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Padé approximant related to inequalities involving the constant e and a generalized Carleman-type inequality

Journal of Inequalities and Applications20172017:205

https://doi.org/10.1186/s13660-017-1479-8

  • Received: 9 June 2017
  • Accepted: 4 August 2017
  • Published:

Abstract

Based on the Padé approximation method, in this paper we determine the coefficients \(a_{j}\) and \(b_{j}\) (\(1\leq j \leq k\)) such that
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}= \frac{x^{k}+a_{1}x^{k-1}+ \cdots +a_{k}}{x^{k}+b_{1}x^{k-1}+\cdots +b_{k}}+O \biggl( \frac{1}{x ^{2k+1}} \biggr) , \quad x\to \infty , $$
where \(k\geq 1\) is any given integer. Based on the obtained result, we establish new upper bounds for \(( 1+1/x ) ^{x}\). As an application, we give a generalized Carleman-type inequality.

Keywords

  • Carleman’s inequality
  • weight coefficient
  • Padé approximant

MSC

  • 26D15
  • 41A60

1 Introduction

Let \(a_{n} \geq 0\) for \(n \in \mathbb{N}:=\{1, 2, \ldots \}\) and \(0<\sum_{n=1}^{\infty }a_{n}<\infty \). Then
$$ \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n}< e \sum_{n=1}^{ \infty }a_{n}. $$
(1.1)
The constant e is the best possible. The inequality (1.1) was presented in 1922 in [1] by Carleman and it is called Carleman’s inequality. Carleman discovered this inequality during his important work on quasi-analytical functions.
Carleman’s inequality (1.1) was generalized by Hardy [2] (see also [3, p.256]) as follows: If \(a_{n} \geq 0\), \(\lambda_{n}>0\), \(\Lambda_{n}=\sum_{m=1}^{n}\lambda_{m}\) for \(n \in \mathbb{N}\), and \(0<\sum_{n=1}^{\infty }\lambda_{n}a_{n}< \infty \), then
$$ \sum_{n=1}^{\infty } \lambda_{n} \bigl(a_{1}^{\lambda_{1}}a_{2}^{\lambda_{2}} \cdots a_{n}^{\lambda_{n}} \bigr)^{1/ \Lambda_{n}}< e \sum _{n=1} ^{ \infty }\lambda_{n}a_{n}. $$
(1.2)
Note that inequality (1.2) is usually referred to as a Carleman-type inequality or weighted Carleman-type inequality. In [2], Hardy himself said that it was Pólya who pointed out this inequality to him.
In [420], some strengthened and generalized results of (1.1) and (1.2) have been given by estimating the weight coefficient \(( 1+1/n ) ^{n}\). For example, Yang [17] proved that, for \(n\in \mathbb{N}\),
$$ e \biggl( 1-\frac{1}{2(n+\frac{5}{6})} \biggr) < \biggl( 1+ \frac{1}{n} \biggr) ^{n}< e \biggl( 1-\frac{1}{2(n+1)} \biggr) , $$
(1.3)
and then used it to obtain the following strengthened Carleman inequality:
$$ \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n}< e \sum_{n=1}^{ \infty } \biggl( 1-\frac{1}{2(n+1)} \biggr) a_{n}. $$
(1.4)
Xie and Zhong [15] proved that, for \(x\geq 1\),
$$ e \biggl( 1-\frac{7}{14x+12} \biggr) < \biggl( 1+ \frac{1}{x} \biggr) ^{x}< e \biggl( 1-\frac{6}{12x+11} \biggr) , $$
(1.5)
and then used it to improve the Carleman-type inequality (1.2) as follows. If \(0<\lambda_{n+1} \leq \lambda_{n}\), \(\Lambda_{n}=\sum_{m=1}^{n}\lambda_{m}\), \(a_{n} \geq 0\) for \(n \in \mathbb{N}\), and \(0<\sum_{n=1}^{\infty }\lambda_{n}a_{n}< \infty \), then
$$\begin{aligned} \sum_{n=1}^{\infty } \lambda_{n+1} \bigl(a_{1}^{\lambda_{1}}a_{2}^{ \lambda_{2}} \cdots a_{n}^{\lambda_{n}} \bigr)^{1/ \Lambda_{n}} < e\sum _{n=1} ^{\infty } \biggl( 1- \frac{6}{12(\frac{\Lambda_{n}}{\lambda_{n}})+11} \biggr) \lambda_{n}a _{n}. \end{aligned}$$
(1.6)
Taking \(\lambda_{n} \equiv 1\) in (1.6) yields
$$\begin{aligned} \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n} < e\sum_{n=1}^{ \infty } \biggl( 1-\frac{6}{12n+11} \biggr) a_{n}, \end{aligned}$$
(1.7)
which improves (1.4).
Recently, Mortici and Hu [14] proved that, for \(x\geq 1\),
$$\begin{aligned} &\frac{x+\frac{5}{12}}{x+\frac{11}{12}}-\frac{5}{288x^{3}}+\frac{343}{8{,}640x^{4}} -\frac{2{,}621}{41{,}472x^{5}} \\ &\quad < \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr)^{x} < \frac{x+\frac{5}{12}}{x+\frac{11}{12}}-\frac{5}{288x^{3}}+\frac{343}{8{,}640x ^{4}}-\frac{2{,}621}{41{,}472x^{5}}+ \frac{300{,}901}{3{,}483{,}648x^{6}}, \end{aligned}$$
