SUMSETS CONTAINING A TERM OF A SEQUENCE

Abstract

Let $S=\{s_{1}, s_{2}, \ldots \}$ be an unbounded sequence of positive integers with $s_{n+1}/s_{n}$ approaching $\alpha $ as $n\rightarrow \infty $ and let $\beta>\max (\alpha , 2)$. We show that for all sufficiently large positive integers l, if $A\subset [0, l]$ with $l\in A$, $\gcd A=1$ and $|A|\geq (2-{k}/{\lambda \beta })l/(\lambda +1)$, where $\lambda =\lceil {k}/{\beta }\rceil $, then $kA\cap S\neq \emptyset $ for $2<\beta \leq 3$ and $k\geq {2\beta }/{(\beta -2)}$ or for $\beta>3$ and $k\geq 3$.

1 Introduction

For two sets $A, B$ of integers and $a\in \mathbb {Z}$ , define

$$ \begin{align*}A+B=\{a+b: a\in A, b\in B\},\end{align*} $$

and

$$ \begin{align*}a-B=\{a-b: b\in B\}.\end{align*} $$

For a positive integer $h\geq 2$ , let

$$ \begin{align*}hA=\{a_{1}+\cdots+a_{h}:a_{1}, \ldots, a_{h}\in A\}.\end{align*} $$

A set A of nonnegative integers is called normal if $0\in A$ and the greatest common divisor of all elements of A is 1.

In 1990, Erdős and Freiman [2] proved a conjecture of Erdős and Freud: if a set $A$ of integers is a subset of $[1,n]$ and $|A|>{n}/{3}$ , then a power of 2 can be written as the sum of elements of A. In 1989, Nathanson and Sárközy [6] improved this result by showing that 3504 elements of A is enough. Finally, Lev [4] obtained the best possible result that a power of 2 is the sum of at most four elements of A. In 2004, Abe [1] extended Lev’s result to a power of m. In 2006, Pan [7] generalised the results of Lev and Abe.

Theorem 1.1 [7, Theorem 1].

Let $k, m, n\geq 2$ be integers. Let A be a normal subset of $[0, n]$ satisfying

$$ \begin{align*}|A|>\frac{1}{l+1}\bigg(\bigg(2-\frac{k}{lm}\bigg)n+2l\bigg),\end{align*} $$

where $l=\lceil k/m\rceil $ . If $m\geq 3$ , or $m=2$ and k is even, then $kA$ contains a power of m.

Pan conjectured that Theorem 1.1 still holds for $m=2$ and k is odd. In 2012, Wu and Chen [8] made some progress towards this conjecture.

In 2010, Kapoor [3] extended Pan’s result for $2A$ to general sequences. He proved the following two results.

Theorem 1.2 [3, Theorem 1].

Let $\{a_{1}, a_{2}, a_{3}, \ldots \}$ be an unbounded sequence of positive integers. Assume that $a_{n+1}/a_{n}$ approaches some limit $\alpha $ as $n\rightarrow \infty $ , and let $\beta>2$ be some real number greater than $\alpha $ . Then for sufficiently large $x\geq 0$ , if A is a set of nonnegative integers less than or equal to x containing $0$ and satisfying

$$ \begin{align*}|A|\geq\bigg(1-\frac{1}{\beta}\bigg)x,\end{align*} $$

then $2A$ contains an element of $\{a_{n}\}$ .

Theorem 1.3 [3, Theorem 2].

Let $\{a_{1}, a_{2}, a_{3}, \ldots \}$ be an unbounded sequence of positive integers such that $a_{n+1}/a_{n}\leq \beta $ for some constant $\beta \geq 2$ . Then for any $x\geq 0$ , if A is a set of nonnegative integers less than or equal to x containing $0$ and satisfying

$$ \begin{align*}|A|>\bigg(1-\frac{1}{\beta}\bigg)x+\frac{1}{\beta}\cdot\bigg\lfloor\frac{a_{1}-1}{2}\bigg\rfloor+1,\end{align*} $$

then $2A$ contains an element of $\{a_{n}\}$ .

