A basic problem of $(p,q)$-Bernstein-type operators

Abstract

In this note, we give an elaboration of a basic problem on convergence theorem of $(p,q)$-analogue of Bernstein-type operators. By some classical analysis techniques, we derive an exact class of $(p_n,q_n)$-integer satisfying ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}=\mathrm{\infty }$ with ${lim}_{n\to \mathrm{\infty }}{p}_n=1$ and ${lim}_{n\to \mathrm{\infty }}{q}_n=1$ under $0<q_n<p_n\le 1$. Our results provide an erratum to corresponding results on $(p,q)$-analogue of Bernstein-type operators that appeared in recent literature.

Introduction

During the last decades, the applications of q-calculus emerged as a new area in the field of approximation theory. The rapid development of q-calculus has led to the discovery of various generalizations of Bernstein polynomials involving q-integers. A detailed review of the results on q-Bernstein polynomials along with an extensive bibliography is given in [1]. The q-Bernstein polynomials are shown to be closely related to the q-deformed binomial distribution [2]. It plays an important role in the q-boson theory giving a q-deformation of the quantum harmonic formalism [3]. The q-analogue of the boson operator calculus has proved to be a powerful tool in theoretical physics. It provides explicit expressions for the representations of the quantum group $S{U}_q(2)$ [4]. Meanwhile, the $(p,q)$-integers were introduced in order to generalize or unify several forms of q-oscillator algebras well known in the earlier physics literature related to the representation theory of single parameter quantum algebras [5].

Recently, $(p,q)$-integers have been introduced into classical linear positive operators to construct new approximation processes. A sequence of $(p,q)$-analogue of Bernstein operators was first introduced by Mursaleen [6, 7]. Besides, $(p,q)$-analogues of Szász-Mirakyan [8], Baskakov Kantorovich [9], Bleimann-Butzer-Hahn [10] and Kantorovich-type Bernstein-Stancu-Schurer [11] operators were also considered, see [12–15]. For further developments, one can also refer to [8, 16–18]. These operators are double parameters corresponding to p and q versus single parameter q-Bernstein-type operators [1, 19, 20]. The aim of these generalizations is to provide appropriate and powerful tools to application areas such as numerical analysis, computer-aided geometric design and solutions of differential equations (see, e.g., [21]).

For example, consider the $(p,q)$-analogue of the Bernstein operators proposed in [7]. Given $f\in C[0,1]$ and $0<q<p\le 1$, operators $B_{n,p,q}$ are defined as follows:

$$\begin{array}{rl}& B_{n,p,q}(f;x)\\ & :=\frac{1}{p^{\frac{n(n-1)}{2}}}\sum _{k=0}^n{\left[\begin{array}{c}n\\ k\end{array}\right]}_{p,q}{p}^{\frac{k(k-1)}{2}}{x}^k\prod _{s=0}^{n-k-1}(p^s-q^{s}x)f\left(\frac{{[k]}_{p,q}}{p^{k-n}{[n]}_{p,q}}\right),n=1,2,\dots ,\end{array}$$ 1.1

where, for any nonnegative integer k and $0<q<p\le 1$, the $(p,q)$-integer ${[k]}_{p,q}$ is defined by

$${[k]}_{p,q}:=p^{k-1}+p^{k-2}q+\cdots +q^{k-1}=\frac{p^k-q^k}{p-q}(k=0,1,2,\dots ),{[0]}_{p,q}:=0,$$

and the $(p,q)$-factorial ${[k]}_{p,q}!$ is defined by

$${[k]}_{p,q}!:={[1]}_{p,q}{[2]}_{p,q}\cdots {[k]}_{p,q}(k=1,2,\dots ),{[0]}_{p,q}!:=1.$$

For integers k, n with $0\le k\le n$, the $(p,q)$-binomial coefficient is defined by

$${\left[\begin{array}{c}n\\ k\end{array}\right]}_{p,q}:=\frac{{[n]}_{p,q}!}{{[k]}_{p,q}!{[n-k]}_{p,q}!}.$$

In general, we expect $B_{n,p,q}(f;x)$ to converge to $f(x)$ as $n\to \mathrm{\infty }$. But we see that, for fixed value of p and q with $q\in (0,1)$ and $p\in (q,1]$,

$${[n]}_{p,q}\to 0\text{or}1/(1-q)\text{as }n\to \mathrm{\infty }.$$

To obtain a sequence of generalized $(p,q)$-analogue Bernstein polynomials which converge, we let $q_n\in (0,1)$ and $p_n\in (q_n,1]$ depend on n. We then choose a sequence $(p_n,q_n)$ such that ${[n]}_{p_n,q_n}\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, to ensure that $B_{n,p,q}(f;x)$ converge to $f(x)$.