(1.8)
and then they used it to establish the following improvement of Carleman’s inequality:
$$\begin{aligned} &\sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n} \\ &\quad < e \sum_{n=1}^{ \infty } \biggl( \frac{12n+5}{12n+11}-\frac{5}{288n^{3}}+\frac{343}{8{,}640n ^{4}}- \frac{2{,}621}{41{,}472n^{5}}+\frac{300{,}901}{3{,}483{,}648n^{6}} \biggr) a _{n}, \end{aligned}$$
which can be written as
$$\begin{aligned} \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n}< e \sum_{n=1}^{ \infty } ( 1-\varepsilon_{n} ) a_{n}, \end{aligned}$$
(1.9)
where
$$\begin{aligned} \varepsilon_{n}=\frac{104{,}509{,}440n^{6}+3{,}628{,}800n^{4}-4{,}971{,}456n^{3}+5{,}603{,}472n ^{2}-5{,}945{,}040n-16{,}549{,}555}{17{,}418{,}240n^{6}(12n+11)}. \end{aligned}$$
(1.10)
For information as regards the history of Carleman-type inequalities, please refer to [2124].
It follows from (1.8) that
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}=\frac{x+\frac{5}{12}}{x+ \frac{11}{12}}+O \biggl( \frac{1}{x^{3}} \biggr) ,\quad x\to \infty . $$
(1.11)
Using the Padé approximation method, in Section 3 we derive (1.11) and the following approximation formula:
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}=\frac{x^{2}+ \frac{87}{100}x+ \frac{37}{240}}{x^{2}+\frac{137}{100}x+ \frac{457}{1{,}200}}+O \biggl( \frac{1}{x ^{5}} \biggr) , \quad x\to \infty . $$
(1.12)
Equation (1.12) motivates us to present the following inequality:
$$ \biggl( 1+\frac{1}{n} \biggr) ^{n}< e \biggl( \frac{n^{2}+\frac{87}{100}n+ \frac{37}{240}}{n^{2}+\frac{137}{100}n+\frac{457}{1{,}200}} \biggr) =e \biggl( 1-\frac{8(75n+34)}{1{,}200n^{2}+1{,}644n+457} \biggr) , \quad n\in \mathbb{N}. $$
(1.13)
Following the same method used in the proof of Theorem 3.2, we can prove the inequality (1.13). We here omit it.
According to Pólya’s proof of (1.1) in [25],
$$ \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n}\leq \sum_{n=1} ^{\infty } \biggl( 1+\frac{1}{n} \biggr) ^{n}a_{n}, $$
(1.14)
and then the following strengthened Carleman’s inequality is derived directly from (1.13):
$$ \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n}< e \sum_{n=1}^{ \infty } \biggl( 1-\frac{8(75n+34)}{1{,}200n^{2}+1{,}644n+457} \biggr) a_{n}, $$
(1.15)
which improves (1.7).
Based on the Padé approximation method, we determine the coefficients \(a_{j}\) and \(b_{j}\) (\(1\leq j\leq k\)) such that
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}=\frac{x^{k}+a_{1}x^{k-1}+ \cdots +a_{k}}{x^{k}+b_{1}x^{k-1}+\cdots +b_{k}}+O \biggl( \frac{1}{x ^{2k+1}} \biggr) ,\quad x\to \infty , $$
(1.16)
where \(k\geq 1\) is any given integer. Based on the obtained result, we establish new upper bounds for \(( 1+1/x ) ^{x}\). As an application, we give a generalization to the Carleman-type inequality.

The numerical values given have been calculated using the computer program MAPLE 13.

2 A useful lemma

For later use, we introduce the following set of partitions of an integer \(n \in \mathbb{N} =\mathbb{N}_{0} \setminus \{ 0 \} := \{1,\,2,\,3,\,\ldots \}\):
$$ \mathcal{A}_{n} := \bigl\{ ( k_{1}, k_{2}, \ldots , k_{n} ) \in \mathbb{N}_{0}^{n} : k_{1}+2k_{2}+ \cdots +nk_{n}=n \bigr\} . $$
(2.1)
In number theory, the partition function \(p(n)\) represents the number of possible partitions of \(n \in \mathbb{N}\) (e.g., the number of distinct ways of representing n as a sum of natural numbers regardless of order). By convention, \(p(0) = 1\) and \(p(n) = 0\) if n is a negative integer. For more information on the partition function \(p(n)\), please refer to [26] and the references therein. The first values of the partition function \(p(n)\) are (starting with \(p(0)=1\)) (see [27]):
$$ 1, \,1,\,2, \,3,\,5,\,7,\,11,\,15,\,22,\,30, \,42,\,\ldots . $$
It is easy to see that the cardinality of the set \(\mathcal{A}_{n}\) is equal to the partition function \(p(n)\). Now we are ready to present a formula which determines the coefficients \(a_{j}\) in (2.2) with the help of the partition function given by the following lemma.

