1 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 \(SU_{q}(2)\) [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 \(0< q< p\leq1\), operators \(B_{n,p,q}\) are defined as follows:

$$ \begin{aligned}[b] &B_{n,p,q}(f;x) \\ &\quad:=\frac{1}{p^{\frac{n(n-1)}{2}}}\sum_{k=0}^{n}\left [ \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right ]_{p,q}p^{\frac{k(k-1)}{2}}x^{k} \prod_{s=0}^{n-k-1} \bigl(p^{s}-q^{s}x \bigr)f \biggl(\frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggr),\quad n=1,2,\ldots, \end{aligned}$$
(1.1)

where, for any nonnegative integer k and \(0< q< p\leq1\), the \((p,q)\)-integer \([k]_{p,q}\) is defined by

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

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

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

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

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

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

$$[n]_{p,q}\to0\quad \text{or}\quad 1/(1-q)\quad \text{as }n\to\infty. $$

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 \([n]_{p_{n},q_{n}}\to \infty\) as \(n\to\infty\), 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 \(0< q_{n}< p_{n}\leq1\) such that \(\lim _{n\rightarrow\infty }p_{n}=1\) and \(\lim _{n\rightarrow\infty}q_{n}=1\). 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 \(\lim _{n\to\infty} [n]_{p_{n},q_{n}}=\infty\); otherwise, these operators do not define approximation processes. However, the claim that both \(\lim _{n\to\infty}p_{n}=1\) and \(\lim _{n\to\infty}q_{n}=1\) with \(0< q_{n}< p_{n}\leq1\) imply that \(\lim _{n\to\infty} [n]_{p_{n},q_{n}}=\infty\), in general, is not true. A counterexample is presented below.

Example 1.2

Let \(p_{n}=1-1/\sqrt{n}\), then \([n]_{p_{n},q_{n}}\to0\) for any sequence \(\{ q_{n}\}\) satisfying \(0<{q_{n}<p_{n}}\). Indeed,

$$ \begin{aligned}[b] 0&\leq[n]_{p_{n},q_{n}}=p_{n}^{n-1}+p_{n}^{n-2}q_{n}+ \cdots +p_{n}q_{n}^{n-2}+q_{n}^{n-1} \\ &\leq np_{n}^{n-1}=n(1-1/\sqrt{n})^{n-1}\sim ne^{-\sqrt{n}}\to0,\quad n\to\infty. \end{aligned}$$
(1.2)

Later, the author [8] presented a more accurate assertion: Let \(q_{n}\), \(p_{n}\) such that \(0< q_{n}< p_{n}\leq1\) and \(q_{n}\to1\), \(p_{n}\to1\), \(q^{n}_{n}\to a\), \(p^{n}_{n}\to b\) (\(a< b\)) as \(n\to\infty\), then \(\lim _{n\to \infty}[n]_{p_{n},q_{n}}\to\infty\). It is natural to ask: What is the class of sequences \((p_{n},q_{n})\) satisfying \(\lim _{n\rightarrow \infty}[n]_{p_{n},q_{n}}=\infty\) when \(\lim _{n\rightarrow\infty }p_{n}=1\) and \(\lim _{n\rightarrow\infty}q_{n}=1\) under \(0< q_{n}< p_{n}\leq1\)? Undoubtedly, this is an important problem. In this note, we will solve this problem in Section 2.

2 Main results

For \(0< q_{n}< p_{n}\leq1\), set \(q_{n}:=1-\alpha_{n}\), \(p_{n}:=1-\beta_{n}\) such that \(0\leq\beta_{n}<\alpha_{n}<1\), \(\alpha_{n}\rightarrow0\), \(\beta _{n}\rightarrow0\) as \(n\rightarrow\infty\). In the sequel, we use notation \(a_{n} \sim b_{n}\text{ $(a_{n},b_{n}>0)$}\Leftrightarrow\lim _{n\to \infty}\frac{b_{n}}{a_{n}}=1\).

First, let us present the following auxiliary proposition.