The convergence theorems for $(p,q)$-analogue Bernstein-type operators were established in some recent papers (see [6], Theorem 3.1 (Remark 3.1), [7], Theorem 1, and further reading [10], Theorem 2.2, [14], Theorem 3.1, [12], Theorem 3, and [11], Remark 2.3, see also [9, 15]). For example, Mursaleen [7] gives the following.

Theorem 1.1

Let $0<q_n<p_n\le 1$ such that ${lim}_{n\to \mathrm{\infty }}{p}_n=1$ and ${lim}_{n\to \mathrm{\infty }}{q}_n=1$. Then, for each $f\in C[0,1]$, $B_{n,p_n,q_n}(f;x)$ converge uniformly to f on $[0,1]$.

All linear positive operators mentioned in the articles cited above require that ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}=\mathrm{\infty }$; otherwise, these operators do not define approximation processes. However, the claim that both ${lim}_{n\to \mathrm{\infty }}{p}_n=1$ and ${lim}_{n\to \mathrm{\infty }}{q}_n=1$ with $0<q_n<p_n\le 1$ imply that ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}=\mathrm{\infty }$, in general, is not true. A counterexample is presented below.

Example 1.2

Let $p_n=1-1/\sqrt{n}$, then ${[n]}_{p_n,q_n}\to 0$ for any sequence $\{q_n\}$ satisfying $0<q_n<p_n$. Indeed,

$$\begin{array}{rl}0& \le {[n]}_{p_n,q_n}=p_n^{n-1}+p_n^{n-2}{q}_n+\cdots +p_n{q}_n^{n-2}+q_n^{n-1}\\ & \le n{p}_n^{n-1}=n{(1-1/\sqrt{n})}^{n-1}\sim n{e}^{-\sqrt{n}}\to 0,n\to \mathrm{\infty }.\end{array}$$ 1.2

Later, the author [8] presented a more accurate assertion: Let $q_n$, $p_n$ such that $0<q_n<p_n\le 1$ and $q_n\to 1$, $p_n\to 1$, $q_n^n\to a$, $p_n^n\to b$ ($a<b$) as $n\to \mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}\to \mathrm{\infty }$. It is natural to ask: What is the class of sequences $(p_n,q_n)$ satisfying ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}=\mathrm{\infty }$ when ${lim}_{n\to \mathrm{\infty }}{p}_n=1$ and ${lim}_{n\to \mathrm{\infty }}{q}_n=1$ under $0<q_n<p_n\le 1$? Undoubtedly, this is an important problem. In this note, we will solve this problem in Section 2.

Main results

For $0<q_n<p_n\le 1$, set $q_n:=1-{\alpha }_n$, $p_n:=1-{\beta }_n$ such that $0\le {\beta }_n<{\alpha }_n<1$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$ as $n\to \mathrm{\infty }$. In the sequel, we use notation $a_n\sim b_n(a_n,b_n>0)\iff {lim}_{n\to \mathrm{\infty }}\frac{b_n}{a_n}=1$.

First, let us present the following auxiliary proposition.