Lemma 2.1

[28]

The following approximation formula holds true:
$$ \biggl( 1+\frac{1}{x} \biggr) ^{x}=e \sum _{j=0}^{\infty } \frac{c _{j}}{x ^{j}} \quad \textit{as}\ x\to \infty , $$
(2.2)
where the coefficients \(c_{j}\) \((j \in \mathbb{N})\) are given by
$$ \begin{aligned}& c_{0}=1 \quad \textit{and}\quad c_{j}=(-1)^{j} \sum_{ ( k_{1}, k_{2}, \ldots , k_{j} ) \in \mathcal{A} _{j}} \frac{1}{k_{1}!k_{2}!\cdots k_{j}!} \biggl( \frac{1}{2} \biggr) ^{k_{1}} \biggl( \frac{1}{3} \biggr) ^{k_{2}}\cdots \biggl( \frac{1}{j+1} \biggr) ^{k_{j}},\end{aligned}$$
(2.3)
where the \(\mathcal{A}_{j}\) \((\textit{for}\ j \in \mathbb{N})\) are given in (2.1).

3 Padé approximant related to asymptotics for the constant e

For later use, we introduce the Padé approximant (see [2934]). Let f be a formal power series
$$\begin{aligned} f(t)=c_{0}+c_{1}t+c_{2}t^{2}+ \cdots . \end{aligned}$$
(3.1)
The Padé approximation of order \((p, q)\) of the function f is the rational function, denoted by
$$\begin{aligned}{} [p/q]_{f}(t)=\frac{\sum_{j=0}^{p}a_{j}t^{j}}{1+\sum_{j=1}^{q}b_{j}t ^{j}}, \end{aligned}$$
(3.2)
where \(p\geq 0\) and \(q\geq 1\) are two given integers, the coefficients \(a_{j}\) and \(b_{j}\) are given by (see [2931, 33, 34])
$$\begin{aligned} \textstyle\begin{cases} a_{0}=c_{0}, \\ a_{1}=c_{0}b_{1}+c_{1}, \\ a_{2}=c_{0}b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{p} = c_{0}b_{p}+\cdots + c_{p-1}b_{1} + c_{p}, \\ 0 = c_{p+1} + c_{p}b_{1} + \cdots + c_{p-q+1}b_{q}, \\ \vdots & \\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots + c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}$$
(3.3)
and the following holds:
$$\begin{aligned}{} [p/q]_{f}(t)- f (t) = O\bigl(t^{p+q+1} \bigr). \end{aligned}$$
(3.4)
Thus, the first \(p + q + 1\) coefficients of the series expansion of \([p/q]_{f}\) are identical to those of f. Moreover, we have (see [32])
$$ \begin{aligned} &[p/q]_{f}(t)= \frac{ \left \vert \begin{matrix}{} t^{q}f_{p-q}(t) & t^{q-1}f_{p-q+1}(t) &\cdots &f_{p}(t) \\ c_{p-q+1} & c_{p-q+2} &\cdots &c_{p+1} \\ \vdots &\vdots &\ddots &\vdots \\ c_{p} & c_{p+1} &\cdots &c_{p+q} \end{matrix} \right \vert } { \left \vert \begin{matrix}{} t^{q} & t^{q-1} &\cdots &1 \\ c_{p-q+1} & c_{p-q+2} &\cdots &c_{p+1} \\ \vdots &\vdots &\ddots &\vdots \\ c_{p} & c_{p+1} &\cdots &c_{p+q} \end{matrix} \right \vert } , \end{aligned} $$
(3.5)
with \(f_{n}(x) = c_{0}+ c_{1}x+ \cdots + c_{n}x^{n}\), the nth partial sum of the series f (\(f_{n}\) is identically zero for \(n < 0\)).
Let
$$\begin{aligned} f(x)=\frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}. \end{aligned}$$
(3.6)
