Skip to content

Advertisement

Open Access

Some inequalities related to two expansions of \((1+1/x)^{x}\)

Journal of Inequalities and Applications20152015:393

https://doi.org/10.1186/s13660-015-0928-5

Received: 5 November 2015

Accepted: 1 December 2015

Published: 14 December 2015

Abstract

We prove the following theorem: Let
$$\begin{aligned}& \biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr)=e \Biggl(1-\sum _{k=1}^{\infty}\frac{d_{k}}{ (\frac{11}{12}+x )^{k}} \Biggr), \\& \sigma_{m}(x)=\sum_{k=1}^{m} \frac{b_{k}}{ (1+x )^{k}} \quad\mbox{and}\quad S_{m}(x)=\sum_{k=1}^{m} \frac{d_{k}}{ (\frac{11}{12}+x )^{k}}. \end{aligned}$$
  1. (1)

    If \(m\geq6\) is even, we have \(S_{m}(x)>\sigma_{m}(x)\) for all \(x>0\).

     
  2. (2)

    If \(m\geq7\) is odd, we have \(S_{m}(x)>\sigma_{m}(x)\) for all \(x>1\).

     
This provides an intuitive explanation for the main result in Mortici and Hu (On an infinite series for \((1+ 1/x)^{x}\), 2014, arXiv:1406.7814 [math.CA]).

Keywords

Carleman inequalityintegral representationseriesnumber e

MSC

42B2542B35

1 Introduction

The Carleman inequality [2]
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty}a_{n}, $$
whenever \(a_{n}\geq0\), \(n=1,2,3,\ldots\) , with \(0<\sum_{n=1}^{\infty}a_{n}<\infty\), has attracted the attention of many authors in the recent past [1, 310].
In [7], Yang proved
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{6}\frac{b_{k}}{ ( n+1 ) ^{k}} \Biggr) a_{n} $$
with \(b_{1}=1/2\), \(b_{2}=1/24\), \(b_{3}=1/48\), \(b_{4}=73/5{,}760\), \(b_{5}=11/1{,}280\), \(b_{6}=1{,}945/580{,}608\), and Yang conjectured that if
$$ \biggl( 1+\frac{1}{x} \biggr) ^{x}=e \Biggl( 1-\sum _{k=1}^{\infty}\frac {b_{k}}{ ( x+1 ) ^{k}} \Biggr) ,\quad x>0, $$
(1.1)
then \(b_{k}>0\), \(k=1,2,3,\ldots\) .
Later, this conjecture was proved by Yang [8], Gylletberg and Yan [11], and Chen [9], respectively. As an application, Yang proved for any positive integer m
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{m }\frac{b_{k}}{ ( n+1 ) ^{k}} \Biggr) a_{n}, $$
(1.2)
whenever \(a_{n}\geq0\), \(n=1,2,3,\ldots\) , and \(0<\sum_{n=1}^{\infty }a_{n}<\infty\), with \(b_{1}=\frac{1}{2}\) and
$$ b_{n+1}=\frac{1}{n+1} \Biggl( \frac{1}{n+2}-\sum _{k=1}^{n}\frac {b_{k}}{n+2-k} \Biggr).$$

In the final part of his paper, Yang [8] remarked that in order to obtain better results, the right-hand side of (1.1) could be replaced by \(e [ 1-\sum_{n=1}^{\infty} ( d_{n}/ ( x+\varepsilon ) ^{n} ) ] \), where \(\varepsilon\in(0,1]\) and \(d_{n}=d_{n} ( \varepsilon ) \), but information about the values of ε are not provided.

Recently, Mortici and Hu [1] proved that \(\varepsilon=11/12\) provides the faster series
$$\sum_{n=1}^{\infty}\frac{d_{n}}{ (x+\varepsilon) ^{n}} $$
and therefore the following inequality is better than (1.2):
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{m }\frac{d_{k}}{ ( n+\frac{11}{12} ) ^{k}} \Biggr) a_{n}. $$
(1.3)

The proof of this conclusion is based on the following theorem [12], which is a powerful tool for measuring the speed of convergence.

Theorem

If \(( \omega_{n} )_{n\geq1}\) is convergent to zero and
$$ \lim_{n\rightarrow\infty}n^{k}(\omega_{n}- \omega_{n+1})=l\in \mathbb{R}, $$
with \(k>1\), then there exists the limit
$$ \lim_{n\rightarrow\infty}n^{k-1}\omega_{n}= \frac{l}{k-1}. $$
The purpose of this paper is to establish some inequalities which explain Mortici’s conclusion in a quantitative way. But our proof is not based on the theorem.

