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 $b_j$ ($1\le j\le k$) such that

$$\frac{1}{e}{(1+\frac{1}{x})}^x=\frac{x^k+a_1{x}^{k-1}+\cdots +a_k}{x^k+b_1{x}^{k-1}+\cdots +b_k}+O\left(\frac{1}{x^{2k+1}}\right),x\to \mathrm{\infty },$$

where $k\ge 1$ 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 $a_n\ge 0$ for $n\in \mathbb{N}:=\{1,2,\dots \}$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_n<\mathrm{\infty }$. Then

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}<e\sum _{n=1}^{\mathrm{\infty }}{a}_n.$$ 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 $a_n\ge 0$, ${\lambda }_n>0$, ${\mathrm{\Lambda }}_n={\sum }_{m=1}^n{\lambda }_m$ for $n\in \mathbb{N}$, and $0<{\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n{a}_n<\mathrm{\infty }$, then

$$\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{(a_1^{{\lambda }_1}{a}_2^{{\lambda }_2}\cdots a_n^{{\lambda }_n})}^{1/{\mathrm{\Lambda }}_n}<e\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{a}_n.$$ 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 [4–20], 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 $n\in \mathbb{N}$,

$$e(1-\frac{1}{2(n+\frac{5}{6})})<{(1+\frac{1}{n})}^n<e(1-\frac{1}{2(n+1)}),$$ 1.3

and then used it to obtain the following strengthened Carleman inequality:

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}<e\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{2(n+1)})a_n.$$ 1.4

Xie and Zhong [15] proved that, for $x\ge 1$,

$$e(1-\frac{7}{14x+12})<{(1+\frac{1}{x})}^x<e(1-\frac{6}{12x+11}),$$ 1.5

and then used it to improve the Carleman-type inequality (1.2) as follows. If $0<{\lambda }_{n+1}\le {\lambda }_n$, ${\mathrm{\Lambda }}_n={\sum }_{m=1}^n{\lambda }_m$, $a_n\ge 0$ for $n\in \mathbb{N}$, and $0<{\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n{a}_n<\mathrm{\infty }$, then

$$\sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n+1}{(a_1^{{\lambda }_1}{a}_2^{{\lambda }_2}\cdots a_n^{{\lambda }_n})}^{1/{\mathrm{\Lambda }}_n}<e\sum _{n=1}^{\mathrm{\infty }}(1-\frac{6}{12(\frac{{\mathrm{\Lambda }}_n}{{\lambda }_n})+11}){\lambda }_n{a}_n.$$ 1.6

Taking ${\lambda }_n\equiv 1$ in (1.6) yields

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}<e\sum _{n=1}^{\mathrm{\infty }}(1-\frac{6}{12n+11})a_n,$$ 1.7

which improves (1.4).

Recently, Mortici and Hu [14] proved that, for $x\ge 1$,

$$\begin{array}{rl}& \frac{x+\frac{5}{12}}{x+\frac{11}{12}}-\frac{5}{288x^3}+\frac{343}{8,640x^4}-\frac{2,621}{41,472x^5}\\ & <\frac{1}{e}{(1+\frac{1}{x})}^x<\frac{x+\frac{5}{12}}{x+\frac{11}{12}}-\frac{5}{288x^3}+\frac{343}{8,640x^4}-\frac{2,621}{41,472x^5}+\frac{300,901}{3,483,648x^6},\end{array}$$ 1.8

and then they used it to establish the following improvement of Carleman’s inequality:

$$\begin{array}{rl}& \sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}\\ & <e\sum _{n=1}^{\mathrm{\infty }}(\frac{12n+5}{12n+11}-\frac{5}{288n^3}+\frac{343}{8,640n^4}-\frac{2,621}{41,472n^5}+\frac{300,901}{3,483,648n^6})a_n,\end{array}$$

which can be written as

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}<e\sum _{n=1}^{\mathrm{\infty }}(1-{\epsilon }_n)a_n,$$ 1.9

where

$${\epsilon }_n=\frac{104,509,440n^6+3,628,800n^4-4,971,456n^3+5,603,472n^2-5,945,040n-16,549,555}{17,418,240n^6(12n+11)}.$$ 1.10

For information as regards the history of Carleman-type inequalities, please refer to [21–24].