We extend Pan’s result for $kA\ (k\geq 3)$ to general sequences that grow like the powers of a real number greater than or equal to 2.

Theorem 1.4. Let $\beta> 2$ be a real number and let $S=\{s_{1}, s_{2}, \ldots \}$ be an unbounded sequence of positive integers such that $\lim _{n\to \infty } {s_{n+1}}/{s_{n}}=\alpha <\beta $ . Let $k\geq 3$ be a positive integer. For large enough l, let A be a normal subset of $[0, l]$ with $l\in A$ such that

$$ \begin{align*}|A|\geq\frac{1}{\lambda+1}\bigg(2-\frac{k}{\lambda\beta}\bigg)l,\end{align*} $$

where $\lambda =\lceil {k}/{\beta }\rceil $ . If $2<\beta \leq 3$ , then $kA\cap S\neq \emptyset $ for all $k\geq {2\beta }/{(\beta -2)}$ . If $\beta>3$ , then $kA\cap S\neq \emptyset $ for all $k\geq 3$ .

Theorem 1.5. Let $\beta>2$ be a real number and let $S=\{s_{1}, s_{2}, \ldots \}$ be an unbounded sequence of positive integers such that $s_{n+1}/s_{n}\leq \beta $ . Let $k\geq 3$ and l be positive integers such that

(1.1) $$ \begin{align} l\bigg(\frac{k}{\beta}-\lambda+1\bigg)\geq\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+1. \end{align} $$

Let A be a normal subset of $[0, l]$ with $l\in A$ satisfying

(1.2) $$ \begin{align} |A|>\frac{2}{ \lambda\beta(\lambda+1)} \bigg(\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\frac{\beta}{2}\bigg)+\frac{1}{\lambda+1} \bigg(\bigg(2-\frac{k}{\lambda\beta}\bigg)l+2\lambda\bigg), \end{align} $$

where $\lambda =\lceil {k}/{\beta }\rceil $ . If $2<\beta <3$ , then $kA\cap S\neq \emptyset $ for all $k\geq {2\beta }/{(\beta -2)}$ . If $\beta \geq 3$ , then $kA\cap S\neq \emptyset $ for all $k\geq 3$ .

2 Lemmas

Lemma 2.1 [5, Corollary 1].

If A is a normal subset of $[0, l]$ with $l\in A$ and $\rho =\lceil (l-1)/(|A|-2)\rceil -1$ , then

$$ \begin{align*} |hA|\geq \begin{cases} B_{h}(|A|) &\text{if } h\leq \rho,\\ B_{\rho}(|A|)+(h-\rho)l &\text{if } h\geq \rho,\\ \end{cases} \end{align*} $$

where $B_{h}(x)=\tfrac 12h(h+1)(x-2)+h+1$ .

Lemma 2.2. Let $\beta \geq 2$ be a real number. Let $S=\{s_{1}, s_{2}, \ldots \}$ be an unbounded sequence of positive integers such that $s_{n+1}/s_{n}\leq \beta $ . Let a, b be two positive integers. Suppose A and B are sets of integers satisfying $A\subseteq [0, a]$ , $B\subseteq [0, b]$ and $0\in A\cap B$ . If

(2.1) $$ \begin{align} |A|+|B|>3+\frac{2}{\beta}\bigg\lfloor \frac{s_{1}-1}{2}\bigg\rfloor+\max\bigg\{a, b, \bigg(1-\frac{1}{\beta}\bigg)(a+b)\bigg\}, \end{align} $$

then $(A+B)\cap S\neq \emptyset $ .