It follows from (2.2) that, as \(x\to \infty \),
$$\begin{aligned} f(x) &=\sum_{j=0}^{\infty } \frac{c_{j}}{x^{j}}=1-\frac{1}{2x}+\frac{11}{24x ^{2}}-\frac{7}{16x^{3}}+ \frac{2{,}447}{5{,}760x^{4}}-\frac{959}{2{,}304x^{5}}+\frac{238{,}043}{580{,}608x ^{6}}-\cdots , \end{aligned}$$
(3.7)
with the coefficients \(c_{j}\) given by (2.3). In what follows, the function f is given in (3.6).
We now give a derivation of equation (1.11). To this end, we consider
$$\begin{aligned}{} [1/1]_{f}(x)=\frac{\sum_{j=0}^{1}a_{j}x^{-j}}{1+\sum_{j=1}^{1}b_{j}x ^{-j}}. \end{aligned}$$
Noting that
$$\begin{aligned} c_{0}=1, \qquad c_{1}=-\frac{1}{2},\qquad c_{2}=\frac{11}{24}, \qquad c _{3}=- \frac{7}{16}, \qquad c_{4}=\frac{2{,}447}{5{,}760} \end{aligned}$$
(3.8)
holds, we have, by (3.3),
$$\begin{aligned} \textstyle\begin{cases} a_{0}=1, \\a_{1}=b_{1}-\frac{1}{2}, \\0 =\frac{11}{24}- \frac{1}{2}b_{1}, \end{cases}\displaystyle \end{aligned}$$
that is,
$$\begin{aligned} a_{0}=1, \qquad a_{1}=\frac{5}{12},\qquad b_{1}=\frac{11}{12}. \end{aligned}$$
We thus obtain
$$ [1/1]_{f}(x)= \frac{1+\frac{5}{12x}}{1+\frac{11}{12x}}=\frac{x+ \frac{5}{12}}{x+\frac{11}{12}}, $$
(3.9)
and we have, by (3.4),
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}-\frac{x+\frac{5}{12}}{x+ \frac{11}{12}}= O \biggl( \frac{1}{x^{3}} \biggr) , \quad x\to \infty . $$
(3.10)
We now give a derivation of equation (1.12). To this end, we consider
$$\begin{aligned}{} [2/2]_{f}(x)=\frac{\sum_{j=0}^{2}a_{j}x^{-j}}{1+\sum_{j=1}^{2}b_{j}x ^{-j}}. \end{aligned}$$
Noting that (3.8) holds, we have, by (3.3),
$$\begin{aligned} \textstyle\begin{cases} a_{0}=1, \\a_{1}=b_{1}-\frac{1}{2}, \\a_{2} =b_{2}-\frac{1}{2}b_{1}+ \frac{11}{24} , \\0 = - \frac{7}{16}+\frac{11}{24}b_{1}-\frac{1}{2}b_{2} , \\0 = \frac{2{,}447}{5{,}760}-\frac{7}{16}b_{1}+\frac{11}{24}b_{2}, \end{cases}\displaystyle \end{aligned}$$
that is,
$$\begin{aligned} a_{0}=1,\qquad a_{1}=\frac{87}{100},\qquad a_{2} = \frac{37}{240}, \qquad b_{1}=\frac{137}{100},\qquad b_{2}=\frac{457}{1{,}200}. \end{aligned}$$
We thus obtain
$$ [2/2]_{f}(x)= \frac{1+\frac{87}{100x}+\frac{37}{240x^{2}}}{1+ \frac{137}{100x}+\frac{457}{1{,}200x^{2}}}=\frac{x^{2}+\frac{87}{100}x+ \frac{37}{240}}{x^{2}+\frac{137}{100}x+\frac{457}{1{,}200}} $$
(3.11)
and we have, by (3.4),
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}-\frac{x^{2}+ \frac{87}{100}x+ \frac{37}{240}}{x^{2}+\frac{137}{100}x+ \frac{457}{1{,}200}}= O \biggl( \frac{1}{x ^{5}} \biggr) , \quad x\to \infty . $$
(3.12)

Using the Padé approximation method and the expansion (3.7), we now present a general result given by Theorem 3.1. As a consequence, we obtain (1.16).