It follows from (1.8) that

$$\frac{1}{e}{(1+\frac{1}{x})}^x=\frac{x+\frac{5}{12}}{x+\frac{11}{12}}+O\left(\frac{1}{x^3}\right),x\to \mathrm{\infty }.$$ 1.11

Using the Padé approximation method, in Section 3 we derive (1.11) and the following approximation formula:

$$\frac{1}{e}{(1+\frac{1}{x})}^x=\frac{x^2+\frac{87}{100}x+\frac{37}{240}}{x^2+\frac{137}{100}x+\frac{457}{1,200}}+O\left(\frac{1}{x^5}\right),x\to \mathrm{\infty }.$$ 1.12

Equation (1.12) motivates us to present the following inequality:

$${(1+\frac{1}{n})}^n<e\left(\frac{n^2+\frac{87}{100}n+\frac{37}{240}}{n^2+\frac{137}{100}n+\frac{457}{1,200}}\right)=e(1-\frac{8(75n+34)}{1,200n^2+1,644n+457}),n\in \mathbb{N}.$$ 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],

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}\le \sum _{n=1}^{\mathrm{\infty }}{(1+\frac{1}{n})}^n{a}_n,$$ 1.14

and then the following strengthened Carleman’s inequality is derived directly from (1.13):

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}<e\sum _{n=1}^{\mathrm{\infty }}(1-\frac{8(75n+34)}{1,200n^2+1,644n+457})a_n,$$ 1.15

which improves (1.7).

Based on the Padé approximation method, we determine the coefficients $a_j$ and $b_j$ ($1\le j\le k$) such that

$$\frac{1}{e}{(1+\frac{1}{x})}^x=\frac{x^k+a_1{x}^{k-1}+\cdots +a_k}{x^k+b_1{x}^{k-1}+\cdots +b_k}+O\left(\frac{1}{x^{2k+1}}\right),x\to \mathrm{\infty },$$ 1.16

where $k\ge 1$ 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 $n\in \mathbb{N}={\mathbb{N}}_0\setminus \{0\}:=\{1,2,3,\dots \}$:

$${\mathcal{A}}_n:=\{(k_1,k_2,\dots ,k_n)\in {\mathbb{N}}_0^n:k_1+2k_2+\cdots +n{k}_n=n\}.$$ 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]):

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

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:

$${(1+\frac{1}{x})}^x=e\sum _{j=0}^{\mathrm{\infty }}\frac{c_j}{x^j}\mathit{\text{as}}x\to \mathrm{\infty },$$ 2.2

where the coefficients $c_j$ $(j\in \mathbb{N})$ are given by

$$\begin{array}{rl}& c_0=1\mathit{\text{and}}{c}_j={(-1)}^j\sum _{(k_1,k_2,\dots ,k_j)\in {\mathcal{A}}_j}\frac{1}{k_1!k_2!\cdots k_j!}{\left(\frac{1}{2}\right)}^{k_1}{\left(\frac{1}{3}\right)}^{k_2}\cdots {\left(\frac{1}{j+1}\right)}^{k_j},\end{array}$$ 2.3

where the ${\mathcal{A}}_j$ $(\mathit{\text{for}}j\in \mathbb{N})$ are given in (2.1).