Lemma 2.1

Let $n\in \mathbb{N}$, then as $n\to \mathrm{\infty }$ we have

$${[n]}_{p_n,q_n}\to \mathrm{\infty }\Rightarrow e^{n{\beta }_n}/n\to 0.$$ 2.1

On the other hand,

$$e^{n{\alpha }_n}/n\to 0\Rightarrow {[n]}_{p_n,q_n}\to \mathrm{\infty }.$$ 2.2

Proof

We note that

$$\begin{array}{rl}{[n]}_{p_n,q_n}& =q_n^{n-1}+p_n{q}_n^{n-2}+\cdots +p_n^{n-2}{q}_n+p_n^{n-1}=q_n^{n-1}(1+p_n/q_n+\cdots +{(p_n/q_n)}^{n-1})\\ & >n{q}_n^{n-1}=n{(1-{\alpha }_n)}^{(-1/{\alpha }_n)(n-1)(-{\alpha }_n)}\sim n/e^{n{\alpha }_n},n\to \mathrm{\infty }.\end{array}$$ 2.3

Similarly, we have

$$\begin{array}{rl}{[n]}_{p_n,q_n}& =q_n^{n-1}+p_n{q}_n^{n-2}+\cdots +p_n^{n-2}{q}_n+p_n^{n-1}=p_n^{n-1}(1+q_n/p_n+\cdots +{(q_n/p_n)}^{n-1})\\ & <n{p}_n^{n-1}=n{(1-{\beta }_n)}^{(-1/{\beta }_n)(n-1)(-{\beta }_n)}\sim n/e^{n{\beta }_n},n\to \mathrm{\infty }.\end{array}$$ 2.4

Therefore, from (2.3) and (2.4), for sufficiently large n, there exist two positive real numbers $C_1$ and $C_2$ satisfying

$$C_1\cdot n/e^{n{\alpha }_n}<{[n]}_{p_n,q_n}<C_2\cdot n/e^{n{\beta }_n}.$$ 2.5

This yields the proof. □

The main result of this work is expressed by the next assertion.

Theorem 2.1

The following statements are true:

($\mathcal{A}$)

If ${lim}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$ and $e^{n{\beta }_n}/n\to 0$, then ${[n]}_{p_n,q_n}\to \mathrm{\infty }$.

($\mathcal{B}$)

If ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$ and $e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\to 0$, then ${[n]}_{p_n,q_n}\to \mathrm{\infty }$.

($\mathcal{C}$)

If ${\underset{\_}{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$, ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$ and $max\{e^{n{\beta }_n}/n,e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\}\to 0$, then ${[n]}_{p_n,q_n}\to \mathrm{\infty }$.

Conversely,

(${\mathcal{B}}^{\prime }$)

If ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$ and ${[n]}_{p_n,q_n}\to \mathrm{\infty }$, then $e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\to 0$.

(${\mathcal{C}}^{\prime }$)

If ${\underset{\_}{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1,{\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$ and ${[n]}_{p_n,q_n}\to \mathrm{\infty }$, then $max\{e^{n{\beta }_n}/n,e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\}\to 0$.

Proof

Case ($\mathcal{A}$):

If ${lim}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$, since ${lim}_{n\to \mathrm{\infty }}{e}^{n{\alpha }_n}/n={lim}_{n\to \mathrm{\infty }}{e}^{n{\beta }_n}/n/e^{n({\beta }_n-{\alpha }_n)}\to 0$ and combined with (2.2) imply ${[n]}_{p_n,q_n}\to \mathrm{\infty }$ (see Remark 2.1).

Case ($\mathcal{B}$) and Case (${\mathcal{B}}^{\prime }$):

If ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$. Note that, for sufficiently large n,

$${[n]}_{p_n,q_n}=p_n^{n-1}\frac{1-{(q_n/p_n)}^n}{1-q_n/p_n}\sim \frac{1}{e^{n{\beta }_n}}\frac{1-{(1-({\alpha }_n-{\beta }_n)/(1-{\beta }_n))}^n}{({\alpha }_n-{\beta }_n)/(1-{\beta }_n)}.$$ 2.6

Since

$$\frac{{\alpha }_n-{\beta }_n}{1-{\beta }_n}\to 0,$$ 2.7

it is not difficult to obtain from (2.7) and ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$ that, for sufficiently large n, there exists $c\in (0,1)$ such that

$${(1-\frac{{\alpha }_n-{\beta }_n}{1-{\beta }_n})}^n\sim e^{n({\beta }_n-{\alpha }_n)}<c.$$ 2.8

Set $d_n:=1-{(1-\frac{{\alpha }_n-{\beta }_n}{1-{\beta }_n})}^n$, then for sufficiently large n, $d_n>1-c$. Thus, from (2.6), (2.7) and (2.8), we have

$${[n]}_{p_n,q_n}\sim \frac{1}{e^{n{\beta }_n}}\cdot \frac{d_n}{{\alpha }_n-{\beta }_n},$$ 2.9

which entails that

$${[n]}_{p_n,q_n}\to \mathrm{\infty }\iff e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\to 0.$$ 2.10

This yields the proof of Case ($\mathcal{B}$) and Case (${\mathcal{B}}^{\prime }$).

