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

Journal of Inequalities and Applications20152015:393

  • Received: 5 November 2015
  • Accepted: 1 December 2015
  • Published:


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]).


  • Carleman inequality
  • integral representation
  • series
  • number e


  • 42B25
  • 42B35

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
$$ \biggl( 1+\frac{1}{x} \biggr) ^{x}=e \Biggl( 1-\sum _{k=1}^{\infty}\frac {b_{k}}{ ( x+1 ) ^{k}} \Biggr) ,\quad x>0, $$
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
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{m }\frac{b_{k}}{ ( n+1 ) ^{k}} \Biggr) a_{n}, $$
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):
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{m }\frac{d_{k}}{ ( n+\frac{11}{12} ) ^{k}} \Biggr) a_{n}. $$

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


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

$$\begin{aligned}& \biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr)=e \Biggl(1-\sum _{k=1}^{\infty}\frac{d_{k}}{ (\frac{11}{12}+x )^{k}} \Biggr), \\& \sigma_{m}(x)=\sum_{k=1}^{m} \frac{b_{k}}{ (1+x )^{k}} \quad\textit{and}\quad S_{m}(x)=\sum_{k=1}^{m} \frac{d_{k}}{ (\frac{11}{12}+x )^{k}}. \end{aligned}$$
  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
$$\biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr). $$
$$\begin{aligned}& b_{k}>0 ,\quad k=1,2 ,\ldots, \end{aligned}$$
$$\begin{aligned}& b_{1}=\frac{1}{2}, \end{aligned}$$
$$\begin{aligned}& b_{n+1}=\frac{1}{n+1} \Biggl(\frac{1}{n+2}-\sum _{j=1}^{n}\frac{b_{j}}{n+2-j} \Biggr),\quad n=1 ,2 ,\ldots, \\& eb_{k}=\int_{0}^{1}g(s)s^{k-2}\,ds,\quad k= 1, 2 , \ldots. \end{aligned}$$


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

Remark 1

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


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

Lemma 2

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


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 \(n\geq4\)
$$\begin{aligned}& \int_{0}^{\frac{1}{12}}\frac{g(s)}{s} (1-12s )^{n-1}\,ds\leq \int_{0}^{\frac{1}{12}}\frac{g(s)}{s} (1-12s )^{3}\,ds, \end{aligned}$$
$$\begin{aligned}& \int_{\frac{1}{12}}^{\frac{1}{6}}\frac{g(s)}{s} (1-12s )^{n-1}\,ds< 0, \end{aligned}$$
$$\begin{aligned}& \int_{\frac{1}{6}}^{1}\frac{g(s)}{s} (1-12s )^{n-1}\,ds\leq \int_{\frac{1}{6}}^{1}\frac{g(s)}{s} (1-12s )^{3}\,ds. \end{aligned}$$
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
$$ (-1 )^{m-1}+ \int_{0}^{1}\frac{g(s)}{1-s} (12s-1 )^{m-1}\,ds>0. $$


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 \(\frac{1}{12}\leq{s}\leq\frac{1}{2}\)
$$ h(s,x)>0. $$


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
$$ \biggl(\frac{(s-\frac{1}{12})(1+x)}{s(\frac{11}{12}+x)} \biggr)^{5} < \frac{s+x}{1-s+x}. $$
To prove (2.12), we define \(h_{1}(s,x)\) as
$$ h_{1}=\ln(1-s+x)-\ln(s+x)+5\ln\biggl(s-\frac{1}{12}\biggr)+5 \ln(1+x)-5\ln{s}-5\ln \biggl(\frac{11}{12}+x\biggr). $$
Easy computations reveal that
$$\begin{aligned}& h_{1}\biggl(\frac{1}{2},0\biggr)=5\ln\biggl(\frac{10}{11} \biggr)< 0, \end{aligned}$$
$$\begin{aligned}& \frac{\partial{h_{1}(s,0)}}{\partial{s}}>0, \end{aligned}$$
$$\begin{aligned}& \frac{\partial{h_{1}(s,x)}}{\partial{x}}< 0. \end{aligned}$$
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 \([0,\frac{1}{12}]\).


It suffices to show that \(\frac{\partial{h(s,x)}}{\partial{s}}>0\). Partial differentiation yields
$$\begin{aligned} \frac{\partial{h(s,x)}}{\partial{s}} \geq{}&\frac{m-1}{(1+x)(1-s+x)} \biggl(\frac{s}{1+x} \biggr)^{m-2} \\ &{}- \frac{m-1}{(11+12x)(s+x)} \biggl(\frac{12s-1}{11+12x} \biggr)^{m-2}. \end{aligned}$$
If \(\frac{1}{2}\leq{s}\leq{1}\), then for all \(x>0\) and \({m}\geq{2}\), we have
$$\begin{aligned}& \frac{m-1}{(1+x)(1-s+x)}>\frac{m-1}{(11+12x)(s+x)},\\& \frac{s}{1+x}>\frac{12s-1}{11+12x}. \end{aligned}$$
$$\frac{\partial{h(s,x)}}{\partial{s}}>0. $$
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 \(x\geq1\)
$$ h\biggl(\frac{1}{2},x\biggr)>\bigl|h(0,x)\bigr|. $$


Because of \(x>1\), (2.18) is equivalent to
$$ \biggl(5+\frac{1+2x}{2+2x} \biggr)^{m-1}-5^{m-1}>1+ \frac{1}{2x}. $$
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


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.



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 (, 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

Henan College of Finance and Taxation, Department of Information Engineering, Zhengzhou, Henan, 451464, China
Department of Foundation, Zhejiang University of Water Resources and Electric Power, Hangzhou, Zhejiang, 310018, China


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