Our main result is the following theorem.

Theorem 1

Let
$$\begin{aligned}& \biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr)=e \Biggl(1-\sum _{k=1}^{\infty}\frac{d_{k}}{ (\frac{11}{12}+x )^{k}} \Biggr), \\& \sigma_{m}(x)=\sum_{k=1}^{m} \frac{b_{k}}{ (1+x )^{k}} \quad\textit{and}\quad S_{m}(x)=\sum_{k=1}^{m} \frac{d_{k}}{ (\frac{11}{12}+x )^{k}}. \end{aligned}$$
  1. (1)

    If \(m\geq6\) is even, then \(S_{m}(x)>\sigma_{m}(x)\) for all \(x>0\).

     
  2. (2)

    If \(m\geq7\) is odd, then \(S_{m}(x)>\sigma_{m}(x)\) for all \(x>1\).

     

2 Lemmas

In order to prove our main results we need the following lemmas, and throughout this paper we set
$$\begin{aligned}& g(s)=\frac{1}{\pi}s^{s}(1-s)^{1-s}\sin(\pi s),\\& h(s,x)=\frac{1}{1-s+x} \biggl(\frac{s}{1+x} \biggr)^{m-1}- \frac{1}{s+x} \biggl(\frac{12s-1}{11+12x} \biggr)^{m-1}. \end{aligned}$$
Here \(0\leq{s}\leq1\), \(x>0\), and \(m\geq1\) is an integer.

Lemma 1

For \(x>0\), let
$$\biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr). $$
Then
$$\begin{aligned}& b_{k}>0 ,\quad k=1,2 ,\ldots, \end{aligned}$$
(2.1)
$$\begin{aligned}& b_{1}=\frac{1}{2}, \end{aligned}$$
(2.2)
$$\begin{aligned}& b_{n+1}=\frac{1}{n+1} \Biggl(\frac{1}{n+2}-\sum _{j=1}^{n}\frac{b_{j}}{n+2-j} \Biggr),\quad n=1 ,2 ,\ldots, \\& eb_{k}=\int_{0}^{1}g(s)s^{k-2}\,ds,\quad k= 1, 2 , \ldots. \end{aligned}$$
(2.3)

Proof

For (2.1) and (2.2), see [7]. For (2.3), see [13]. □

Remark 1

By (2.3) of Lemma 1, we have
$$\int_{0}^{1}g(s) (s)^{n-2}\,ds= \int_{0}^{1}g(s) (1-s)^{n-2}\,ds=eb_{n}\quad (n=2,3,\ldots). $$

Example

$$\begin{aligned}& \int_{0}^{1}g(s)\,ds=eb_{2}= \frac{e}{24},\qquad \int_{0}^{1}g(s)s\,ds=eb_{3}= \frac{e}{48},\\& \begin{aligned}[b] \int_{0}^{1}\frac{1}{s}g(s)\,ds&= \int_{0}^{1}\frac{1}{1-s}g(s)\,ds \\ &= \int_{0}^{1}\bigl(1+s+s^{2}+\dotsm \bigr)g(s)\,ds \\ &=e\sum_{n=2}^{\infty}b_{n} =e\sum _{n=1}^{\infty}b_{n}-eb_{1} \\ &=e\biggl(1-\frac{1}{e}\biggr)-\frac{e}{2} =\frac{e}{2}-1. \end{aligned} \end{aligned}$$

Lemma 2

For \(x>0\), let
$$\biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{d_{k}}{ (\frac{11}{12}+x )^{k}} \Biggr). $$
Then
$$\begin{aligned} &d_{1}=\frac{1}{2}, \end{aligned}$$
(2.4)
$$\begin{aligned} &d_{n+1}=\frac{ (-1 )^{n+1}}{12^{n-1}e} \biggl(1+ \int_{0}^{1}g(s)\frac{ (12s-1 )^{n-1}}{s}\,ds \biggr),\quad n=1,2,\ldots, \\ &d_{n}>0 ,\quad n=4,5,\ldots, \end{aligned}$$
(2.5)
$$\begin{aligned} &b_{n}>d_{n} ,\quad n=2,3,\ldots. \end{aligned}$$
(2.6)

Proof

For (2.4), see [1].

Now, we prove (2.5). If n is an odd, then \(d_{n+1}\) is obviously positive.