Proof. We may assume that $a\leq b$ . If $0\leq b<s_{1}$ , put $x_{0}=\lfloor (s_{1}-1)/2\rfloor $ . If $b<x_{0}$ , then

$$ \begin{align*} |A|+|B|&>2+\frac{2}{\beta}\bigg\lfloor \frac{s_{1}-1}{2}\bigg\rfloor+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &>2+\frac{1}{\beta}(a+b)+\bigg(1-\frac{1}{\beta}\bigg)(a+b) =2+a+b, \end{align*} $$

which is a contradiction since $|A|+|B|\leq a+b+2$ . Thus, $x_{0}\leq b<s_{1}$ . If $a+b<s_{1}$ , then

$$ \begin{align*} |A|+|B|&>3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &\geq 3+\frac{1}{\beta}(s_{1}-2)+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &\geq3+\frac{1}{\beta}(a+b-1)+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &\geq3-\frac{1}{\beta}+a+b > a+b+2, \end{align*} $$

which is again impossible. Thus, $a+b\geq s_{1}$ . Then,

$$ \begin{align*} |A|+|B|&>3+\frac{1}{\beta}(s_{1}-2)+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &\geq3+\frac{1}{\beta}s_{1}-\frac{2}{\beta}+\bigg(1-\frac{1}{\beta}\bigg)s_{1}\\ &=3-\frac{2}{\beta}+s_{1} \geq s_{1}+2. \end{align*} $$

Since $B, s_{1}-A\subseteq [0, s_{1}]$ , we must have $s_{1}\in A+B$ .

Next, we consider $b\geq s_{1}$ . Choose r such that $s_{r}\leq b<s_{r+1}$ . We proceed by induction on $a+b$ .

If $a+b=s_{1}+1$ , then $A\subseteq [0, 1]$ and $B\subseteq [0, s_{1}]$ . Since $s_{1}-A$ , $B\subseteq [0,s_{1}]$ and

$$ \begin{align*} |A|+|B|&>3+\frac{1}{\beta}(s_{1}-2)+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &=3-\frac{3}{\beta}+s_{1}+1 >s_{1}+2, \end{align*} $$

we have $s_{1}\in A+B$ .

Now, assume that $a+b>s_{1}+1$ and that the lemma holds for the sets $A{'}\subseteq [0, a_{1}]$ and $B{'}\subseteq [0, b_{1}]$ with $a_{1}+b_{1}<a+b$ . Suppose that $(A+B)\cap S=\emptyset $ .

Case 1: $a<s_{r}$ . Write $A_{1}=s_{r}-A$ . Thus, $A_{1}\subseteq [s_{r}-a, s_{r}]\subseteq [0, b]$ . If $|A_{1}|+|B|>b+1$ , then

$$ \begin{align*} |A_{1}\cap B|=|A_{1}|+|B|-|A_{1}\cup B| \geq b+2-(b+1) =1. \end{align*} $$

Thus, $s_{r}\in A+B$ , which is impossible. Hence, $|A|+|B|=|A_{1}|+|B|\leq b+1$ , which contradicts the hypothesis (2.1).

Case 2: $a\geq s_{r}$ and $a+b>s_{r+1}$ . Write

$$ \begin{align*}A_{1}=[0, b]\cap(s_{r+1}-A), \;B_{1}=[s_{r+1}-a, b]\cap B, \end{align*} $$

$$ \begin{align*}A_{2}=[0, s_{r+1}-b-1]\cap A,\; B_{2}=[0, s_{r+1}-a-1]\cap B. \end{align*} $$

Then,

(2.2) $$ \begin{align} |B|=|B_{1}|+|B_{2}|\end{align} $$

(2.3) $$ \begin{align} |A|&=|A_{2}|+|[s_{r+1}-b, a]\cap A|\nonumber\\ &=|A_{2}|+|s_{r+1}-([s_{r+1}-b, a]\cap A)| \nonumber \\ &=|A_{2}|+|[s_{r+1}-a, b]\cap(s_{r+1}-A)|\nonumber\\ &=|A_{1}|+|A_{2}|. \end{align} $$

Since $A_{1}, B_{1}\subseteq [s_{r+1}-a,b]$ and $s_{r+1}\notin A_{1}+B_{1}$ ,

(2.4) $$ \begin{align} |A_{1}|+|B_{1}|\leq b-s_{r+1}+a+1.\end{align} $$

If $b=s_{r+1}-1$ , then $|A_{2}|=1$ and $|B_{2}|\leq s_{r+1}-a$ . By (2.2)–(2.4),

$$ \begin{align*}|A|+|B|\leq s_{r+1}-a+1+b-s_{r+1}+a+1=b+2,\end{align*} $$

which contradicts the hypothesis (2.1). Thus, $b\leq s_{r+1}-2$ .