Case ($\mathcal{C}$) and Case (${\mathcal{C}}^{\prime }$):

If ${\underset{\_}{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$, ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$. Since the sequence $0<x_n:=e^{n({\beta }_n-{\alpha }_n)}<1$ is bounded, set $E:=\{x|x\text{ is a limit point of }\{x_n\},n\ge 1\}$, then $supE=1$ from ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$. Now, we are going to extract a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

1. ${lim}_{k\to \mathrm{\infty }}{x}_{n_k}=1$, and

2. $E^1:=\{x|x\text{ is a limit point of }\{x_n\}\setminus \{x_{n_k}\}\}$, with $sup{E}^1<1$.

We verify that it is possible to extract such a subsequence $\{x_{n_k}\}$. Since 1 is a limit point of $A:=\{x_n,n\ge 1\}$, take a subsequence $\{x_{n_k^{(1)}}\}$ of A such that ${lim}_{k\to \mathrm{\infty }}{x}_{n_k^{(1)}}=1$, set $A^{(1)}:=\{x_{n_k^{(1)}},k\ge 1\}$ and let $A_1:=A\setminus A^{(1)}$. If 1 is also a limit point of $A_1$, take a subsequence $\{x_{n_k^{(2)}}\}$ of $A_1$ such that ${lim}_{k\to \mathrm{\infty }}{x}_{n_k^{(2)}}=1$, set $A^{(2)}:=\{x_{n_k^{(2)}},k\ge 1\}$ and let $A_2:=A_1\setminus A^{(2)}$. Continuing this process, we obtain a series of sequences, i.e., $A^{(1)},A^{(2)},\dots$ , set $A_f:=A\setminus {\bigcup }_{s\in \mathbb{I}\subseteq \mathbb{N}}{A}^{(s)}$. Since A is a countable set, this process will stop until 1 is not a limit point of $A_f$ after finite or countable steps; otherwise, we will see that 1 is the only limit point of A, which contradicts to the assumptions of Case ($\mathcal{C}$). Then we can take the subsequence $\{x_{n_k}\}={\bigcup }_{s\in \mathbb{N}}{A}^{(s)}$ which satisfies (a) and (b).

Set $\{x_{n_k^{\prime }}\}:=\{x_n\}\setminus \{x_{n_k}\}$, it is obvious that $\{n_k,k\ge 0\}\cup \{n_k^{\prime },k\ge 0\}=\mathbb{N}$. Then from ($\mathcal{A}$) and (a), we have seen that ${[n]}_{p_{n_k},q_{n_k}}\to \mathrm{\infty }\iff e^{n_k{\beta }_{n_k}}/n_k\to 0$ as $k\to \mathrm{\infty }$. And since ${\overline{lim}}_{n\to \mathrm{\infty }}{x}_{n_k^{\prime }}<1$ from (b), we also have seen from ($\mathcal{B}$) that ${[n]}_{p_{n_k^{\prime }},q_{n_k^{\prime }}}\to \mathrm{\infty }\iff e^{n_k^{\prime }{\beta }_{n_k^{\prime }}}({\alpha }_{n_k^{\prime }}-{\beta }_{n_k^{\prime }})\to 0$ as $k\to \mathrm{\infty }$. In summary,