Lemma 2.1

Let \(n\in\mathbb{N}\), then as \(n\to\infty\) we have

$$ [n]_{p_{n},q_{n}}\to\infty\quad\Rightarrow\quad e^{n\beta_{n}}/n\to0. $$
(2.1)

On the other hand,

$$ e^{n\alpha_{n}}/n\to0\quad\Rightarrow\quad[n]_{p_{n},q_{n}}\to\infty. $$
(2.2)

Proof

We note that

$$ \begin{aligned}[b] [n]_{p_{n},q_{n}}&=q_{n}^{n-1}+p_{n}q_{n}^{n-2}+\cdots+p_{n}^{n-2}q_{n}+p_{n}^{n-1} =q_{n}^{n-1}\bigl(1+p_{n}/q_{n}+\cdots+(p_{n}/q_{n})^{n-1} \bigr) \\ &>nq_{n}^{n-1}=n(1-\alpha_{n})^{(-1/{\alpha_{n}})(n-1)(-\alpha_{n})} \sim n/e^{n\alpha_{n}},\quad n\to\infty. \end{aligned}$$
(2.3)

Similarly, we have

$$ \begin{aligned}[b] [n]_{p_{n},q_{n}}&=q_{n}^{n-1}+p_{n}q_{n}^{n-2}+\cdots+p_{n}^{n-2}q_{n}+p_{n}^{n-1} =p_{n}^{n-1}\bigl(1+q_{n}/p_{n}+\cdots+(q_{n}/p_{n})^{n-1} \bigr) \\ &< np_{n}^{n-1}=n(1-\beta_{n})^{(-1/{\beta_{n}})(n-1)(-\beta_{n})} \sim n/e^{n\beta_{n}},\quad n\to\infty. \end{aligned}$$
(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 \(\lim _{n\to\infty}e^{n(\beta _{n}-\alpha_{n})}=1\) and \(e^{n\beta_{n}}/n\to0\), then \([n]_{p_{n},q_{n}}\to \infty\).

(\(\mathcal{B}\)):

If \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\) and \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\to 0\), then \([n]_{p_{n},q_{n}}\to\infty\).

(\(\mathcal{C}\)):

If \(\varliminf _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}=1\) and \(\max\{e^{n\beta_{n}}/n,e^{n\beta _{n}}(\alpha_{n}-\beta_{n})\}\to0\), then \([n]_{p_{n},q_{n}}\to\infty\).

Conversely,

(\(\mathcal{B'}\)):

If \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\) and \([n]_{p_{n},q_{n}}\to\infty\), then \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\to0\).

(\(\mathcal{C'}\)):

If \(\varliminf _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1,\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}=1\) and \([n]_{p_{n},q_{n}}\to\infty\), then \(\max\{e^{n\beta_{n}}/n, e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\}\to0\).

Proof

Case (\(\mathcal{A}\)):

If \(\lim _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}=1\), since \(\lim _{n\to\infty}e^{n\alpha_{n}}/n=\lim _{n\to \infty}e^{n\beta_{n}}/n/e^{n(\beta_{n}-\alpha_{n})}\to0\) and combined with (2.2) imply \([n]_{p_{n},q_{n}}\to\infty\) (see Remark 2.1).

Case (\(\mathcal{B}\)) and Case (\(\mathcal{B'}\)):

If \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}<1\). 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

$$ \frac{\alpha_{n}-\beta_{n}}{1-\beta_{n}}\to0, $$
(2.7)

it is not difficult to obtain from (2.7) and \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}<1\) that, for sufficiently large n, there exists \(c\in(0,1)\) such that

$$ \biggl(1-\frac{\alpha_{n}-\beta_{n}}{1-\beta_{n}} \biggr)^{n}\sim e^{n(\beta_{n}-\alpha_{n})}< c. $$
(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

$$ [n]_{p_{n},q_{n}}\to\infty\quad\Leftrightarrow\quad e^{n\beta_{n}}( \alpha_{n}-\beta _{n})\to0. $$
(2.10)

This yields the proof of Case (\(\mathcal{B}\)) and Case (\(\mathcal{B'}\)).

Case (\(\mathcal{C}\)) and Case (\(\mathcal{C'}\)):

If \(\varliminf _{n\to\infty}e^{n(\beta_{n}-\alpha _{n})}<1\), \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}=1\). Since the sequence \(0< x_{n}:=e^{n(\beta_{n}-\alpha_{n})}<1\) is bounded, set \(E:=\{x|x\text{ is a limit point of }\{x_{n}\},n\geq1\}\), then \(\sup E=1\) from \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha _{n})}=1\). Now, we are going to extract a subsequence \(\{x_{n_{k}}\}\) of \(\{ x_{n}\}\) such that

  1. (a)

    \(\lim _{k\to\infty}x_{n_{k}}=1\), and

  2. (b)