Padé approximant related to asymptotics for the constant e

For later use, we introduce the Padé approximant (see [29–34]). Let f be a formal power series

$$f(t)=c_0+c_{1}t+c_2{t}^2+\cdots .$$ 3.1

The Padé approximation of order $(p,q)$ of the function f is the rational function, denoted by

$${[p/q]}_f(t)=\frac{{\sum }_{j=0}^p{a}_j{t}^j}{1+{\sum }_{j=1}^q{b}_j{t}^j},$$ 3.2

where $p\ge 0$ and $q\ge 1$ are two given integers, the coefficients $a_j$ and $b_j$ are given by (see [29–31, 33, 34])

$$\{\begin{array}{c}{a}_0=c_0,\hfill \\ a_1=c_0{b}_1+c_1,\hfill \\ a_2=c_0{b}_2+c_1{b}_1+c_2,\hfill \\ ⋮\hfill \\ a_p=c_0{b}_p+\cdots +c_{p-1}{b}_1+c_p,\hfill \\ 0=c_{p+1}+c_p{b}_1+\cdots +c_{p-q+1}{b}_q,\hfill \\ ⋮\hfill & \hfill \\ 0=c_{p+q}+c_{p+q-1}{b}_1+\cdots +c_p{b}_q,\hfill \end{array}$$ 3.3

and the following holds:

$${[p/q]}_f(t)-f(t)=O\left(t^{p+q+1}\right).$$ 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])

$$\begin{array}{rl}& {[p/q]}_f(t)=\frac{\left|\begin{array}{cccc}{t}^q{f}_{p-q}(t)& t^{q-1}{f}_{p-q+1}(t)& \cdots & f_p(t)\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|}{\left|\begin{array}{cccc}{t}^q& t^{q-1}& \cdots & 1\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|},\end{array}$$ 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

$$f(x)=\frac{1}{e}{(1+\frac{1}{x})}^x.$$ 3.6

It follows from (2.2) that, as $x\to \mathrm{\infty }$,

$$f(x)=\sum _{j=0}^{\mathrm{\infty }}\frac{c_j}{x^j}=1-\frac{1}{2x}+\frac{11}{24x^2}-\frac{7}{16x^3}+\frac{2,447}{5,760x^4}-\frac{959}{2,304x^5}+\frac{238,043}{580,608x^6}-\cdots ,$$ 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

$${[1/1]}_f(x)=\frac{{\sum }_{j=0}^1{a}_j{x}^{-j}}{1+{\sum }_{j=1}^1{b}_j{x}^{-j}}.$$

Noting that

$$c_0=1,c_1=-\frac{1}{2},c_2=\frac{11}{24},c_3=-\frac{7}{16},c_4=\frac{2,447}{5,760}$$ 3.8

holds, we have, by (3.3),

$$\{\begin{array}{c}{a}_0=1,\hfill \\ a_1=b_1-\frac{1}{2},\hfill \\ 0=\frac{11}{24}-\frac{1}{2}{b}_1,\hfill \end{array}$$

that is,

$$a_0=1,a_1=\frac{5}{12},b_1=\frac{11}{12}.$$

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),

$$\frac{1}{e}{(1+\frac{1}{x})}^x-\frac{x+\frac{5}{12}}{x+\frac{11}{12}}=O\left(\frac{1}{x^3}\right),x\to \mathrm{\infty }.$$ 3.10

We now give a derivation of equation (1.12). To this end, we consider

$${[2/2]}_f(x)=\frac{{\sum }_{j=0}^2{a}_j{x}^{-j}}{1+{\sum }_{j=1}^2{b}_j{x}^{-j}}.$$

Noting that (3.8) holds, we have, by (3.3),

$$\{\begin{array}{c}{a}_0=1,\hfill \\ a_1=b_1-\frac{1}{2},\hfill \\ a_2=b_2-\frac{1}{2}{b}_1+\frac{11}{24},\hfill \\ 0=-\frac{7}{16}+\frac{11}{24}{b}_1-\frac{1}{2}{b}_2,\hfill \\ 0=\frac{2,447}{5,760}-\frac{7}{16}{b}_1+\frac{11}{24}{b}_2,\hfill \end{array}$$

that is,

$$a_0=1,a_1=\frac{87}{100},a_2=\frac{37}{240},b_1=\frac{137}{100},b_2=\frac{457}{1,200}.$$

