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

## 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.

## 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  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  (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 , Hardy himself said that it was Pólya who pointed out this inequality to him.

In , 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  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  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  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 .

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 ,

$$\sum_{n=1}^{\infty }(a_{1}a_{2} \cdots a_{n})^{1/n}\leq \sum_{n=1} ^{\infty } \biggl( 1+\frac{1}{n} \biggr) ^{n}a_{n},$$
(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.

## 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  and the references therein. The first values of the partition function $$p(n)$$ are (starting with $$p(0)=1$$) (see ):

$$1, \,1,\,2, \,3,\,5,\,7,\,11,\,15,\,22,\,30, \,42,\,\ldots .$$

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



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).

## Padé approximant related to asymptotics for the constant e

For later use, we introduce the Padé approximant (see ). 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 )

\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)

## 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  (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 , $$c_{k}^{\lambda_{n}}=\frac{( \Lambda_{n+1})^{\Lambda_{n}}}{(\Lambda_{n})^{\Lambda_{n-1}}}$$ 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  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).

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