Proof. Let $\epsilon>0$ be a real number. Since $\lim _{n\rightarrow \infty }{\log n}/{n}=0$ by L’Hopital’s rule, there exists $n_0\in \mathbb {N}$ such that

$$ \begin{align*}\frac{\log x_{i}}{x_{i}}<\epsilon \quad \mbox{for all } i\geq n_0.\end{align*} $$

Therefore, for each $t> n_0$ ,

$$ \begin{align*}\dfrac{1}{t}\sum_{i=1}^{t}\dfrac{\log x_i}{x_i}\leq\dfrac{1}{t} \sum_{i\leq n_0} \dfrac{\log x_i}{x_i}+\dfrac{1}{t}\sum_{n_0<i\leq t}\dfrac{\log x_i}{x_i}<\dfrac{1}{t} \sum_{i\leq n_0} \dfrac{\log x_i}{x_i}+\frac{t-n_0}{t}\epsilon\end{align*} $$

and so

$$ \begin{align*}\lim_{t\rightarrow \infty}\dfrac{1}{t}\sum_{i=1}^{t}\dfrac{\log x_i}{x_i} \leq \epsilon.\end{align*} $$

Since $\epsilon $ is an arbitrary positive real number,

$$ \begin{align*} \lim_{t\rightarrow \infty}\dfrac{1}{t}\sum_{i=1}^{t}\dfrac{\log x_i}{x_i}=0. \\[-44pt] \end{align*} $$

Proof. We prove the result by contradiction. If possible, assume that for each $\alpha>1$ and for each $k\in \mathbb {N}$ , there exists $n(k,\alpha )>k$ such that

(3.1) $$ \begin{align} \dfrac{a_{n(k,\alpha)+1}}{a_{n(k,\alpha)}} \leq \alpha. \end{align} $$

Let $n_{0} = 1$ . For every $t\in \mathbb {N}$ , define $\alpha _{t} = 1+{1}/{t} \ \text {and} \ n_{t} = n(n_{t-1}, \alpha _{t}).$ Since $(a_{i})^{\infty }_{i=1}$ is an increasing sequence, by (3.1), we get a strictly increasing sequence $(n_{t})_{t=1}^{\infty }$ such that ${1 {\kern-1pt}\leq{\kern-1pt} {a_{n_{t}+1}}/{a_{n_{t}}} {\kern-1pt}\leq{\kern-1pt} 1{\kern-1pt}+{\kern-1pt} {1}/{t}}$ . This implies $\lim _{t\rightarrow \infty }{a_{n_{t}+1}}/{a_{n_{t}}} {\kern-1pt}={\kern-1pt}1$ and so ${\liminf _{n\rightarrow \infty } {a_{n+1}}/{a_{n}} {\kern-1pt}\leq{\kern-1pt} 1}$ , which is a contradiction.

Proof. Since $|A\cap [1,m]|> |A\cap (m,ln]|$ , we have $A\cap [1,m]\neq \varnothing $ . Let u be the largest element in the set $A\cap [1,m]$ . From the hypothesis $n\geq 2m$ , we have ${(n,ln]\subseteq u+(m,ln].}$ This gives

$$ \begin{align*}(n,ln]\subseteq ( A+((m,ln]\setminus A))\cup( u+((m,ln]\cap A)),\end{align*} $$

as $u\in A$ . Therefore, to prove the proposition, we need to show that

(4.1) $$ \begin{align} u+X\subseteq A+((m,ln]\setminus A), \end{align} $$

where $X:=\{x\in (m,ln]\cap A : u+x\in (n,ln] \}.$

Now, we prove (4.1). If $X = \varnothing $ , we are done. Otherwise, let $y\in X$ and define

$$ \begin{align*}U_y:=\{u+y-a: a\in A\cap [1,m]\}.\end{align*} $$

Since $|A\cap [1,m]|> |A\cap (m,ln]|$ , we have

(4.2) $$ \begin{align} |U_y|> |A\cap (m,ln]|. \end{align} $$

Since $y\in X$ and $u=\max (A\cap [1,m])$ ,

$$ \begin{align*}m<y\leq y+u-a<y+u \leq ln \quad\mbox{for all } a\in A\cap [1,m].\end{align*} $$

This gives

(4.3) $$ \begin{align} U_y\subseteq (m, ln]. \end{align} $$

By combining (4.2) and (4.3),

$$ \begin{align*}U_y\cap ((m,ln]\setminus A)\neq \varnothing.\end{align*} $$

Let $v\in U_y$ with $v\in (m,ln]\setminus A$ . Then,

$$ \begin{align*}u+y=b+v\in A+ (m,ln]\setminus A,\end{align*} $$

for some $b\in A$ . This proves (4.1) and completes the proof of the proposition.

