1 Introduction
The Carleman inequality [2]
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, 3‐10].
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty}a_{n}, $$
In [7], Yang proved
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
then \(b_{k}>0\), \(k=1,2,3,\ldots\) .
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{6}\frac{b_{k}}{ ( n+1 ) ^{k}} \Biggr) a_{n} $$
$$ \biggl( 1+\frac{1}{x} \biggr) ^{x}=e \Biggl( 1-\sum _{k=1}^{\infty}\frac {b_{k}}{ ( x+1 ) ^{k}} \Biggr) ,\quad x>0, $$
(1.1)
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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
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
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{m }\frac{b_{k}}{ ( n+1 ) ^{k}} \Biggr) a_{n}, $$
(1.2)
$$ b_{n+1}=\frac{1}{n+1} \Biggl( \frac{1}{n+2}-\sum _{k=1}^{n}\frac {b_{k}}{n+2-k} \Biggr).$$
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
and therefore the following inequality is better than (1.2):
$$\sum_{n=1}^{\infty}\frac{d_{n}}{ (x+\varepsilon) ^{n}} $$
$$ \sum_{n=1}^{\infty} ( a_{1}a_{2} \cdots a_{n} ) ^{1/n}< e\sum_{n=1}^{\infty} \Biggl( 1-\sum_{k=1}^{m }\frac{d_{k}}{ ( n+\frac{11}{12} ) ^{k}} \Biggr) a_{n}. $$
(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.
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Theorem
If
\(( \omega_{n} )_{n\geq1}\)
is convergent to zero and
with
\(k>1\), then there exists the limit
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.
$$ \lim_{n\rightarrow\infty}n^{k}(\omega_{n}- \omega_{n+1})=l\in \mathbb{R}, $$
$$ \lim_{n\rightarrow\infty}n^{k-1}\omega_{n}= \frac{l}{k-1}. $$
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)
If
\(m\geq6\)
is even, then
\(S_{m}(x)>\sigma_{m}(x)\)
for all
\(x>0\).
(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
Here \(0\leq{s}\leq1\), \(x>0\), and \(m\geq1\) is an integer.
$$\begin{aligned}& g(s)=\frac{1}{\pi}s^{s}(1-s)^{1-s}\sin(\pi s),\\& h(s,x)=\frac{1}{1-s+x} \biggl(\frac{s}{1+x} \biggr)^{m-1}- \frac{1}{s+x} \biggl(\frac{12s-1}{11+12x} \biggr)^{m-1}. \end{aligned}$$
Lemma 1
For
\(x>0\), let
Then
$$\biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr). $$
$$\begin{aligned}& b_{k}>0 ,\quad k=1,2 ,\ldots, \end{aligned}$$
(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)
Remark 1
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
If n is an even, then we have for all \(n\geq4\)
From (2.7), (2.8), (2.9), and Remark 1, we get
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). □
$$\begin{aligned}& \int_{0}^{\frac{1}{12}}\frac{g(s)}{s} (1-12s )^{n-1}\,ds\leq \int_{0}^{\frac{1}{12}}\frac{g(s)}{s} (1-12s )^{3}\,ds, \end{aligned}$$
(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)
$$\begin{aligned} \int_{0}^{1}\frac{g(s)}{s} (1-12s )^{n-1}\,ds \leq{}& \int_{0}^{1} \biggl(\frac{1}{s}-36+432s-1{,}728s^{2} \biggr)g(s)\,ds\\ &{}+ \int_{\frac{1}{12}}^{\frac{1}{6}} \bigl(1{,}728s^{2}-432s+30 \bigr)\,ds\\ \leq{}& \biggl(\frac{e}{2}-1 \biggr)-\frac{3e}{2}+9e-21.9e+6 \int_{0}^{1}g(s)\,ds\\ ={}&{-}1-13.9e+\frac{e}{4}< -1. \end{aligned}$$
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
To prove (2.12), we define \(h_{1}(s,x)\) as
