a_i=i$ for all integers i with $1\le i\le 6$ , $a_7=9$ and $a_8=16$ . So $a_9\ge 17$ , $a_{10}\ge 18$ and $a_{11}\ge 19$ which implies that $\mathrm {lcm}(a_5, a_9)=\mathrm {lcm}(5, a_9)\ge 20$ with equality if and only if $a_9=20$ , $\mathrm {lcm}(a_6, a_{10})=\mathrm {lcm}(6, a_{10})\ge 18$ with equality if and only if $a_{10}=18$ and $\mathrm {lcm}(a_7, a_{11})=\mathrm {lcm}(9, a_{11})\ge 27$ with equality if and only if $a_{11}=27$ . One then deduces that

$$ \begin{align*} S_9 & =\frac{389}{720}+\frac{1}{\mathrm{lcm}(a_5, a_9)} \le\frac{389}{720}+\frac{1}{20}=\frac{425}{720}<\frac{537}{840}, \\[4pt] S_{10}& =\frac{389}{720}+\sum_{i=1}^2\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} <\frac{389}{720}+\frac{1}{20}+\frac{1}{18}=\frac{465}{720}<\frac{607}{840}, \\[4pt] S_{11} & =\frac{389}{720}+\sum_{i=1}^3\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} \le\frac{389}{720}+\frac{1}{20}+\frac{1}{18} +\frac{1}{27}=\frac{465}{720}+\frac{1}{27}<\frac{667}{840} \end{align*} $$

as desired.

If $\Box _2 = {453}/{840}$ , then by Lemma 2.4, we must have $a_i=i$ for all integers i with $1\le i\le 8$ . So $a_9\ge 9$ which implies that $\mathrm {lcm}(a_5, a_9)\ge 10$ with equality if and only if $a_9=10$ . Furthermore, $\mathrm {lcm}(a_6, a_{10})\ge 12$ with equality if and only if $a_{10}=12$ and $\mathrm {lcm}(a_7, a_{11})\ge 14$ with equality if and only if $a_{11}=14$ . Thus

(3.1) $$ \begin{align} \hspace{-74pt}S_9 =\frac{453}{840}+\frac{1}{\mathrm{lcm}(a_5, a_9)} \le\frac{453}{840}+\frac{1}{10}=\frac{537}{840}, \end{align} $$

(3.2) $$ \begin{align} \hspace{-34pt}\ \ \ S_{10} =\frac{453}{840}+\sum_{i=1}^2\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} \le\frac{453}{840}+\frac{1}{10}+\frac{1}{12}=\frac{607}{840}, \end{align} $$

(3.3) $$ \begin{align} S_{11} & =\frac{453}{840}+\sum_{i=1}^3\frac{1}{\mathrm{lcm}(a_{4+i}, a_{8+i})} \le\frac{453}{840}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}=\frac{667}{840}, \end{align} $$

where each equality in (3.1) to (3.3) holds if and only if $a_i=i$ for all integers i with $1\le i\le 8$ , $a_9=10$ , $a_{10}=12$ and $a_{11}=14$ . So part (v) is true when $9\le n\le 11$ .

In what follows, we always assume that $n\ge 12$ . Then we can write $n=4m$ or $n=4m+r$ for some integers m and r with $m\ge 3$ and $1\le r\le 3$ . For any integer i with $1\le i\le 4$ , we define

$$ \begin{align*} S_m^{(i)}:=\sum_{j=1}^{m-2}\frac{1}{\mathrm{lcm}(a_{4j+i}, a_{4j+4+i})}. \end{align*} $$

Then

(3.4) $$ \begin{align} S_{4m}&=\sum_{i=1}^4\bigg(\frac{1}{\mathrm{lcm}(a_i, a_{i+4})}+S_m^{(i)}\bigg) \end{align} $$

and

(3.5) $$ \begin{align} S_{4m+r}=S_{4m}+\sum_{i=1}^r\frac{1}{\mathrm{lcm}(a_{4m-4+i}, a_{4m+i})}. \end{align} $$

For any integer i with $1\le i\le 4$ , applying Lemma 2.1 to the subsequence $\{a_{i+4}, a_{i+8},\ldots , a_{i+4(m-1)}\}$ yields

(3.6) $$ \begin{align} S_m^{(i)}=\sum_{j=1}^{m-2}\frac{1}{\mathrm{lcm}(a_{i+4j}, a_{i+4j+4})} \le \frac{1}{a_{i+4}}\bigg(1-\frac{1}{2^{m-2}}\bigg) \end{align} $$

with equality in (3.6) if and only if $a_{i+4j}=a_{i+4}\times 2^{j-1}$ for all integers j with $1\le j\le m-1$ . Further, for any integer i with $1\le i\le r$ , applying Lemma 2.1 to the subsequence $\{a_{4+i}, a_{8+i},\ldots , a_{4m+i}\}$ gives