Proof. The corollary follows from Proposition 4.2 by taking $x_1 = q, x_2 =4q, m=2q, n=2q$ and $B = (q,4q]\setminus A$ .

Proof of Theorem 2.6.

Since $\liminf _{n\rightarrow \infty } (a_{n+1}/a_{n})>1$ , by Lemma 3.3, there exists a natural number $n_{0}$ and a real number $\alpha>1$ such that

(4.4) $$ \begin{align} a_{n+1}\geq \alpha a_n \quad\mbox{for all } n\geq n_0. \end{align} $$

Since $\alpha>1$ , we can choose $r\in \mathbb {N}$ such that

(4.5) $$ \begin{align} \alpha^r\geq 4. \end{align} $$

Let p be a positive integer such that $p>\max \{a_{n_0}, a_{2r+1}\}$ and let q be a positive integer such that $q\geq p$ . Then, there exist positive integers $u_q$ and $m_q$ such that $u_q \geq n_0$ and

$$ \begin{align*} A\cap (q, 4q]=\{a_{u_q+1},a_{u_q+2},\ldots ,a_{u_q+m_q}\}. \end{align*} $$

This, together with (4.4) and (4.5), gives

$$ \begin{align*}a_{u_q+r+1}\geq \alpha^{r}a_{u_q+1}> 4q.\end{align*} $$

This implies that $a_{u_q+r+1}\not \in A\cap (q,4q]$ and so $|A\cap (q,4q]|=m_q\leq r$ . Therefore,

(4.6) $$ \begin{align} |A\cap (x,4x]|\leq r \quad \mbox{for all } x\geq p. \end{align} $$

Since $ p>a_{2r+1}$ , we have $|A\cap [1,x)|\geq 2r+1$ for every $x\geq p$ . This, along with inequality (4.6), implies

(4.7) $$ \begin{align} |A\cap (x,4x]|<|A \cap [1, x)| -1 \quad\mbox{for all } x\geq p. \end{align} $$

By Proposition 4.1,

$$ \begin{align*}(2x,4x]\subset A+ ((x,4x]\setminus A) \quad\mbox{for all } x\geq p.\end{align*} $$

From this expression, (4.7) and Corollary 4.3, for every $m\geq p$ , there exists a set ${S_m \subseteq (m,4m]\setminus A }$ such that

(4.8) $$ \begin{align} (2m,4m]\subset A+ S_m \end{align} $$

and

$$ \begin{align*} |S_m|\leq C m \dfrac{\log (|A\cap [1,m]|-|A\cap (m,4m]|)}{|A\cap [1,m]|-|A\cap (m,4m]|}. \end{align*} $$

Further, using (4.6) and the fact that ${\log x}/{x}$ is a decreasing function on $(1,\infty )$ ,

$$ \begin{align*} |S_m|\leq C m \dfrac{\log (|A\cap [1,m]|-r)}{|A\cap [1,m]|-r} \quad\text{for every } m\geq p. \end{align*} $$

Since $p>a_{2r+1}$ , we have $|A\cap [1,m]|\geq 2r+1$ for every $m\geq p$ . Therefore,

(4.9) $$ \begin{align} |S_m|\leq2 C m \dfrac{\log (|A\cap [1,m]|)}{|A\cap [1,m]|} \quad\text{for every } m\geq p. \end{align} $$

Define

$$ \begin{align*}B:=\bigcup_{i=2+\lfloor\log_2 (p)\rfloor}^{\infty} S_{2^i}.\end{align*} $$

By taking m as a power of $2$ in (4.8) and (4.9), we see that $S_{2^{l-1}}\subset (2^{l-1},2^{l+1}]\setminus A$ and $(2^l,2^{l+1}] \subset A+ S_{2^{l-1}}$ for every positive integer $l>1+\lfloor \log _2 (p)\rfloor $ . This proves that B is an additive complement of A in the complement of A.