• (i) ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}=\mathrm{\infty }\iff$ for any subsequence $\{N_k\}$ of a natural number set such that ${lim}_{k\to \mathrm{\infty }}{N}_k=\mathrm{\infty }$, we have ${lim}_{k\to \mathrm{\infty }}{[n]}_{p_{N_k},q_{N_k}}=\mathrm{\infty }$.
• (ii)
  Since ${\{n_k\}}_{k\ge 0}$ and ${\{n_k^{\prime }\}}_{k\ge 0}$ are two subsequences of a natural number set such that ${lim}_{k\to \mathrm{\infty }}{n}_k=\mathrm{\infty }$ and ${lim}_{k\to \mathrm{\infty }}{n}_k^{\prime }=\mathrm{\infty }$, thus as $k\to \mathrm{\infty }$

  $$\underset{n\to \mathrm{\infty }}{lim}{[n]}_{p_n,q_n}=\mathrm{\infty }\Rightarrow \{\begin{array}{c}{[n]}_{p_{n_k},q_{n_k}}\to \mathrm{\infty }\iff e^{n_k{\beta }_{n_k}}/n_k\to 0,\hfill \\ {[n]}_{p_{n_k^{\prime }},q_{n_k^{\prime }}}\to \mathrm{\infty }\iff e^{n_k^{\prime }{\beta }_{n_k^{\prime }}}({\alpha }_{n_k^{\prime }}-{\beta }_{n_k^{\prime }})\to 0.\hfill \end{array}$$

We assert that the inverse proposition of (ii) also holds. Indeed, for each sufficiently large $N>0$, there exists a positive integer $K_1$ such that for every natural number $k>K_1$, we have ${[n]}_{p_{n_k},q_{n_k}}>N$; meanwhile, for the previous $N>0$, there exists a positive integer $K_2$ such that for every natural number $k>K_2$, we have ${[n]}_{p_{n_k^{\prime }},q_{n_k^{\prime }}}>N$; take $n_0=max\{n_{K_1},n_{K_2}^{\prime }\}$, for every natural number $n>n_0$, we have ${[n]}_{p_n,q_n}>N$, i.e., ${lim}_{n\to \mathrm{\infty }}{[n]}_{p_n,q_n}=\mathrm{\infty }$. Now we have proved that as $k\to \mathrm{\infty }$

$$\underset{n\to \mathrm{\infty }}{lim}{[n]}_{p_n,q_n}=\mathrm{\infty }\iff \mathrm{\Delta }:=\{\begin{array}{cc}{e}^{n_k{\beta }_{n_k}}/n_k\to 0,\hfill & \text{(c)}\hfill \\ e^{n_k^{\prime }{\beta }_{n_k^{\prime }}}({\alpha }_{n_k^{\prime }}-{\beta }_{n_k^{\prime }})\to 0.\hfill & \text{(d)}\hfill \end{array}$$ 2.11

Next, we infer that as $k\to \mathrm{\infty }$

$$\mathrm{\Delta }\iff \underset{n\to \mathrm{\infty }}{lim}max\{e^{n{\beta }_n}/n,e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\}=0.$$ 2.12

‘⇐’ of (2.12) is straightforward. Now we show ‘⇒’. On the one hand, by Remark (2.2), we know from (d) of Δ that $e^{n_k^{\prime }{\beta }_{n_k^{\prime }}}/n_k^{\prime }\to 0$, and combined with (c) $e^{n_k{\beta }_{n_k}}/n_k\to 0$, we can show $e^{n{\beta }_n}/n\to 0$ by using a similar method as in the previous paragraph. On the other hand, $e^{n_k{\beta }_{n_k}}({\alpha }_{n_k}-{\beta }_{n_k})\to 0$ is straightforward (since ${lim}_{k\to \mathrm{\infty }}{x}_{n_k}={lim}_{k\to \mathrm{\infty }}{e}^{n_k({\beta }_{n_k}-{\alpha }_{n_k})}=1$, and note (c) in Δ), and combined with $e^{n_k^{\prime }{\beta }_{n_k^{\prime }}}({\alpha }_{n_k^{\prime }}-{\beta }_{n_k^{\prime }})\to 0$, we can also deduce that $e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\to 0$. This yields the proof of ‘⇒’ in (2.12).