If n is an even, then we have for all \(n\geq4\)
$$\begin{aligned}& \int_{0}^{\frac{1}{12}}\frac{g(s)}{s} (1-12s )^{n-1}\,ds\leq \int_{0}^{\frac{1}{12}}\frac{g(s)}{s} (1-12s )^{3}\,ds, \end{aligned}$$
(2.7)
$$\begin{aligned}& \int_{\frac{1}{12}}^{\frac{1}{6}}\frac{g(s)}{s} (1-12s )^{n-1}\,ds< 0, \end{aligned}$$
(2.8)
$$\begin{aligned}& \int_{\frac{1}{6}}^{1}\frac{g(s)}{s} (1-12s )^{n-1}\,ds\leq \int_{\frac{1}{6}}^{1}\frac{g(s)}{s} (1-12s )^{3}\,ds. \end{aligned}$$
(2.9)
From (2.7), (2.8), (2.9), and Remark 1, we get
$$\begin{aligned} \int_{0}^{1}\frac{g(s)}{s} (1-12s )^{n-1}\,ds \leq{}& \int_{0}^{1} \biggl(\frac{1}{s}-36+432s-1{,}728s^{2} \biggr)g(s)\,ds\\ &{}+ \int_{\frac{1}{12}}^{\frac{1}{6}} \bigl(1{,}728s^{2}-432s+30 \bigr)\,ds\\ \leq{}& \biggl(\frac{e}{2}-1 \biggr)-\frac{3e}{2}+9e-21.9e+6 \int_{0}^{1}g(s)\,ds\\ ={}&{-}1-13.9e+\frac{e}{4}< -1. \end{aligned}$$
Thus from this and (2.4), we have \(d_{n+1}>0\). This proves (2.5). The proof of (2.6) is similar to (2.5). □

Remark 2

By Lemma 2, it is not obvious that \(S_{m}(x)>\sigma_{m}(x)\).

Lemma 3

Let \(m\geq3\) be an integer, we have
$$ (-1 )^{m-1}+ \int_{0}^{1}\frac{g(s)}{1-s} (12s-1 )^{m-1}\,ds>0. $$
(2.10)

Proof

The proof is similar to the proof of (2.5). □

Lemma 4

Let \(x>0\), and \(m\geq6\) be an integer. Then we have for all \(\frac{1}{12}\leq{s}\leq\frac{1}{2}\)
$$ h(s,x)>0. $$
(2.11)

Proof

Noting that \(\frac{s(\frac{11}{12}+x)}{(s-\frac{1}{12})(1+x)}>1\) for all \(x>0\), the inequality (2.11) is equivalent to
$$ \biggl(\frac{(s-\frac{1}{12})(1+x)}{s(\frac{11}{12}+x)} \biggr)^{5} < \frac{s+x}{1-s+x}. $$
(2.12)
To prove (2.12), we define \(h_{1}(s,x)\) as
$$ h_{1}=\ln(1-s+x)-\ln(s+x)+5\ln\biggl(s-\frac{1}{12}\biggr)+5 \ln(1+x)-5\ln{s}-5\ln \biggl(\frac{11}{12}+x\biggr). $$
(2.13)
Easy computations reveal that
$$\begin{aligned}& h_{1}\biggl(\frac{1}{2},0\biggr)=5\ln\biggl(\frac{10}{11} \biggr)< 0, \end{aligned}$$
(2.14)
$$\begin{aligned}& \frac{\partial{h_{1}(s,0)}}{\partial{s}}>0, \end{aligned}$$
(2.15)
$$\begin{aligned}& \frac{\partial{h_{1}(s,x)}}{\partial{x}}< 0. \end{aligned}$$
(2.16)
Thus from (2.14), (2.15), and (2.16), we have
$${h_{1}(s,x)}\leq{h_{1}(s,0)}\leq{h_{1}\biggl( \frac{1}{2},0\biggr)}< 0, $$
which implies
$$h(s,x)>0. $$
 □

Lemma 5

Let \(x>0\), and \(m\geq2\) be an integer, then \(h(s,x)\) is a monotonic increasing function of s on \([\frac{1}{2},1]\). If m is an odd, then \(h(s,x)\) is a monotonic increasing function of s on \([0,\frac{1}{12}]\).