Easy computations reveal that
Thus from (2.14), (2.15), and (2.16), we have
which implies
□
$$ \biggl(\frac{(s-\frac{1}{12})(1+x)}{s(\frac{11}{12}+x)} \biggr)^{5} < \frac{s+x}{1-s+x}. $$
(2.12)
$$ h_{1}=\ln(1-s+x)-\ln(s+x)+5\ln\biggl(s-\frac{1}{12}\biggr)+5 \ln(1+x)-5\ln{s}-5\ln \biggl(\frac{11}{12}+x\biggr). $$
(2.13)
$$\begin{aligned}& h_{1}\biggl(\frac{1}{2},0\biggr)=5\ln\biggl(\frac{10}{11} \biggr)< 0, \end{aligned}$$
(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)
$${h_{1}(s,x)}\leq{h_{1}(s,0)}\leq{h_{1}\biggl( \frac{1}{2},0\biggr)}< 0, $$
$$h(s,x)>0. $$
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
If \(\frac{1}{2}\leq{s}\leq{1}\), then for all \(x>0\) and \({m}\geq{2}\), we have
Thus
If m is an odd, then for \(0<{s}<\frac{1}{12}\), we have
From this and (2.17), we get
This completes the proof of Lemma 5. □
$$\begin{aligned} \frac{\partial{h(s,x)}}{\partial{s}} \geq{}&\frac{m-1}{(1+x)(1-s+x)} \biggl(\frac{s}{1+x} \biggr)^{m-2} \\ &{}- \frac{m-1}{(11+12x)(s+x)} \biggl(\frac{12s-1}{11+12x} \biggr)^{m-2}. \end{aligned}$$
(2.17)
$$\begin{aligned}& \frac{m-1}{(1+x)(1-s+x)}>\frac{m-1}{(11+12x)(s+x)},\\& \frac{s}{1+x}>\frac{12s-1}{11+12x}. \end{aligned}$$
$$\frac{\partial{h(s,x)}}{\partial{s}}>0. $$
$$(12s-1)^{m-2}< 0. $$
$$\frac{\partial{h(s,x)}}{\partial{s}}>0. $$
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)
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
By Lemma 3, it suffices to show that
Let first \(m\geq6\) be even. From Lemma 4 and Lemma 5, for all \(x>0\), we have
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 \(x>0\).
$$\begin{aligned} S_{m}(x)-\sigma_{m}(x) ={}&\frac{1}{(11+12x)^{m}(1+x)} \biggl((-1)^{m-1}+ \int_{0}^{1}\frac {g(s)}{1-s}(12s-1)^{m-1} \biggr) \\ &{}+\frac{1}{1+x} \int_{0}^{1}g(s) \biggl(\frac{1}{1-s+x} \biggl( \frac{s}{1+x} \biggr)^{m-1}-\frac{1}{x+s} \biggl( \frac {12s-1}{11+12x} \biggr)^{m-1} \biggr)\,ds \\ ={}&\frac{1}{(11+12x)^{m}(1+x)} \biggl((-1)^{m-1}+ \int_{0}^{1}\frac {g(s)}{1-s}(12s-1)^{m-1} \biggr) \\ &{}+\frac{1}{1+x} \int_{0}^{1}g(s)h(s,x)\,ds. \end{aligned}$$
$$\int_{0}^{1}g(s)h(s,x)\,ds>0. $$
$$\int_{0}^{1}g(s)h(s,x)\,ds= \int_{0}^{\frac{1}{12}}g(s)h(s,x)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds+ \int_{\frac {1}{2}}^{1}g(s)h(s,x)\,ds>0. $$
Now let \(m\geq7\) be odd. From Lemma 4, Lemma 5, and Lemma 6 for all \(x\geq1\) we have
This completes the proof of Theorem 1. □
$$\begin{aligned} \int_{0}^{1}g(s)h(s,x)\,ds&= \int_{0}^{\frac{1}{12}}g(s)h(s,x)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds+ \int_{\frac {1}{2}}^{1}g(s)h(s,x)\,ds \\ &\geq \int_{0}^{\frac{1}{12}}g(s)h(0,x)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds+ \int_{\frac {1}{2}}^{1}g(s)h\biggl(\frac{1}{2},x \biggr)\,ds \\ &\geq\biggl(h(0,x)+h\biggl(\frac{1}{2},x\biggr)\biggr) \int_{0}^{\frac{1}{12}}g(s)\,ds + \int_{\frac{1}{12}}^{\frac{1}{2}}g(s)h(s,x)\,ds>0. \end{aligned}$$
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.
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.
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Authors’ contributions
The authors completed the paper together. They each read and approved the final manuscript.