Theorem 3.1

The Padé approximation of order \((p, q)\) of the asymptotic formula of the function \(f(x)=\frac{1}{e} ( 1+\frac{1}{x} ) ^{x}\) (at the point \(x=\infty \)) is the following rational function:
$$\begin{aligned}{} [p/q]_{f}(x)=\frac{1+\sum_{j=1}^{p}a_{j}x^{-j}}{1+\sum_{j=1}^{q}b_{j}x ^{-j}}=x^{q-p} \biggl( \frac{x^{p}+a_{1}x^{p-1}+\cdots +a_{p}}{x^{q}+b _{1}x^{q-1}+\cdots +b_{q}} \biggr) , \end{aligned}$$
(3.13)
where \(p\geq 1\) and \(q\geq 1\) are two given integers, the coefficients \(a_{j}\) and \(b_{j}\) are given by
$$\begin{aligned} \textstyle\begin{cases} a_{1}=b_{1}+c_{1}, \\ a_{2}=b_{2}+c_{1}b_{1}+c_{2}, \\ \vdots \\ a_{p} = b_{p}+\cdots + c_{p-1}b_{1} + c_{p}, \\ 0 = c_{p+1} + c_{p}b_{1} + \cdots + c_{p-q+1}b_{q}, \\ \vdots & \\ 0 = c_{p+q} + c_{p+q-1}b_{1} + \cdots + c_{p}b_{q}, \end{cases}\displaystyle \end{aligned}$$
(3.14)
\(c_{j}\) is given in (2.3), and the following holds:
$$\begin{aligned} f (x) -[p/q]_{f}(x) = O \biggl( \frac{1}{x^{p+q+1}} \biggr) ,\quad x\to \infty . \end{aligned}$$
(3.15)
Moreover, we have
$$ \begin{aligned} &[p/q]_{f}(x)= \frac{ \left \vert \begin{matrix}{} \frac{1}{x^{q}}f_{p-q}(x) & \frac{1}{x^{q-1}}f_{p-q+1}(x) &\cdots &f _{p}(x) \\c_{p-q+1} & c_{p-q+2} &\cdots &c_{p+1} \\ \vdots &\vdots &\ddots &\vdots \\ c_{p} & c_{p+1} &\cdots &c_{p+q} \end{matrix} \right \vert } { \left \vert \begin{matrix}{} \frac{1}{x^{q}} & \frac{1}{x^{q-1}} &\cdots &1 \\c_{p-q+1} & c_{p-q+2} &\cdots &c_{p+1} \\ \vdots &\vdots &\ddots &\vdots \\ c_{p} & c_{p+1} &\cdots &c_{p+q} \end{matrix} \right \vert } , \end{aligned} $$
(3.16)
with \(f_{n}(x)=\sum_{j=0}^{n}\frac{c_{j}}{x^{j}}\), the nth partial sum of the asymptotic series (3.7).

Remark 3.1

Using (3.16), we can also derive (3.9) and (3.11). Indeed, we have
$$ \begin{aligned}{} [1/1]_{f}(x) &=\frac{ \left \vert \begin{matrix}{} \frac{1}{x}f_{0}(x) &f_{1}(x) \\c_{1} &c_{2} \\\end{matrix} \right \vert } { \left \vert \begin{matrix}{} \frac{1}{x} &1 \\c_{1} &c_{2} \\\end{matrix} \right \vert } = \frac{ \left \vert \begin{matrix}{} \frac{1}{x} &1-\frac{1}{2x} \\-\frac{1}{2} &\frac{11}{24} \\\end{matrix} \right \vert } { \left \vert \begin{matrix}{} \frac{1}{x} &1 \\-\frac{1}{2} &\frac{11}{24} \\\end{matrix} \right \vert } \\ & =\frac{x+\frac{5}{12}}{x+\frac{11}{12}} \end{aligned} $$
and
$$ \begin{aligned}{} [2/2]_{f}(x) &=\frac{ \left \vert \begin{matrix}{} \frac{1}{x^{2}}f_{0}(x) & \frac{1}{x}f_{1}(x) &f_{2}(x) \\c_{1} &c_{2} &c_{3} \\c_{2} &c_{3} &c_{4} \\\end{matrix} \right \vert } { \left \vert \begin{matrix}{} \frac{1}{x^{2}} &\frac{1}{x} &1 \\c_{1} &c_{2} &c_{3} \\c_{2} &c_{3} &c_{4} \\\end{matrix} \right \vert } = \frac{ \left \vert \begin{matrix}{} \frac{1}{x^{2}} & \frac{1}{x} ( 1-\frac{1}{2x} ) &1- \frac{1}{2x}+\frac{11}{24x^{2}} \\-\frac{1}{2} &\frac{11}{24} &-\frac{7}{16} \\\frac{11}{24} &-\frac{7}{16} &\frac{2{,}447}{5{,}760} \\\end{matrix} \right \vert } { \left \vert \begin{matrix}{} \frac{1}{x^{2}} &\frac{1}{x} &1 \\-\frac{1}{2} &\frac{11}{24} &-\frac{7}{16} \\\frac{11}{24} &-\frac{7}{16} &\frac{2{,}447}{5{,}760} \\\end{matrix} \right \vert } \\ &=\frac{x^{2}+\frac{87}{100}x+\frac{37}{240}}{x^{2}+\frac{137}{100}x+ \frac{457}{1{,}200}}. \end{aligned} $$

Remark 3.2

Setting \((p, q)=(k, k)\) in (3.15), we obtain (1.16).