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),

$$\frac{1}{e}{(1+\frac{1}{x})}^x-\frac{x^2+\frac{87}{100}x+\frac{37}{240}}{x^2+\frac{137}{100}x+\frac{457}{1,200}}=O\left(\frac{1}{x^5}\right),x\to \mathrm{\infty }.$$ 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 $x=\mathrm{\infty }$) is the following rational function:

$${[p/q]}_f(x)=\frac{1+{\sum }_{j=1}^p{a}_j{x}^{-j}}{1+{\sum }_{j=1}^q{b}_j{x}^{-j}}=x^{q-p}\left(\frac{x^p+a_1{x}^{p-1}+\cdots +a_p}{x^q+b_1{x}^{q-1}+\cdots +b_q}\right),$$ 3.13

where $p\ge 1$ and $q\ge 1$ are two given integers, the coefficients $a_j$ and $b_j$ are given by

$$\{\begin{array}{c}{a}_1=b_1+c_1,\hfill \\ a_2=b_2+c_1{b}_1+c_2,\hfill \\ ⋮\hfill \\ a_p=b_p+\cdots +c_{p-1}{b}_1+c_p,\hfill \\ 0=c_{p+1}+c_p{b}_1+\cdots +c_{p-q+1}{b}_q,\hfill \\ ⋮\hfill & \hfill \\ 0=c_{p+q}+c_{p+q-1}{b}_1+\cdots +c_p{b}_q,\hfill \end{array}$$ 3.14

$c_j$ is given in (2.3), and the following holds:

$$f(x)-{[p/q]}_f(x)=O\left(\frac{1}{x^{p+q+1}}\right),x\to \mathrm{\infty }.$$ 3.15

Moreover, we have

$$\begin{array}{rl}& {[p/q]}_f(x)=\frac{\left|\begin{array}{cccc}\frac{1}{x^q}{f}_{p-q}(x)& \frac{1}{x^{q-1}}{f}_{p-q+1}(x)& \cdots & f_p(x)\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|}{\left|\begin{array}{cccc}\frac{1}{x^q}& \frac{1}{x^{q-1}}& \cdots & 1\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|},\end{array}$$ 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

$$\begin{array}{rl}{[1/1]}_f(x)& =\frac{\left|\begin{array}{cc}\frac{1}{x}{f}_0(x)& f_1(x)\\ c_1& c_2\end{array}\right|}{\left|\begin{array}{cc}\frac{1}{x}& 1\\ c_1& c_2\end{array}\right|}=\frac{\left|\begin{array}{cc}\frac{1}{x}& 1-\frac{1}{2x}\\ -\frac{1}{2}& \frac{11}{24}\end{array}\right|}{\left|\begin{array}{cc}\frac{1}{x}& 1\\ -\frac{1}{2}& \frac{11}{24}\end{array}\right|}\\ & =\frac{x+\frac{5}{12}}{x+\frac{11}{12}}\end{array}$$

and

$$\begin{array}{rl}{[2/2]}_f(x)& =\frac{\left|\begin{array}{ccc}\frac{1}{x^2}{f}_0(x)& \frac{1}{x}{f}_1(x)& f_2(x)\\ c_1& c_2& c_3\\ c_2& c_3& c_4\end{array}\right|}{\left|\begin{array}{ccc}\frac{1}{x^2}& \frac{1}{x}& 1\\ c_1& c_2& c_3\\ c_2& c_3& c_4\end{array}\right|}=\frac{\left|\begin{array}{ccc}\frac{1}{x^2}& \frac{1}{x}(1-\frac{1}{2x})& 1-\frac{1}{2x}+\frac{11}{24x^2}\\ -\frac{1}{2}& \frac{11}{24}& -\frac{7}{16}\\ \frac{11}{24}& -\frac{7}{16}& \frac{2,447}{5,760}\end{array}\right|}{\left|\begin{array}{ccc}\frac{1}{x^2}& \frac{1}{x}& 1\\ -\frac{1}{2}& \frac{11}{24}& -\frac{7}{16}\\ \frac{11}{24}& -\frac{7}{16}& \frac{2,447}{5,760}\end{array}\right|}\\ & =\frac{x^2+\frac{87}{100}x+\frac{37}{240}}{x^2+\frac{137}{100}x+\frac{457}{1,200}}.\end{array}$$