Since $(A+B)\cap S=\emptyset $ , we have $(A_2+B_2)\cap S=\emptyset $ . Noting that

$$ \begin{align*}s_{r+1}-b-1+s_{r+1}-a-1<2(a+b)-a-b-2<a+b,\end{align*} $$

it follows from the hypothesis that if

$$ \begin{align*} \max\bigg\{s_{r+1}-b-1, s_{r+1}-a-1, \bigg(1-\frac{1}{\beta}\bigg)(2s_{r+1}-a-b-2)\bigg\}=s_{r+1}-a-1, \end{align*} $$

then

(2.5) $$ \begin{align} |A_{2}|+|B_{2}|\leq s_{r+1}-a-1+3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor. \end{align} $$

By (2.2)–(2.5),

$$ \begin{align*} |A|+|B| \leq b-s_{r+1}+a+1+s_{r+1}-a-1+3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor = b+3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor, \end{align*} $$

which contradicts (2.1). If

$$ \begin{align*} \max\bigg\{s_{r+1}-b-1, s_{r+1} -a-1, \ & \bigg(1-\frac{1}{\beta}\bigg)(2s_{r+1}-a-b-2)\bigg\} \\ & =\bigg(1-\frac{1}{\beta}\bigg)(2s_{r+1}-a-b-2), \end{align*} $$

then

(2.6) $$ \begin{align} |A_{2}|+|B_{2}|\leq\bigg(1-\frac{1}{\beta}\bigg)(2s_{r+1}-a-b-2)+3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor. \end{align} $$

By (2.2)–(2.4) and (2.6),

$$ \begin{align*} |A|+|B|&\leq b-s_{r+1}+a+1+\bigg(1-\frac{1}{\beta}\bigg)(2s_{r+1}-a-b-2)+3+\frac{2}{\beta} \bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor\\ &= s_{r+1}-\frac{2}{\beta}s_{r+1}+\frac{1}{\beta}(a+b)+\frac{2}{\beta}+2+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor\\ &\leq a+b-\frac{2}{\beta}(a+b)+\frac{1}{\beta}(a+b)+\frac{2}{\beta}+2+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor\\ &= \bigg(1-\frac{1}{\beta}\bigg)(a+b)+\frac{2}{\beta}+2+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor\\ &\leq \bigg(1-\frac{1}{\beta}\bigg)(a+b)+3+\frac{2}{\beta}\bigg\lfloor\frac{a_{1}-1}{2}\bigg\rfloor, \end{align*} $$

which again contradicts (2.1).

Case 3: $a\geq s_{r}$ and $a+b\leq s_{r+1}$ . Write

$$ \begin{align*}A_{1}=[0, s_{r}]\cap A,\; B_{1}=[0, s_{r}]\cap B,\end{align*} $$

$$ \begin{align*}A_{2}=(s_{r}, a]\cap A,\; B_{2}=(s_{r}, b]\cap B.\end{align*} $$

Since $(A+B)\cap S=\emptyset $ , it follows that $(A_{1}+B_{1})\cap S=\emptyset $ and so $|s_{r}-A_{1}|+|B_{1}|\leq s_{r}+1$ . Thus,

$$ \begin{align*} |A|+|B|&\leq a+b-2s_{r}+s_{r}+1 = a+b-s_{r}+1. \end{align*} $$

However, by (2.1),

$$ \begin{align*} |A|+|B|&> 2+\bigg(1-\frac{1}{\beta}\bigg)(a+b)\\ &= a+b-\frac{1}{\beta}(a+b)+2\\ &\geq a+b-\frac{1}{\beta}s_{r+1}+2\\ &\geq a+b-s_{r}+2, \end{align*} $$

which is a contradiction. Hence, we have $(A+B)\cap S\neq \emptyset $ .