Now, we claim that the density of B is zero. Observe that

$$ \begin{align*}B\cap [1,n]\subset \bigcup_{i=\gamma}^{\beta_n} S_{2^i} \quad\mbox{for all } n\in \mathbb{N},\end{align*} $$

where $\gamma =2+\lfloor \log _2 (p)\rfloor $ and $\beta _n=\max \{j:2^j<n\}.$ This gives

$$ \begin{align*}|B\cap [1,n]|\leq \sum_{i=\gamma}^{\beta_n} 2 C 2^i \dfrac{\log (|A\cap [1,2^i]|)}{|A\cap [1,2^i]|}\leq 4C\sum_{i=\gamma}^{\beta_n} \sum_{j=2^{i-1}+1}^{2^i} \dfrac{\log (|A\cap [1,j]|)}{|A\cap [1,j]|},\end{align*} $$

which implies that

$$ \begin{align*}|B\cap [1,n]|\leq 4C \sum_{j=p}^{n} \dfrac{\log (|A\cap [1,j]|)}{|A\cap [1,j]|}.\end{align*} $$

Thus, by Lemma 3.2,

$$ \begin{align*}\lim_{n\rightarrow\infty}\dfrac{|B\cap [1,n]|}{n}=0.\end{align*} $$

Therefore, the density of the set B is zero.

Proof. We have

$$ \begin{align*} \sum_{i\in B} 1_{A+i}(n)&=| A \cap (n-B)| =|A\cap [n-b,n-a)|-|A\cap (n-((a,b]\setminus B))|\\ &\geq |A\cap [n-b,n-a)|-|(a,b]\setminus B|. \\[-2.9pc] \end{align*} $$

Proof. Since every translate $A+x$ , where $x\in B$ , contains at most r elements of R,

$$ \begin{align*}\sum_{t\in R} 1_{A+b}(t)\leq r \quad\mbox{for all } b\in B.\end{align*} $$

Therefore,

$$ \begin{align*} \sum_{b\in B}\sum_{t\in R} 1_{A+b}(t)\leq \sum_{b\in B}r= r|B|.\\[-45pt] \end{align*} $$

Proof of Proposition 4.2.

Define $f: \mathbb {N}\rightarrow \mathcal {P}(\mathbb {N})$ by

$$ \begin{align*}f(x):=(A+x)\cap (m, m+n] \quad\text{for all } x\in \mathbb{N}.\end{align*} $$

Since $(m, m+n]\subset \bigcup _{i\in B} (A+ i)$ , by induction, we can choose a sequence $(b_i)_{i=1}^t$ of distinct elements of B such that for every $j\in [1,t]$ ,

(5.1) $$ \begin{align} &\bigg|f(b_j)\setminus \bigg(\bigcup_{i=1}^{j-1}f(b_i)\bigg) \bigg|= \max\bigg\{ \bigg|f(r) \setminus \bigg(\bigcup_{i=1}^{j-1}f(b_i)\bigg)\bigg|: r\in B\setminus \bigcup_{i=1}^{j-1}\{b_i\}\bigg\}, \end{align} $$

where t is the smallest positive integer such that

$$ \begin{align*} (m,m+n] \subseteq \bigcup_{i=1}^{t} f(b_i). \end{align*} $$

Let $S=\bigcup _{i=1}^t \{b_i\}$ . Then, $(m,m+n] \subseteq A+S$ .

To complete the proof of the proposition, we shall now compute the cardinality of the set S. Let $q:=|f(b_1)|$ . For every $j \in \mathbb {N}$ , define

(5.2) $$ \begin{align} T_j:=\bigg\{r\in [1,t]: \bigg|f(b_r)\setminus \bigg(\bigcup_{i=1}^{r-1}f(b_i)\bigg) \bigg|=j\bigg\} \quad \text{and}\quad K_j :=|T_j|. \end{align} $$

From (5.1), $T_j=\varnothing $ and $ K_j=0 $ for each integer $j>q$ . Also, from (5.1) and (5.2), we obtain a decreasing sequence $(n_i)_{i=0}^q$ of positive integers such that $n_0=t$ , $n_q=1$ , and

$$ \begin{align*} T_q=[1,n_{q-1}], \quad T_{q-1}=(n_{q-1},n_{q-2}], \ldots, T_{j}=(n_j,n_{j-1}], \ldots, T_1=(n_{1}, t]. \end{align*} $$

Therefore,

(5.3) $$ \begin{align} |S|=&\bigg|\bigcup_{x=1}^{q} T_x\bigg|=\sum_{x=1}^{q}K_x \end{align} $$

and computation of $K_x$ will help us to compute $|S|$ .