Remark 3.2

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

Setting

$$(p,q)=(3,3)\text{and}(p,q)=(4,4),$$

respectively, we obtain by Theorem 3.1, as $x\to \mathrm{\infty }$,

$$\frac{1}{e}{(1+\frac{1}{x})}^x=\frac{x^3+\frac{162,713}{121,212}{x}^2+\frac{13,927}{26,936}x+\frac{41,501}{786,240}}{x^3+\frac{223,319}{121,212}{x}^2+\frac{237,551}{242,424}x+\frac{3,950,767}{29,090,880}}+O\left(\frac{1}{x^7}\right)$$ 3.17

and

$$\begin{array}{rl}\frac{1}{e}{(1+\frac{1}{x})}^x=& \frac{x^4+\frac{1,157,406,727}{634,301,284}{x}^3+\frac{8,452,872,239}{7,611,615,408}{x}^2+\frac{81,587,251,465}{319,687,847,136}x+\frac{15,842,677}{924,376,320}}{x^4+\frac{1,474,557,369}{634,301,284}{x}^3+\frac{13,811,559,391}{7,611,615,408}{x}^2+\frac{170,870,679,559}{319,687,847,136}x+\frac{1,724,393,461,793}{38,362,541,656,320}}\\ & +O\left(\frac{1}{x^9}\right).\end{array}$$ 3.18

Equations (3.17) and (3.18) motivate us to establish the following theorem.

Theorem 3.2

For $x>0$,

$${(1+\frac{1}{x})}^x<e\left(\frac{x^3+\frac{162,713}{121,212}{x}^2+\frac{13,927}{26,936}x+\frac{41,501}{786,240}}{x^3+\frac{223,319}{121,212}{x}^2+\frac{237,551}{242,424}x+\frac{3,950,767}{29,090,880}}\right)$$ 3.19

and

$$\begin{array}{rl}& {(1+\frac{1}{x})}^x\\ & <e\left(\frac{x^4+\frac{1,157,406,727}{634,301,284}{x}^3+\frac{8,452,872,239}{7,611,615,408}{x}^2+\frac{81,587,251,465}{319,687,847,136}x+\frac{15,842,677}{924,376,320}}{x^4+\frac{1,474,557,369}{634,301,284}{x}^3+\frac{13,811,559,391}{7,611,615,408}{x}^2+\frac{170,870,679,559}{319,687,847,136}x+\frac{1,724,393,461,793}{38,362,541,656,320}}\right).\end{array}$$ 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

$$F(x)<0\text{for}x>0,$$

where

$$\begin{array}{rl}F(x)=& xln(1+\frac{1}{x})-1\\ & -ln\left(\frac{x^4+\frac{1,157,406,727}{634,301,284}{x}^3+\frac{8,452,872,239}{7,611,615,408}{x}^2+\frac{81,587,251,465}{319,687,847,136}x+\frac{15,842,677}{924,376,320}}{x^4+\frac{1,474,557,369}{634,301,284}{x}^3+\frac{13,811,559,391}{7,611,615,408}{x}^2+\frac{170,870,679,559}{319,687,847,136}x+\frac{1,724,393,461,793}{38,362,541,656,320}}\right).\end{array}$$