Setting
$$ (p, q)=(3, 3) \quad \text{and}\quad (p, q)=(4, 4), $$
respectively, we obtain by Theorem 3.1, as \(x\to \infty \),
$$ \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}=\frac{x^{3}+ \frac{162{,}713}{121{,}212}x^{2}+\frac{13{,}927}{26{,}936}x+\frac{41{,}501}{786{,}240}}{x ^{3}+\frac{223{,}319}{121{,}212}x^{2}+\frac{237{,}551}{242{,}424}x+ \frac{3{,}950{,}767}{29{,}090{,}880}}+O \biggl( \frac{1}{x^{7}} \biggr) $$
(3.17)
and
$$\begin{aligned} \frac{1}{e} \biggl( 1+\frac{1}{x} \biggr) ^{x}={}&\frac{x^{4}+ \frac{1{,}157{,}406{,}727}{634{,}301{,}284}x^{3}+\frac{8{,}452{,}872{,}239}{7{,}611{,}615{,}408}x^{2}+ \frac{81{,}587{,}251{,}465}{319{,}687{,}847{,}136}x+\frac{15{,}842{,}677}{924{,}376{,}320}}{x^{4}+ \frac{1{,}474{,}557{,}369}{634{,}301{,}284}x^{3}+\frac{13{,}811{,}559{,}391}{7{,}611{,}615{,}408}x^{2}+ \frac{170{,}870{,}679{,}559}{319{,}687{,}847{,}136}x+ \frac{1{,}724{,}393{,}461{,}793}{38{,}362{,}541{,}656{,}320}} \\ & {}+O \biggl( \frac{1}{x^{9}} \biggr) . \end{aligned}$$
(3.18)
Equations (3.17) and (3.18) motivate us to establish the following theorem.

Theorem 3.2

For \(x>0\),
$$\begin{aligned} \biggl( 1+\frac{1}{x} \biggr) ^{x} &< e \biggl( \frac{x^{3}+ \frac{162{,}713}{121{,}212}x^{2}+\frac{13{,}927}{26{,}936}x+\frac{41{,}501}{786{,}240}}{x ^{3}+\frac{223{,}319}{121{,}212}x^{2}+\frac{237{,}551}{242{,}424}x+ \frac{3{,}950{,}767}{29{,}090{,}880}} \biggr) \end{aligned}$$
(3.19)
and
$$\begin{aligned} &\biggl( 1+\frac{1}{x} \biggr) ^{x} \\ &\quad < e \biggl( \frac{x^{4}+ \frac{1{,}157{,}406{,}727}{634{,}301{,}284}x^{3}+\frac{8{,}452{,}872{,}239}{7{,}611{,}615{,}408}x^{2}+ \frac{81{,}587{,}251{,}465}{319{,}687{,}847{,}136}x+\frac{15{,}842{,}677}{924{,}376{,}320}}{x^{4}+ \frac{1{,}474{,}557{,}369}{634{,}301{,}284}x^{3}+\frac{13{,}811{,}559{,}391}{7{,}611{,}615{,}408}x^{2}+ \frac{170{,}870{,}679{,}559}{319{,}687{,}847{,}136}x+ \frac{1{,}724{,}393{,}461{,}793}{38{,}362{,}541{,}656{,}320}} \biggr) . \end{aligned}$$
(3.20)