Define $H: [0,q]\rightarrow \mathcal {P}(\mathbb {N})$ by

$$ \begin{align*}H(x):=(m,m+n]\setminus \bigg(\bigcup_{i=x+1}^{q}\bigcup_{r\in T_i} f(b_r)\bigg).\end{align*} $$

One can easily see that $H(x-1)\subset H(x)$ for $x\geq 1$ . This implies that

$$ \begin{align*} |H(x)|-|H(x-1)|&= |H(x)\setminus H(x-1)|\\ &=\bigg|\bigg(\bigcup_{i=x}^{q}\bigcup_{r\in T_x} f(b_r)\bigg)\setminus \bigg(\bigcup_{i=x+1}^{q}\bigcup_{r\in T_i} f(b_r)\bigg)\bigg|\\ &=\bigg|\bigg(\bigcup_{r\in T_x} f(b_r)\bigg)\setminus \bigg(\bigcup_{i=x+1}^{q}\bigcup_{r\in T_i} f(b_r)\bigg)\bigg|\\ &=\bigg|\bigg(\bigcup_{r=n_x+1}^{n_{x-1}}f(b_r)\bigg)\setminus \bigg(\bigcup_{r=1}^{n_x} f(b_r)\bigg)\bigg|\\ &=\bigg|\bigg(\bigcup_{r=n_x+1}^{n_{x-1}}(f(b_r)\setminus \bigcup_{i=n_x+1}^{r-1}f(b_i))\bigg)\setminus \bigg(\bigcup_{r=1}^{n_x} f(b_r)\bigg)\bigg|\\ &=\bigg|\bigcup_{r=n_x+1}^{n_{x-1}}\bigg(f(b_r)\setminus \bigcup_{i=1}^{r-1}f(b_i)\bigg)\bigg|=\bigg|\bigcup_{r\in T_x}\bigg( f(b_r)\setminus \bigg(\bigcup_{i=1}^{r-1} f(b_i)\bigg)\bigg)\bigg|\\ &=\displaystyle\sum_{r\in T_x}\bigg|f(b_r)\setminus \bigg(\bigcup_{i=1}^{r-1} f(b_i)\bigg)\bigg|=x|T_x|. \end{align*} $$

Therefore,

(5.4) $$ \begin{align} |H(x)|-|H(x-1)|=xK_x. \end{align} $$

From this equation, we can determine $K_x$ by computing the cardinality of $H(x)$ .

We compute the cardinality of $H(x)$ by computing the lower and upper bounds of $\sum _{i\in B}\sum _{t\in H(x)} 1_{A+i}(t)$ . To get the upper bound, we count the number of elements of the set $f(u)$ inside the set $H(x)$ for every $x\geq 1$ and for every $u\in B$ . Observe that for every $u\in B$ and $x\in [0,q]$ ,

$$ \begin{align*} |f(u)\cap H(x)|&=\bigg|f(u)\setminus \bigg(\bigcup_{i=x+1}^{q}\bigcup_{b\in T_i} f(b)\bigg)\bigg| =\bigg|f(u)\setminus \bigg(\bigcup_{i=1}^{n_x} f(b_i)\bigg)\bigg|. \end{align*} $$

From (5.1), this gives

$$ \begin{align*} |f(u)\cap H(x)|\leq \bigg|f(b_{n_x+1})\setminus \bigg(\bigcup_{i=1}^{n_x} f(b_i)\bigg)\bigg|\leq x, \end{align*} $$

because $n_x+1\in T_j$ for some $j\in [1,x]$ . Therefore,

$$ \begin{align*} |f(u)\cap H(x)|\leq x \quad\mbox{for all } u\in B, x\in [0,q]. \end{align*} $$

This, together with Lemma 5.2, gives

(5.5) $$ \begin{align} \sum_{i\in B}\sum_{t\in H(x)} 1_{A+i}(t)\leq x|B| \quad\mbox{for all } x\in [0,q]. \end{align} $$

We use Lemma 5.1 to get a lower bound for $\sum _{i\in B}\sum _{t\in H(x)} 1_{A+i}(t)$ . Since $B \subseteq (x_{1}, x_{2}]$ , Lemma 5.1 implies that

$$ \begin{align*} \sum_{i\in B}\sum_{t\in H(x)}1_{A+i}(t)&=\sum_{t\in H(x)}\sum_{i\in B} 1_{A+i}(t)\\ &\geq \sum_{t\in H(x)}(|A \cap [t-x_2,t-x_1)|- (x_2-x_1-|B|)) \end{align*} $$

for every $x \in [0,q]$ . Note that $t-x_2\leq 0$ for every $t\in H(x)$ , because $H(x)\subset (m,m+n]$ and $m+n\leq x_2$ . Thus, for every $x\in [0,q]$ ,