Differentiation yields

$$F^{\prime }(x)=ln(1+\frac{1}{x})-\frac{P_8(x)}{P_9(x)},$$

where

$$\begin{array}{rl}{P}_8(x)=& 4,534,960,145,139,175,220,907,601+89,156,435,404,854,709,617,164,400x\\ & +753,611,422,427,554,143,580,166,880x^2\\ & +3,400,732,641,706,885,239,015,784,320x^3\\ & +8,959,898,009,119,992,740,647,591,680x^4\\ & +14,212,846,466,921,911,377,490,790,400x^5\\ & +13,355,464,865,044,929,241,744,281,600x^6\\ & +6,842,437,276,900,714,847,214,796,800x^7\\ & +1,471,684,602,332,887,248,995,942,400x^8\end{array}$$

and

$$\begin{array}{rl}{P}_9(x)=& (38,362,541,656,320x^4+69,999,958,848,960x^3+42,602,476,084,560x^2\\ & +9,790,470,175,800x+657,486,938,177\left)\right(38,362,541,656,320x^4\\ & +89,181,229,677,120x^3+69,610,259,330,640x^2+20,504,481,547,080x\\ & +1,724,393,461,793)(x+1).\end{array}$$

Differentiating $F^{\prime }(x)$, we find

$$F^{″}(x)=-\frac{Q_8(x)}{Q_{19}(x)},$$

where

$$\begin{array}{rl}{Q}_8(x)=& 1,285,425,745,031,439,744,924,351,944,181,267,498,830,297,392,321\\ & +28,378,097,964,665,213,870,448,253,775,917,974,735,833,555,915,520x\\ & +247,639,239,538,550,650,618,428,925,475,351,177,418,903,828,519,360x^2\\ & +1,131,116,309,072,948,249,686,419,776,599,013,563,965,352,036,853,760x^3\\ & +2,998,129,273,934,033,621,834,452,343,529,577,599,070,175,646,117,120x^4\\ & +4,775,194,702,079,256,668,486,950,292,217,012,539,098,845,384,867,840x^5\\ & +4,503,188,365,939,207,771,317,966,173,833,346,921,724,385,791,590,400x^6\\ & +2,315,562,242,935,704,170,341,114,308,201,588,127,064,283,807,744,000x^7\\ & +500,009,489,498,922,911,594,629,442,997,057,334,195,586,408,448,000x^8\end{array}$$

and

$$\begin{array}{rl}{Q}_{19}(x)=& x(38,362,541,656,320x^4+69,999,958,848,960x^3+42,602,476,084,560x^2\\ & +9,790,470,175,800x+657,486,938,177{)}^2(38,362,541,656,320x^4\\ & +89,181,229,677,120x^3+69,610,259,330,640x^2+20,504,481,547,080x\\ & +1,724,393,461,793{)}^2{(x+1)}^2.\end{array}$$

Hence, $F^{″}(x)<0$ for $x>0$, and we have

$$F^{\prime }(x)>\underset{t\to \mathrm{\infty }}{lim}{F}^{\prime }(t)=0⟹F(x)<\underset{t\to \mathrm{\infty }}{lim}F(t)=0\text{for}x>0.$$

The proof is complete. □

The inequality (3.20) can be written as

$${(1+\frac{1}{x})}^x<e(1-\mathcal{E}(x)),x>0,$$ 3.21

where

$$\begin{array}{rl}\mathcal{E}(x)=& 48(399,609,808,920x^3+562,662,150,960x^2\\ & +223,208,570,235x+22,227,219,242)/(38,362,541,656,320x^4\\ & +89,181,229,677,120x^3+69,610,259,330,640x^2+20,504,481,547,080x\\ & +1,724,393,461,793).\end{array}$$ 3.22