Proof

It suffices to show that \(\frac{\partial{h(s,x)}}{\partial{s}}>0\). Partial differentiation yields
$$\begin{aligned} \frac{\partial{h(s,x)}}{\partial{s}} \geq{}&\frac{m-1}{(1+x)(1-s+x)} \biggl(\frac{s}{1+x} \biggr)^{m-2} \\ &{}- \frac{m-1}{(11+12x)(s+x)} \biggl(\frac{12s-1}{11+12x} \biggr)^{m-2}. \end{aligned}$$
(2.17)
If \(\frac{1}{2}\leq{s}\leq{1}\), then for all \(x>0\) and \({m}\geq{2}\), we have
$$\begin{aligned}& \frac{m-1}{(1+x)(1-s+x)}>\frac{m-1}{(11+12x)(s+x)},\\& \frac{s}{1+x}>\frac{12s-1}{11+12x}. \end{aligned}$$
Thus
$$\frac{\partial{h(s,x)}}{\partial{s}}>0. $$
If m is an odd, then for \(0<{s}<\frac{1}{12}\), we have
$$(12s-1)^{m-2}< 0. $$
From this and (2.17), we get
$$\frac{\partial{h(s,x)}}{\partial{s}}>0. $$
This completes the proof of Lemma 5. □

Lemma 6

Let \(m\geq3\) be an integer, then we have for all \(x\geq1\)
$$ h\biggl(\frac{1}{2},x\biggr)>\bigl|h(0,x)\bigr|. $$
(2.18)

Proof

Because of \(x>1\), (2.18) is equivalent to
$$ \biggl(5+\frac{1+2x}{2+2x} \biggr)^{m-1}-5^{m-1}>1+ \frac{1}{2x}. $$
(2.19)
The equality (2.19) follows immediately from
$$(m-1)5^{m-2}\frac{1+2x}{2+2x}>(m-1)5^{m-2}\frac{3}{4}> \frac{3}{2}. $$
 □

3 Proof of Theorem 1

Proof

By Lemma 1 and Lemma 2, we get for \(m\geq2\)
$$\begin{aligned}& \begin{aligned}[b] \sigma_{m}(x)&= \frac{e/2}{1+x}+\sum _{k=2}^{m} \int_{0}^{1}\frac{g(s)}{s^{2}} \biggl( \frac {s}{1+x} \biggr)^{k}\,ds \\ &=\frac{e}{2(1+x)}+ \int_{0}^{1}\frac{g(s)}{s^{2}}\sum _{k=2}^{m} \biggl(\frac{s}{1+x} \biggr)^{k}\,ds \\ &=\frac{e}{2(1+x)}+ \int_{0}^{1} \frac{g(s)}{(1+x)(1+x-s)} \biggl(1- \biggl( \frac{s}{1+x} \biggr)^{m-1} \biggr)\,ds \\ &=\frac{e}{2(1+x)}+ \int_{0}^{1}\frac{g(s)}{(1+x)(x+s)}\,ds- \int_{0}^{1}\frac{g(s)}{(1+x)(1-s+x)} \biggl( \frac{s}{1+x} \biggr)^{m-1}\,ds, \end{aligned}\\& \begin{aligned}[b] S_{m}(x)={}&\frac{e}{2(1+x)}+ \int_{0}^{1}\frac{g(s)}{(x+s)(1+x)} +\frac{(-1)^{m-1}}{(11+12x)^{m}(1+x)}\\ &{}- \int_{0}^{1}\frac{g(s)}{(1+x)(x+s)} \biggl( \frac{12s-1}{11+12x} \biggr)^{m-1}\,ds\\ &{}+ \int_{0}^{1}\frac{g(s)}{(11+12x)(1+x)(1-s)} \biggl( \frac {12s-1}{11+12x} \biggr)^{m-1}\,ds. \end{aligned} \end{aligned}$$
To prove our result, we consider
$$\begin{aligned} S_{m}(x)-\sigma_{m}(x) ={}&\frac{1}{(11+12x)^{m}(1+x)} \biggl((-1)^{m-1}+ \int_{0}^{1}\frac {g(s)}{1-s}(12s-1)^{m-1} \biggr) \\ &{}+\frac{1}{1+x} \int_{0}^{1}g(s) \biggl(\frac{1}{1-s+x} \biggl( \frac{s}{1+x} \biggr)^{m-1}-\frac{1}{x+s} \biggl( \frac {12s-1}{11+12x} \biggr)^{m-1} \biggr)\,ds \\ ={}&\frac{1}{(11+12x)^{m}(1+x)} \biggl((-1)^{m-1}+ \int_{0}^{1}\frac {g(s)}{1-s}(12s-1)^{m-1} \biggr) \\ &{}+\frac{1}{1+x} \int_{0}^{1}g(s)h(s,x)\,ds. \end{aligned}$$
By Lemma 3, it suffices to show that
$$\int_{0}^{1}g(s)h(s,x)\,ds>0. $$
Let first \(m\geq6\) be even. From Lemma 4 and Lemma 5, for all \(x>0\), we have
$$\int_{0}^{1}g(s)h(s,x)\,ds= \int_{0}^{\frac{1}{12}}g(s)h(s,x)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds+ \int_{\frac {1}{2}}^{1}g(s)h(s,x)\,ds>0. $$
Here we used the fact that if m is an even and \(0\leq s\leq 1/12\), then \(h(s,x)>0\) for any \(x>0\).
Now let \(m\geq7\) be odd. From Lemma 4, Lemma 5, and Lemma 6 for all \(x\geq1\) we have
$$\begin{aligned} \int_{0}^{1}g(s)h(s,x)\,ds&= \int_{0}^{\frac{1}{12}}g(s)h(s,x)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds+ \int_{\frac {1}{2}}^{1}g(s)h(s,x)\,ds \\ &\geq \int_{0}^{\frac{1}{12}}g(s)h(0,x)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds+ \int_{\frac {1}{2}}^{1}g(s)h\biggl(\frac{1}{2},x \biggr)\,ds \\ &\geq\biggl(h(0,x)+h\biggl(\frac{1}{2},x\biggr)\biggr) \int_{0}^{\frac{1}{12}}g(s)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds>0. \end{aligned}$$
This completes the proof of Theorem 1. □