(5.6) $$ \begin{align} \sum_{i\in B}\sum_{t\in H(x)}1_{A+i}(t)&\geq \sum_{t\in H(x)}(|A \cap [1,t-x_1)|- (x_2-x_1 -|B|))\nonumber\\ & \geq |H(x)| (|A \cap [1,m-x_1)|- (x_2-x_1 -|B|)). \end{align} $$

Combining (5.5) and (5.6) gives

$$ \begin{align*} |H(x)| (|A \cap [1,m-x_1)|- (x_2-x_1 -|B|))\leq x|B|. \end{align*} $$

Since $|A \cap [1,m-x_1)|> (x_2-x_1-|B|)$ , for every $x\in [0,q]$ , we obtain

(5.7) $$ \begin{align} |H(x)|\leq \dfrac{x|B|}{|A \cap [1,m-x_1)|- (x_2-x_1-|B|)}. \end{align} $$

Now, we can compute the upper bound of the cardinality of the set S. Let $q_0$ be an arbitrary positive integer. From (5.3),

(5.8) $$ \begin{align} |S|=\begin{cases} \displaystyle\sum_{x=1}^{q_0}K_x+\sum_{x=q_0+1}^{q}K_{x} & \mbox{for } q_0<q, \\ \displaystyle\sum_{x=1}^{q}K_x & \mbox{for }q_0\geq q. \end{cases} \end{align} $$

Since $f(b_i)\subset (m,m+n]$ for each $i\in \bigcup _{x=1}^{q} T_x$ ,

$$ \begin{align*} n\geq \bigg|\bigcup_{x=q_0+1}^{q}\bigcup_{j\in T_x}\bigg(f(b_{j})\setminus \bigg(\bigcup_{i=1}^{j-1} f(b_i)\bigg)\bigg)\bigg|&=\sum_{x=q_0+1}^{q}\sum_{j\in T_x}\bigg|f(b_{j})\setminus \bigg(\bigcup_{i=1}^{j-1} f(b_i)\bigg)\bigg|\\ &=\sum_{x=q_0+1}^{q}\sum_{j\in T_x}x =\sum_{x=q_0+1}^{q}xK_x\\ &>q_0\sum_{x=q_0+1}^{q}K_x. \end{align*} $$

Therefore,

(5.9) $$ \begin{align} \sum_{x=q_0+1}^{q}K_{x}<\dfrac{n}{q_0}. \end{align} $$

From (5.4),

$$ \begin{align*} \sum_{x=1}^{q'}K_{x} &= \sum_{x=1}^{q'}\dfrac{|H(x)|-|H(x-1)|}{x} = \sum_{x=1}^{q'-1}\dfrac{|H(x)|}{x(x+1)} + \dfrac{1}{q'}|H(q')| \quad\mbox{for all } q'\leq q. \end{align*} $$

From this equation and the upper bound for $|H(x)|$ in (5.7),

(5.10) $$ \begin{align} \sum_{x=1}^{q'}K_{x} \leq \frac{|B|}{(|A \cap [1,m-x_1)|- (x_2-x_1-|B|))}\bigg(\sum_{x=1}^{q'}\dfrac{1}{x}\bigg) \quad\mbox{for all } q'\leq q. \end{align} $$

By combining (5.8), (5.9) and (5.10),

$$ \begin{align*} |S|\leq \frac{|B|}{|A \cap [1,m-x_1)|- (x_2-x_1-|B|)}\bigg(\sum_{x=1}^{q_0}\dfrac{1}{x}\bigg)+\frac{n}{q_0}. \end{align*} $$

Since $\sum _{i=1}^{j} {1}/{i}<C\log j$ for every $j\in \mathbb {N}$ , where C is a constant independent of j, this gives

(5.11) $$ \begin{align} |S|\leq \frac{C|B|\log q_0}{|A \cap [1,m-x_1)|- (x_2-x_1-|B|)}+\frac{n}{q_0}. \end{align} $$

Since $q_0$ is an arbitrary positive integer, we choose

$$ \begin{align*} q_0=\bigg\lfloor\dfrac{|A \cap [1,m-x_1)|- (x_2-x_1-|B|)}{\log({|A \cap [1,m-x_1)|- (x_2-x_1-|B|)})}\bigg\rfloor. \end{align*} $$

Putting this value of $q_0$ in (5.11),

$$ \begin{align*} |S| \leq (C|B|+n) \dfrac{\log(|A \cap [1,m-x_1)|- (x_2-x_1-|B|))}{|A \cap [1,m-x_1)|- (x_2-x_1-|B|)}.\\[-41pt] \end{align*} $$