A generalized Carleman-type inequality

Theorem 4.1

Let $0<{\lambda }_{n+1}\le {\lambda }_n$, ${\mathrm{\Lambda }}_n={\sum }_{m=1}^n{\lambda }_m$ $({\mathrm{\Lambda }}_n\ge 1)$, $a_n\ge 0$ $(n\in \mathbb{N})$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n{a}_n<\mathrm{\infty }$. Then, for $0<p\le 1$,

$$\begin{array}{rl}& \sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n+1}{(a_1^{{\lambda }_1}{a}_2^{{\lambda }_2}\cdots a_n^{{\lambda }_n})}^{1/{\mathrm{\Lambda }}_n}\\ & <\frac{e^p}{p}\sum _{n=1}^{\mathrm{\infty }}{(1-\mathcal{E}\left(\frac{{\mathrm{\Lambda }}_n}{{\lambda }_n}\right))}^p{\lambda }_n{a}_n^p{\mathrm{\Lambda }}_n^{p-1}{\left(\sum _{k=1}^n{\lambda }_k{(c_k{a}_k)}^p\right)}^{(1-p)/p},\end{array}$$ 4.1

where $\mathcal{E}(x)$ is given in (3.22) and

$$c_n^{{\lambda }_n}=\frac{{({\mathrm{\Lambda }}_{n+1})}^{{\mathrm{\Lambda }}_n}}{{({\mathrm{\Lambda }}_n)}^{{\mathrm{\Lambda }}_{n-1}}}.$$

Proof

The inequality

$$\begin{array}{rl}& \sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n+1}{(a_1^{{\lambda }_1}{a}_2^{{\lambda }_2}\cdots a_n^{{\lambda }_n})}^{1/{\mathrm{\Lambda }}_n}\\ & \le \frac{1}{p}\sum _{m=1}^{\mathrm{\infty }}{(1+\frac{1}{{\mathrm{\Lambda }}_m/{\lambda }_m})}^{p{\mathrm{\Lambda }}_m/{\lambda }_m}{\lambda }_m{a}_m^p{\mathrm{\Lambda }}_m^{p-1}{\left(\sum _{k=1}^m{\lambda }_k{(c_k{a}_k)}^p\right)}^{(1-p)/p}\end{array}$$ 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], $c_k^{{\lambda }_n}=\frac{{({\mathrm{\Lambda }}_{n+1})}^{{\mathrm{\Lambda }}_n}}{{({\mathrm{\Lambda }}_n)}^{{\mathrm{\Lambda }}_{n-1}}}$ should be $c_n^{{\lambda }_n}=\frac{{({\mathrm{\Lambda }}_{n+1})}^{{\mathrm{\Lambda }}_n}}{{({\mathrm{\Lambda }}_n)}^{{\mathrm{\Lambda }}_{n-1}}}$; see [9, p.44, line 3]. Likewise, $c_s^{{\lambda }_n}=\frac{{({\mathrm{\Lambda }}_{n+1})}^{{\mathrm{\Lambda }}_n}}{{({\mathrm{\Lambda }}_n)}^{{\mathrm{\Lambda }}_{n-1}}}$ in Theorem 3.1 of [11] should be $c_n^{{\lambda }_n}=\frac{{({\mathrm{\Lambda }}_{n+1})}^{{\mathrm{\Lambda }}_n}}{{({\mathrm{\Lambda }}_n)}^{{\mathrm{\Lambda }}_{n-1}}}$; see [11, p.96, equation (9)].

Remark 4.2

Taking $p=1$ in (4.1) yields

$$\sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n+1}{(a_1^{{\lambda }_1}{a}_2^{{\lambda }_2}\cdots a_n^{{\lambda }_n})}^{1/{\mathrm{\Lambda }}_n}<e\sum _{n=1}^{\mathrm{\infty }}(1-\mathcal{E}\left(\frac{{\mathrm{\Lambda }}_n}{{\lambda }_n}\right)){\lambda }_n{a}_n,$$ 4.3

which improves (1.6). Taking ${\lambda }_n\equiv 1$ in (4.3) yields

$$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}<e\sum _{n=1}^{\mathrm{\infty }}(1-\mathcal{E}(n))a_n,$$ 4.4

which improves (1.9).