    \(E^{1}:=\{x|x\text{ is a limit point of }\{x_{n}\}\setminus\{x_{n_{k}}\}\}\), 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 \(A:=\{x_{n},n\geq1\}\), take a subsequence \(\{x_{n^{(1)}_{k}}\}\) of A such that \(\lim _{k\to \infty}x_{n^{(1)}_{k}}=1\), set \(A^{(1)}:=\{x_{n^{(1)}_{k}},k\geq1\}\) and let \(A_{1}:=A\setminus A^{(1)}\). If 1 is also a limit point of \(A_{1}\), take a subsequence \(\{ x_{n^{(2)}_{k}}\}\) of \(A_{1}\) such that \(\lim _{k\to\infty}x_{n^{(2)}_{k}}=1\), set \(A^{(2)}:=\{x_{n^{(2)}_{k}},k\geq1\}\) and let \(A_{2}:=A_{1}\setminus A^{(2)}\). Continuing this process, we obtain a series of sequences, i.e., \(A^{(1)},A^{(2)},\ldots\) , 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 \(\{x_{n'_{k}}\}:=\{x_{n}\}\setminus\{x_{n_{k}}\}\), it is obvious that \(\{ n_{k},k\geq0\}\cup\{n'_{k},k\geq0\}=\mathbb{N}\). Then from (\(\mathcal{A}\)) and (a), we have seen that \([n]_{p_{n_{k}},q_{n_{k}}}\to\infty\Leftrightarrow e^{n_{k}\beta _{n_{k}}}/n_{k}\to0\) as \(k\to\infty\). And since \(\varlimsup _{n\to\infty}x_{n'_{k}}<1\) from (b), we also have seen from (\(\mathcal{B}\)) that \([n]_{p_{n'_{k}},q_{n'_{k}}}\to\infty\Leftrightarrow e^{n'_{k}\beta _{n'_{k}}}(\alpha_{n'_{k}}-\beta_{n'_{k}})\to0\) as \(k\to\infty\). In summary,

  1. (i)

    \(\lim _{n\to\infty}[n]_{p_{n},q_{n}}=\infty\Leftrightarrow\) for any subsequence \(\{N_{k}\}\) of a natural number set such that \(\lim _{k\to\infty}N_{k}=\infty\), we have \(\lim _{k\to \infty}[n]_{p_{N_{k}},q_{N_{k}}}=\infty\).

  2. (ii)

    Since \(\{n_{k}\}_{k\geq0}\) and \(\{n'_{k}\}_{k\geq0}\) are two subsequences of a natural number set such that \(\lim _{k\to\infty}n_{k}=\infty\) and \(\lim _{k\to \infty}n'_{k}=\infty\), thus as \(k\to\infty\)

    $$\lim _{n\to\infty}[n]_{p_{n},q_{n}}=\infty\quad\Rightarrow\quad \textstyle\begin{cases} [n]_{p_{n_{k}},q_{n_{k}}}\to\infty\quad\Leftrightarrow\quad e^{n_{k}\beta_{n_{k}}}/n_{k}\to0, \\ [n]_{p_{n'_{k}},q_{n'_{k}}}\to\infty\quad\Leftrightarrow\quad e^{n'_{k}\beta _{n'_{k}}}(\alpha_{n'_{k}}-\beta_{n'_{k}})\to0. \end{cases} $$

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 \([n]_{p_{n'_{k}},q_{n'_{k}}}>N\); take \(n_{0}=\max\{ n_{K_{1}},n'_{K_{2}}\}\), for every natural number \(n>n_{0}\), we have \([n]_{p_{n},q_{n}}>N\), i.e., \(\lim _{n\to\infty }[n]_{p_{n},q_{n}}=\infty\). Now we have proved that as \(k\to\infty\)

$$ \lim _{n\to\infty}[n]_{p_{n},q_{n}}=\infty\quad\Leftrightarrow\quad\Delta:= \textstyle\begin{cases} e^{n_{k}\beta_{n_{k}}}/n_{k}\to0, &\text{(c)} \\ e^{n'_{k}\beta_{n'_{k}}}(\alpha_{n'_{k}}-\beta_{n'_{k}})\to0. &\text{(d)} \end{cases} $$
(2.11)