This completes the proof of Lemma 2.2.

Remark 2.3. If $s_{1}$ is odd, this lower bound can be improved to

$$ \begin{align*} |A|+|B|>2+\frac{2}{\beta}\bigg\lfloor \frac{s_{1}-1}{2}\bigg\rfloor+\max\bigg\{a, b, \bigg(1-\frac{1}{\beta}\bigg)(a+b)\bigg\}. \end{align*} $$

3 Proof of Theorem 1.5

Let $k_{1}=\lfloor {k}/{2}\rfloor $ and $k_{2}=\lceil {k}/{2}\rceil $ . Then,

$$ \begin{align*}k_{1}A\subseteq\bigg[0, \bigg\lfloor\frac{k}{2}\bigg\rfloor l\bigg], \quad k_{2}A\subseteq\bigg[0, \bigg\lceil\frac{k}{2}\bigg\rceil l\bigg].\end{align*} $$

Since $\beta>2$ , if $k\geq {\beta }/{(\beta -2)}$ , then

$$ \begin{align*}\bigg\lceil\frac{k}{2}\bigg\rceil\leq(\beta-1)\bigg\lfloor\frac{k}{2}\bigg\rfloor.\end{align*} $$

In particular, if $\beta \geq 3$ , then

$$ \begin{align*}k\geq3\geq1+\frac{2}{\beta-2}=\frac{\beta}{\beta-2}. \end{align*} $$

Hence, if $2<\beta <3$ and $k\geq {\beta }/{(\beta -2)}$ , or $\beta \geq 3$ , then $\lceil {k}/{2}\rceil \leq (\beta -1)\lfloor {k}/{2}\rfloor $ . So,

(3.1) $$ \begin{align} \bigg\lfloor\frac{k}{2}\bigg\rfloor\leq\bigg\lceil\frac{k}{2}\bigg\rceil \leq\bigg(1-\frac{1}{\beta}\bigg)\bigg(\bigg\lceil\frac{k}{2}\bigg\rceil+\bigg\lfloor\frac{k}{2}\bigg\rfloor\bigg). \end{align} $$

Since $0\in A$ ,

$$ \begin{align*}k_{1}A+k_{2}A=\bigg(\bigg\lfloor\frac{k}{2}\bigg\rfloor+\bigg\lceil\frac{k}{2}\bigg\rceil\bigg)A=kA.\end{align*} $$

To show $kA\cap S\neq \emptyset $ , by Lemma 2.2 and (3.1), it is sufficient to show that

$$ \begin{align*}|k_{1}A|+|k_{2}A|>3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\bigg(1-\frac{1}{\beta}\bigg)kl.\end{align*} $$

By (1.2),

(3.2) $$ \begin{align} (\lambda+1)(|A|-2)>\frac{2}{\lambda\beta }\bigg(\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\frac{\beta}{2}\bigg)+\bigg(2-\frac{k}{\lambda\beta }\bigg)l-2. \end{align} $$

Noting that $\lambda \geq {k}/{\beta }$ ,

(3.3) $$ \begin{align} (\lambda+1)(|A|-2)>\bigg(2-\frac{k}{\lambda\beta}\bigg)l-2\geq l-2. \end{align} $$

Write

$$ \begin{align*}\rho=\lceil(l-1)/(|A|-2)\rceil-1.\end{align*} $$

Then by (3.3),

(3.4) $$ \begin{align} \rho<\frac{l-1}{|A|-2}\leq \lambda+1. \end{align} $$

If $\beta>2$ and $k\geq {2\beta }/{(\beta -2)}$ , then

$$ \begin{align*}k\bigg(\frac{1}{2}-\frac{1}{\beta}\bigg)\geq1>\bigg\lceil\frac{k}{\beta}\bigg\rceil-\frac{k}{\beta}\end{align*} $$

and so ${k}/{2}>\lceil {k}/{\beta }\rceil $ . If $\beta \geq 3$ and $3\leq k<2\beta $ , then $\lceil {k}/{\beta }\rceil \leq {k}/{2}$ . If $\beta \geq 3$ and $k\geq 2\beta $ , then

$$ \begin{align*}\lambda=\bigg\lceil\frac{k}{\beta}\bigg\rceil<\frac{k}{\beta}+1=\frac{k}{2}-\frac{(\beta-2)k}{2\beta}+1\leq\frac{k}{2}.\end{align*} $$