(3.7) $$ \begin{align} S_m^{(i)}+\frac{1}{\mathrm{lcm}(a_{4m-4+i}, a_{4m+i})} =\sum_{j=1}^{m-1}\frac{1}{\mathrm{lcm}(a_{4j+i}, a_{4j+4+i})} \le \frac{1}{a_{4+i}}\bigg(1-\frac{1}{2^{m-1}}\bigg) \end{align} $$

with equality in (3.7) if and only if $a_{4j+i}=a_{4+i}\times 2^{j-1}$ for all integers j with $1\le j\le m$ . Then by (3.4) and (3.6),

(3.8) $$ \begin{align} S_{4m} & \le \sum_{i=1}^4\bigg(\frac{1}{\mathrm{lcm}(a_i, a_{i+4})} +\frac{1}{a_{i+4}}\bigg(1-\frac{1}{2^{m-2}}\bigg)\bigg)=\Box_m \end{align} $$

with equality in (3.8) if and only if $a_{4j+i}=a_{4+i}\times 2^{j-1}$ for all integers i and j with $1\le j\le m-1$ and $1\le i\le 4$ . By (3.5), (3.6) and (3.7),

(3.9) $$ \begin{align} S_{4m+r}&=\sum_{i=1}^4\frac{1}{\mathrm{lcm}(a_i, a_{i+4})} +\sum_{i=1}^r\bigg(S_m^{(i)}+\frac{1}{\mathrm{lcm}(a_{4m-4+i}, a_{4m+i})}\bigg)+\sum_{i=r+1}^4 S_m^{(i)}\nonumber\\ &\le \sum_{i=1}^4\frac{1}{\mathrm{lcm}(a_i, a_{i+4})} +\sum_{i=1}^r \frac{1}{a_{4+i}}\bigg(1-\frac{1}{2^{m-1}}\bigg) +\sum_{i=r+1}^4 \frac{1}{a_{4+i}}\bigg(1-\frac{1}{2^{m-2}}\bigg)\nonumber \\ &=\sum_{i=1}^4\bigg(\frac{1}{\mathrm{lcm}(a_i, a_{i+4})}+\frac{1}{a_{i+4}}\bigg(1-\frac{1}{2^{m-2}}\bigg)\bigg) +\frac{1}{2^{m-1}}\sum_{i=1}^r \frac{1}{a_{4+i}}\nonumber\\ &=\Box_m+\frac{1}{2^{m-1}}\sum_{i=1}^r\frac{1}{a_{4+i}}, \end{align} $$

and equality in (3.9) holds if and only if $a_{4j+i}=a_{4+i}\times 2^{j-1}$ for all integers i and j with $1\le j\le m-1$ and $1\le i\le 4$ and $a_{4m+i}=a_{4+i}\times 2^{m-1}$ for all integers i with $1\le i\le r$ . Now by Lemma 2.4, if $m\ge 3$ , then

(3.10) $$ \begin{align} \Box_m\le &\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{21} +\bigg(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\bigg) \bigg(1-\frac{1}{2^{m-2}}\bigg)\notag\\[4pt] =&\frac{493}{420}-\frac{533}{105}\cdot\frac{1}{2^{m+1}}:=S_0, \end{align} $$

with equality in (3.10) if and only if $a_i=i$ for all integers i with $1\le i\le 8$ . Notice that

(3.11) $$ \begin{align} \sum_{i=1}^r\frac{1}{a_{4+i}}\le\sum_{i=1}^r\frac{1}{4+i} \end{align} $$

with equality in (3.11) if and only if $a_{4+i}=4+i$ for all $1\le i\le r$ . Therefore, by (3.8) and (3.10), $S_{4m}\le S_0$ with equality if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4\}$ and $a_{4j+i}=(4+i)\times 2^{j-1}$ for all integers i and j with $1\le j\le m-1$ and $1\le i\le 4$ . It follows from (3.9) and (3.11) that

(3.12) $$ \begin{align} S_{4m+r}\le S_0+\frac{1}{2^{m-1}}\sum_{i=1}^r\frac{1}{4+i} =\frac{493}{420}-\frac{533}{105}\cdot\frac{1}{2^{m+1}} +\frac{1}{2^{m-1}}\sum_{i=1}^r\frac{1}{4+i}, \end{align} $$

with equality in (3.12) if and only if $a_i=i$ for all $i\in \{1, 2, 3, 4\}$ , $a_{4j+i}=(4+i)\times 2^{j-1}$ for all integers i and j with $1\le j\le m-1$ and $1\le i\le 4$ and $a_{4m+i}=(4+i)\times 2^{m-1}$ for $1\le i\le r$ . So part (v) is proved when $n\ge 12$ .

This completes the proof of Theorem 1.1.