Next, we infer that as \(k\to\infty\)

$$ \Delta\quad\Leftrightarrow\quad\lim _{n\to\infty}\max\bigl\{ e^{n\beta _{n}}/n,e^{n\beta_{n}}( \alpha_{n}-\beta_{n})\bigr\} = 0. $$
(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 \(e^{n'_{k}\beta_{n'_{k}}}/n'_{k} \to0\), and combined with (c) \(e^{n_{k}\beta_{n_{k}}}/n_{k}\to0\), we can show \(e^{n\beta_{n}}/n\to0\) by using a similar method as in the previous paragraph. On the other hand, \(e^{n_{k}\beta_{n_{k}}}(\alpha_{n_{k}}-\beta_{n_{k}})\to0\) is straightforward (since \(\lim _{k\to\infty}x_{n_{k}}=\lim _{k\to\infty}e^{n_{k}(\beta_{n_{k}}-\alpha_{n_{k}})}=1\), and note (c) in Δ), and combined with \(e^{n'_{k}\beta_{n'_{k}}}(\alpha _{n'_{k}}-\beta_{n'_{k}})\to0\), we can also deduce that \(e^{n\beta _{n}}(\alpha_{n}-\beta_{n})\to0\). This yields the proof of ‘⇒’ in (2.12).

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

Remark 2.1

In general, from (2.1) we have only \([n]_{p_{n},q_{n}}\to\infty \Rightarrow e^{n\beta_{n}}/n\to0\). However, if \(\lim _{n\to \infty}e^{n(\beta_{n}-\alpha_{n})}=1\) holds, then we have the equivalent relation \(e^{n\beta_{n}}/n\to0\Leftrightarrow[n]_{p_{n},q_{n}}\to\infty\) from (\(\mathcal{A}\)). Thus, we have:

(\(\mathcal{A}_{0}\)):

If \(\lim _{n\to\infty}e^{n(\beta _{n}-\alpha_{n})}=1\), then \(e^{n\beta_{n}}/n\to0\Leftrightarrow [n]_{p_{n},q_{n}}\to\infty\).

Similarly, from (\(\mathcal{B}\)) and (\(\mathcal {B'}\)), (\(\mathcal{C}\)) and (\(\mathcal{C'}\)) we have

(\(\mathcal{B}_{0}\)):

If \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), then \(e^{n\beta_{n}}(\alpha_{n}-\beta _{n})\to0\Leftrightarrow[n]_{p_{n},q_{n}}\to\infty\).

(\(\mathcal{C}_{0}\)):

If \(\varliminf _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}=1\), then \(\max\{e^{n\beta_{n}}/n,e^{n\beta _{n}}(\alpha_{n}-\beta_{n})\}\to0\Leftrightarrow[n]_{p_{n},q_{n}}\to\infty\).

Remark 2.2

In Case (\(\mathcal{B}\)), we can also deduce directly from \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}<1\) and \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\to0\) that \(e^{n\beta_{n}}/n\to 0\) as \(n\to\infty\). Indeed, since \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), and combined with the classical inequality on upper (lower) limit

$$\begin{aligned} \varliminf _{n\to\infty}e^{n(\alpha_{n}-\beta_{n})} \cdot \varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}\geq \varliminf _{n\to\infty} \bigl(e^{n(\alpha_{n}-\beta_{n})} \cdot e^{n(\beta_{n}-\alpha_{n})} \bigr)=1, \end{aligned}$$

we have \(\varliminf _{n\to\infty}e^{n(\alpha_{n}-\beta _{n})}>1\). Thus, for sufficiently large n, there exists \(c_{0}>1\) such that \(e^{n(\alpha_{n}-\beta_{n})}>c_{0}\). This means that \(0<(\log c_{0}) e^{n\beta_{n}}/n<e^{n\beta_{n}}/n\cdot(n(\alpha_{n}-\beta_{n}))\to0\), and we have seen that \(e^{n\beta_{n}}/n\to0\).

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 \(e^{n\beta_{n}}/n=e^{\sqrt{n}}/n \nrightarrow0\) as \(n\to \infty\), thus \([n]_{p_{n},q_{n}}\nrightarrow\infty\).

For \(p_{n}=1\), \(\beta_{n}=0\), \(q_{n}=1-\alpha_{n}\), then \(e^{n\beta_{n}}/n=1/n\to0\) and \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})=\alpha_{n}\to0\). 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 \([n]_{q_{n}}\to\infty \Leftrightarrow q_{n}\to1\) as \(n\to\infty\). See [19], Theorem 2, and [22], formula (2.7).

3 Conclusion

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