Therefore, ($\mathcal{C}$) and (${\mathcal{C}}^{\prime }$) follow from (2.11) and (2.12). □

Remark 2.1

In general, from (2.1) we have only ${[n]}_{p_n,q_n}\to \mathrm{\infty }\Rightarrow e^{n{\beta }_n}/n\to 0$. However, if ${lim}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$ holds, then we have the equivalent relation $e^{n{\beta }_n}/n\to 0\iff {[n]}_{p_n,q_n}\to \mathrm{\infty }$ from ($\mathcal{A}$). Thus, we have:

(${\mathcal{A}}_0$)

If ${lim}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$, then $e^{n{\beta }_n}/n\to 0\iff {[n]}_{p_n,q_n}\to \mathrm{\infty }$.

Similarly, from ($\mathcal{B}$) and (${\mathcal{B}}^{\prime }$), ($\mathcal{C}$) and (${\mathcal{C}}^{\prime }$) we have

(${\mathcal{B}}_0$)

If ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$, then $e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\to 0\iff {[n]}_{p_n,q_n}\to \mathrm{\infty }$.

(${\mathcal{C}}_0$)

If ${\underset{\_}{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$, ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}=1$, then $max\{e^{n{\beta }_n}/n,e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\}\to 0\iff {[n]}_{p_n,q_n}\to \mathrm{\infty }$.

Remark 2.2

In Case ($\mathcal{B}$), we can also deduce directly from ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$ and $e^{n{\beta }_n}({\alpha }_n-{\beta }_n)\to 0$ that $e^{n{\beta }_n}/n\to 0$ as $n\to \mathrm{\infty }$. Indeed, since ${\overline{lim}}_{n\to \mathrm{\infty }}{e}^{n({\beta }_n-{\alpha }_n)}<1$, and combined with the classical inequality on upper (lower) limit

$$\underset{n\to \mathrm{\infty }}{\underset{\_}{lim}}{e}^{n({\alpha }_n-{\beta }_n)}\cdot \underset{n\to \mathrm{\infty }}{\overline{lim}}{e}^{n({\beta }_n-{\alpha }_n)}\ge \underset{n\to \mathrm{\infty }}{\underset{\_}{lim}}(e^{n({\alpha }_n-{\beta }_n)}\cdot e^{n({\beta }_n-{\alpha }_n)})=1,$$

we have ${\underset{\_}{lim}}_{n\to \mathrm{\infty }}{e}^{n({\alpha }_n-{\beta }_n)}>1$. Thus, for sufficiently large n, there exists $c_0>1$ such that $e^{n({\alpha }_n-{\beta }_n)}>c_0$. This means that $0<(log{c}_0)e^{n{\beta }_n}/n<e^{n{\beta }_n}/n\cdot (n({\alpha }_n-{\beta }_n))\to 0$, and we have seen that $e^{n{\beta }_n}/n\to 0$.

Remark 2.3

Now we utilize Theorem 2.1 (Remark 2.1) to elaborate Example 1.2 again. In the example, ${\beta }_n=1/\sqrt{n}$, while $e^{n{\beta }_n}/n=e^{\sqrt{n}}/n\nrightarrow 0$ as $n\to \mathrm{\infty }$, thus ${[n]}_{p_n,q_n}\nrightarrow \mathrm{\infty }$.

For $p_n=1$, ${\beta }_n=0$, $q_n=1-{\alpha }_n$, then $e^{n{\beta }_n}/n=1/n\to 0$ and $e^{n{\beta }_n}({\alpha }_n-{\beta }_n)={\alpha }_n\to 0$. Thus any case of (${\mathcal{A}}_0$)-(${\mathcal{C}}_0$) is straightforward. In this case, $(p_n,q_n)$-integer reduces to $q_n$-integer, and it is known that ${[n]}_{q_n}\to \mathrm{\infty }\iff q_n\to 1$ as $n\to \mathrm{\infty }$. See [19], Theorem 2, and [22], formula (2.7).

Conclusion

In this note, we mainly obtain the sufficient and necessary conditions for $(p,q)$-integer ${[n]}_{p,q}$ tending to infinity as $n\to \mathrm{\infty }$. The conclusion guarantees the $(p,q)$-analogue of Bernstein-type operators to be approximation processes as $n\to \mathrm{\infty }$.