Remark 3

By using computer simulation, we find \(S_{m}(x)>\sigma_{m}(x)\) for all \(x>0\) and all \(m\geq1\), but we leave as an open problem the rigorous proof of this fact.

4 Conclusions

In this paper, we have established some inequalities which explain Mortici’s result in a quantitative way. The authors believe that the present analysis will lead to a significant contribution toward the study of the Carleman inequality.

Declarations

Acknowledgements

The authors are very grateful to the anonymous referees and the Editor for their insightful comments and suggestions. The authors are grateful to Professor Hongwei Chen, Christopher Newport University, USA, for his kind help and valuable suggestions in the preparation of this paper. Supported by Foundation Lead-edga Technologies Research Project of Henan Province, No. 122300410061.

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)
Henan College of Finance and Taxation, Department of Information Engineering, Zhengzhou, China
(2)
Department of Foundation, Zhejiang University of Water Resources and Electric Power, Hangzhou, China

References

  1. Mortici, C, Hu, Y: On an infinite series for \((1+ 1/x)^{x}\) (2014). arXiv:1406.7814 [math.CA]
  2. Hardy, GH, Littlewood, JE, Polya, G: Inequalities. Cambridge University Press, London (1952) MATHGoogle Scholar
  3. Bicheng, Y, Debnath, L: Some inequalities involving the constant e and an application to Carleman’s inequality. J. Math. Anal. Appl. 223, 347-353 (1998) View ArticleMathSciNetMATHGoogle Scholar
  4. Mortici, C, Hu, Y: On some convergences to the constant e and improvements of Carleman’s inequality. Carpath. J. Math. 31, 249-254 (2015) MathSciNetGoogle Scholar
  5. Mortici, C, Yang, X: Estimates of \((1+ 1/x)^{x}\) involved in Carleman’s inequality and Keller’s limit. Filomat 7, 1535-1539 (2015) View ArticleMathSciNetGoogle Scholar
  6. Ping, Y, Guozheng, S: A strengthened Carleman’s inequality. J. Math. Anal. Appl. 240, 290-293 (1999) View ArticleMathSciNetMATHGoogle Scholar
  7. Yang, X: On Carleman’s inequality. J. Math. Anal. Appl. 253, 691-694 (2001) View ArticleMathSciNetMATHGoogle Scholar
  8. Yang, X: Approximations for constant e and their applications. J. Math. Anal. Appl. 262, 651-659 (2001) View ArticleMathSciNetMATHGoogle Scholar
  9. Chen, H: On an infinite series for \((1+\frac{1}{x})^{x}\) and its application. Int. J. Math. Math. Sci. 11, 675-680 (2002) View ArticleMathSciNetMATHGoogle Scholar
  10. Liu, H, Zhu, L: New strengthened Carleman’s inequality and Hardy’s inequality. J. Inequal. Appl. 2007, Article ID 84104 (2007) View ArticleMATHGoogle Scholar
  11. Gyllenberg, M, Yan, P: On a conjecture by Yang. J. Math. Anal. Appl. 264, 687-690 (2001) View ArticleMathSciNetMATHGoogle Scholar
  12. Mortici, C: Product approximations via asymptotic integration. Am. Math. Mon. 117(5), 434-441 (2010) View ArticleMathSciNetMATHGoogle Scholar
  13. Hu, Y, Mortici, C: On the coefficients of an expansion of \((1+\frac{1}{x})^{x}\) related to Carleman’s inequality (2014). arXiv:1401.2236 [math.CA]

Copyright

© Ren and Li 2015

Advertisement