Thus, $\lambda \leq {k}/{2}$ for $\beta \geq 3$ or $2<\beta <3$ and $k\geq {2\beta }/{(\beta -2)}$ . Hence, by (3.4),

$$ \begin{align*}\rho\leq\lambda\leq\bigg\lfloor\frac{k}{2}\bigg\rfloor=k_{1}.\end{align*} $$

By Lemma 2.1,

(3.5) $$ \begin{align} |k_{i}A|\geq B_{\rho}(|A|)+(k_{i}-\rho)l \quad\mbox{for } i=1, 2. \end{align} $$

If $\lambda =\rho $ , then by (3.2) and (3.5),

$$ \begin{align*} |k_{1}A|+|k_{2}A|&\geq \lambda(\lambda+1)(|A|-2)+2\lambda+2+(k-2\lambda)l\\ &> \lambda\bigg(\frac{2}{\lambda\beta}\bigg(\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\frac{\beta}{2}\bigg)+\bigg(2-\frac{k}{\lambda\beta}\bigg)l-2\bigg) +2\lambda+2+(k-2\lambda)l\\ &= 3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\bigg(1-\frac{1}{\beta}\bigg)kl. \end{align*} $$

Now suppose that $\rho \leq \lambda -1$ . Then, $0\leq \rho \leq \lceil {k}/{\beta }\rceil -1$ . Hence, by (1.1),

$$ \begin{align*} |k_{1}A|+|k_{2}A|&\geq \rho(\rho+1)(|A|-2)+2\rho+2+(k-2\rho)l\\ &\geq \rho(l-1)+2\rho+2+(k-2\rho)l\\ &= kl-\rho(l-1)+2\\ &> kl-\bigg(\frac{k}{\beta}+\bigg\lceil\frac{k}{\beta}\bigg\rceil-\frac{k}{\beta}-1\bigg)l+2\\ &= kl-\frac{k}{\beta}l+\bigg(\frac{k}{\beta}-\bigg\lceil\frac{k}{\beta}\bigg\rceil+1\bigg)l+2\\ &\geq \bigg(1-\frac{1}{\beta}\bigg)kl+3+\frac{2}{\beta}\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor. \end{align*} $$

This completes the proof of Theorem 1.5.

4 Proof of Theorem 1.4

Let $\beta ^{-}$ be some constant satisfying $\alpha <\beta ^{-}<\beta $ , and assume that

(4.1) $$ \begin{align} \frac{s_{n+1}}{s_{n}}\leq \beta^{-} \quad\mbox{for all } n\geq1. \end{align} $$

Then for any l so large that

$$ \begin{align*} \frac{1}{\lambda+1}\bigg(2-\frac{k}{\lambda\beta}\bigg)l \geq \frac{1}{\lambda+1}\bigg(\bigg(2-\frac{k}{\lambda\beta^{-}}\bigg)l+2\lambda\bigg) +\frac{2}{\lambda(\lambda+1)\beta^{-}} \bigg(\bigg\lfloor\frac{s_{1}-1}{2}\bigg\rfloor+\frac{\beta^{-}}{2}\bigg), \end{align*} $$

we see that Theorem 1.5, using the constant $\beta ^{-}$ , gives the conclusion of Theorem 1.4. If (4.1) does not hold for all $n\geq 1$ , then as ${s_{n+1}}/{s_{n}}\leq \beta ^{-} $ for sufficiently large n, a simple relabelling of the terms of the sequence, omitting finitely many terms at the beginning, would suffice.

This completes the proof of Theorem 1.4.