Proof

We only prove the inequality (3.20). The proof of (3.19) is analogous. In order to prove (3.20), it suffices to show that
$$\begin{aligned} F(x)< 0 \quad \text{for}\ x>0, \end{aligned}$$
where
$$\begin{aligned} F(x)={}&x\ln \biggl( 1+\frac{1}{x} \biggr) -1 \\ &{}-\ln \biggl( \frac{x^{4}+ \frac{1{,}157{,}406{,}727}{634{,}301{,}284}x^{3}+\frac{8{,}452{,}872{,}239}{7{,}611{,}615{,}408}x^{2}+ \frac{81{,}587{,}251{,}465}{319{,}687{,}847{,}136}x+\frac{15{,}842{,}677}{924{,}376{,}320}}{x^{4}+ \frac{1{,}474{,}557{,}369}{634{,}301{,}284}x^{3}+\frac{13{,}811{,}559{,}391}{7{,}611{,}615{,}408}x^{2}+ \frac{170{,}870{,}679{,}559}{319{,}687{,}847{,}136}x+ \frac{1{,}724{,}393{,}461{,}793}{38{,}362{,}541{,}656{,}320}} \biggr) . \end{aligned}$$
Differentiation yields
$$\begin{aligned} F'(x)=\ln \biggl( 1+\frac{1}{x} \biggr) -\frac{P_{8}(x)}{P_{9}(x)}, \end{aligned}$$
where
$$\begin{aligned} P_{8}(x) ={}&4{,}534{,}960{,}145{,}139{,}175{,}220{,}907{,}601+89{,}156{,}435{,}404{,}854{,}709{,}617{,}164{,}400x \\ & +753{,}611{,}422{,}427{,}554{,}143{,}580{,}166{,}880x^{2} \\ & +3{,}400{,}732{,}641{,}706{,}885{,}239{,}015{,}784{,}320x^{3} \\ & +8{,}959{,}898{,}009{,}119{,}992{,}740{,}647{,}591{,}680x^{4} \\ & +14{,}212{,}846{,}466{,}921{,}911{,}377{,}490{,}790{,}400x^{5} \\ & +13{,}355{,}464{,}865{,}044{,}929{,}241{,}744{,}281{,}600x^{6} \\ & +6{,}842{,}437{,}276{,}900{,}714{,}847{,}214{,}796{,}800x^{7} \\ & +1{,}471{,}684{,}602{,}332{,}887{,}248{,}995{,}942{,}400x^{8} \end{aligned}$$
and
$$\begin{aligned} P_{9}(x) ={}&\bigl(38{,}362{,}541{,}656{,}320x^{4}+69{,}999{,}958{,}848{,}960x^{3}+42{,}602{,}476{,}084{,}560x ^{2} \\ &{}+9{,}790{,}470{,}175{,}800x +657{,}486{,}938{,}177\bigr) \bigl(38{,}362{,}541{,}656{,}320x^{4} \\ &{}+89{,}181{,}229{,}677{,}120x^{3}+69{,}610{,}259{,}330{,}640x ^{2} +20{,}504{,}481{,}547{,}080x \\ &{}+1{,}724{,}393{,}461{,}793\bigr) (x+1). \end{aligned}$$
Differentiating \(F'(x)\), we find
$$\begin{aligned} F''(x)=-\frac{Q_{8}(x)}{Q_{19}(x)}, \end{aligned}$$
where
$$\begin{aligned} Q_{8}(x) ={}&1{,}285{,}425{,}745{,}031{,}439{,}744{,}924{,}351{,}944{,}181{,}267{,}498{,}830{,}297{,}392{,}321 \\ & +28{,}378{,}097{,}964{,}665{,}213{,}870{,}448{,}253{,}775{,}917{,}974{,}735{,}833{,}555{,}915{,}520x \\ & +247{,}639{,}239{,}538{,}550{,}650{,}618{,}428{,}925{,}475{,}351{,}177{,}418{,}903{,}828{,}519{,}360x^{2} \\ & +1{,}131{,}116{,}309{,}072{,}948{,}249{,}686{,}419{,}776{,}599{,}013{,}563{,}965{,}352{,}036{,}853{,}760x^{3} \\ & +2{,}998{,}129{,}273{,}934{,}033{,}621{,}834{,}452{,}343{,}529{,}577{,}599{,}070{,}175{,}646{,}117{,}120x^{4} \\ & +4{,}775{,}194{,}702{,}079{,}256{,}668{,}486{,}950{,}292{,}217{,}012{,}539{,}098{,}845{,}384{,}867{,}840x^{5} \\ & +4{,}503{,}188{,}365{,}939{,}207{,}771{,}317{,}966{,}173{,}833{,}346{,}921{,}724{,}385{,}791{,}590{,}400x^{6} \\ & +2{,}315{,}562{,}242{,}935{,}704{,}170{,}341{,}114{,}308{,}201{,}588{,}127{,}064{,}283{,}807{,}744{,}000x^{7} \\ & +500{,}009{,}489{,}498{,}922{,}911{,}594{,}629{,}442{,}997{,}057{,}334{,}195{,}586{,}408{,}448{,}000x^{8} \end{aligned}$$
and
$$\begin{aligned} Q_{19}(x) ={}&x\bigl(38{,}362{,}541{,}656{,}320x^{4}+69{,}999{,}958{,}848{,}960x^{3}+42{,}602{,}476{,}084{,}560x ^{2} \\ &+9{,}790{,}470{,}175{,}800x +657{,}486{,}938{,}177\bigr)^{2}\bigl(38{,}362{,}541{,}656{,}320x^{4} \\ &+89{,}181{,}229{,}677{,}120x^{3}+69{,}610{,}259{,}330{,}640x ^{2} +20{,}504{,}481{,}547{,}080x \\ &+1{,}724{,}393{,}461{,}793\bigr)^{2}(x+1)^{2}. \end{aligned}$$
Hence, \(F''(x)<0\) for \(x>0\), and we have
$$\begin{aligned} F'(x)>\lim_{t\to \infty }F'(t)=0 \quad \Longrightarrow \quad F(x)< \lim_{t\to \infty }F(t)=0 \quad \text{for}\ x>0. \end{aligned}$$
The proof is complete. □
The inequality (3.20) can be written as
$$\begin{aligned} \biggl( 1+\frac{1}{x} \biggr) ^{x}< e \bigl(1- \mathcal{E}(x) \bigr),\quad x>0, \end{aligned}$$
(3.21)
where
$$\begin{aligned} \mathcal{E}(x)={}& 48\bigl(399{,}609{,}808{,}920x^{3}+562{,}662{,}150{,}960x^{2} \\ &{}+223{,}208{,}570{,}235x+22{,}227{,}219{,}242\bigr) /\bigl(38{,}362{,}541{,}656{,}320x^{4} \\ &{}+89{,}181{,}229{,}677{,}120x^{3}+69{,}610{,}259{,}330{,}640x^{2}+20{,}504{,}481{,}547{,}080x \\ &{}+1{,}724{,}393{,}461{,}793\bigr). \end{aligned}$$
(3.22)

4 A generalized Carleman-type inequality

Theorem 4.1

Let \(0<\lambda_{n+1} \leq \lambda_{n}\), \(\Lambda_{n}=\sum_{m=1}^{n} \lambda_{m}\) \((\Lambda_{n}\geq 1)\), \(a_{n} \geq 0\) \((n \in \mathbb{N})\) and \(0<\sum_{n=1}^{\infty }\lambda_{n}a_{n}<\infty \). Then, for \(0< p \leq 1\),
$$\begin{aligned} & \sum_{n=1}^{\infty } \lambda_{n+1}\bigl(a_{1}^{\lambda_{1}}a_{2}^{\lambda_{2}} \cdots a_{n}^{\lambda_{n}}\bigr)^{1/ \Lambda_{n}} \\ &\quad{} < \frac{e^{p}}{p} \sum_{n=1}^{\infty } \biggl( 1-\mathcal{E} \biggl( \frac{\Lambda_{n}}{ \lambda _{n}} \biggr) \biggr) ^{p} \lambda_{n}a_{n}^{p} \Lambda_{n}^{p-1} \Biggl( \sum_{k=1}^{n} \lambda_{k}(c_{k}a_{k})^{p} \Biggr) ^{(1-p)/p}, \end{aligned}$$
(4.1)
where \(\mathcal{E}(x)\) is given in (3.22) and
$$ c_{n}^{\lambda_{n}}=\frac{(\Lambda_{n+1})^{\Lambda_{n}}}{(\Lambda_{n})^{ \Lambda_{n-1}}}. $$

Proof

The inequality
$$\begin{aligned} & \sum_{n=1}^{\infty } \lambda_{n+1}\bigl(a_{1}^{\lambda_{1}}a_{2}^{\lambda_{2}} \cdots a_{n}^{\lambda_{n}}\bigr)^{1/ \Lambda_{n}} \\ &\quad \leq \frac{1}{p}\sum_{m=1}^{\infty } \biggl( 1+\frac{1}{\Lambda_{m}/ \lambda_{m}} \biggr) ^{p \Lambda_{m} / \lambda_{m}} \lambda_{m}a_{m} ^{p}\Lambda_{m}^{p-1} \Biggl( \sum _{k=1}^{m}\lambda_{k}(c_{k}a_{k})^{p} \Biggr) ^{(1-p)/p} \end{aligned}$$
(4.2)
has been proved in Theorem 2.2 of [9] (see also [11, p.96]). From the above inequality and (3.20), we obtain (4.1). The proof is complete. □

Remark 4.1

In Theorem 2.2 of [9], \(c_{k}^{\lambda_{n}}=\frac{( \Lambda_{n+1})^{\Lambda_{n}}}{(\Lambda_{n})^{\Lambda_{n-1}}}\) should be \(c_{n}^{\lambda_{n}}=\frac{(\Lambda_{n+1})^{\Lambda_{n}}}{(\Lambda_{n})^{ \Lambda_{n-1}}}\); see [9, p.44, line 3]. Likewise, \(c_{s}^{\lambda_{n}}=\frac{(\Lambda_{n+1})^{\Lambda_{n}}}{(\Lambda_{n})^{ \Lambda_{n-1}}}\) in Theorem 3.1 of [11] should be \(c_{n}^{\lambda_{n}}=\frac{(\Lambda_{n+1})^{\Lambda_{n}}}{(\Lambda_{n})^{ \Lambda_{n-1}}}\); see [11, p.96, equation (9)].

Remark 4.2

Taking \(p=1\) in (4.1) yields
$$\begin{aligned} \sum_{n=1}^{\infty } \lambda_{n+1}\bigl(a_{1}^{\lambda_{1}}a_{2}^{\lambda_{2}} \cdots a_{n}^{\lambda_{n}}\bigr)^{1/ \Lambda_{n}} < e\sum _{n=1}^{ \infty } \biggl( 1-\mathcal{E} \biggl( \frac{\Lambda_{n}}{\lambda_{n}} \biggr) \biggr) \lambda_{n}a_{n}, \end{aligned}$$
(4.3)
which improves (1.6). Taking \(\lambda_{n} \equiv 1\) in (4.3) yields
$$\begin{aligned} \sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n} < e\sum_{n=1}^{ \infty } \bigl(1-\mathcal{E}(n) \bigr)a_{n}, \end{aligned}$$
(4.4)
which improves (1.9).

Declarations

Acknowledgements

The authors thank the referees for helpful comments.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454000, China

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