Open Access

On the Keller limit and generalization

Journal of Inequalities and Applications20162016:97

https://doi.org/10.1186/s13660-016-1042-z

Received: 9 January 2016

Accepted: 12 March 2016

Published: 25 March 2016

Abstract

Let c be any real number and let
$$u_{n}(c)=(n+1) \biggl(1+\frac{1}{n+c} \biggr)^{n+c}-n \biggl(1+\frac {1}{n+c-1} \biggr)^{n+c-1}-e. $$
In this note, we establish an integral expression of \(u_{n}(c)\), which provides a direct proof of Theorem 1 in (Mortici and Jang in Filomat 7:1535-1539, 2015).

Keywords

Keller’s limit constant e integral expression

MSC

35B05 35B10

1 Introduction motivation

The limit
$$\lim_{n\rightarrow\infty} \biggl(\frac{(n+1)^{n+1}}{n^{n}}-\frac{n^{n}}{(n-1)^{n-1}} \biggr)=e $$
is well known in the literature as the Keller’s limit, see [2]. Such a limit is very useful in many mathematical contexts and contributes as a tool for establishing some interesting inequalities [36].

In the recent paper [1], Mortici et al. have constructed a new proof of the limit and have discovered the following new results which generalize the Keller limit.

Theorem 1

Let c be any real number and let
$$u_{n}(c)=(n+1) \biggl(1+\frac{1}{n+c} \biggr)^{n+c}-n \biggl(1+\frac {1}{n+c-1} \biggr)^{n+c-1}-e. $$
Then
$$\begin{aligned}& \lim_{n\rightarrow\infty}u_{n}(c)=0, \end{aligned}$$
(1.1)
$$\begin{aligned}& \lim_{n\rightarrow\infty}n^{2}u_{n}(c)= \frac{e}{24}(1-12c), \end{aligned}$$
(1.2)
$$\begin{aligned}& \lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)=\frac{5e}{144}. \end{aligned}$$
(1.3)
The proof of Theorem 1 given in [1] is based on the following double inequality for every x in \(0< x\leq1\):
$$a(x)< (1+x)^{1/x}< b(x), $$
where
$$a(x)=e-\frac{e}{2x}+\frac{11ex^{2}}{24}-\frac{21ex^{3}}{48}+ \frac {2{,}447ex^{4}}{5{,}760}-\frac{959ex^{5}}{2{,}304} $$
and
$$b(x)=a(x)+\frac{959ex^{5}}{2{,}304}. $$

But, this proof has a major objection, namely, for the reader it is very difficult to observe the behavior of \(u_{n}(c)\) as \(n\rightarrow\infty\).

In this note, we will establish an integral expression of \(u_{n}(c)\), which tells us that Theorem 1 is a natural result.

2 Main results

To establish an integral expression of \(u_{n}(c)\), we first recall the following result we obtained in [7].

Theorem 2

Let \(h(s)=\frac{\sin(\pi s)}{\pi}s^{s}(1-s)^{1-s}\), \(0\leq s\leq 1\). Then for every \(x>0\), we have
$$ \biggl(1+\frac{1}{x} \biggr)^{x} =e \Biggl(1-\sum _{j=1}^{\infty}\frac{b_{j}}{(1+x)^{j}} \Biggr), $$
(2.1)
where
$$\begin{aligned}& b_{1}=\frac{1}{2}, \end{aligned}$$
(2.2)
$$\begin{aligned}& b_{j}=\frac{1}{e} \int_{0}^{1}s^{j-2}h(s)\,ds \quad(j=2,3, \ldots). \end{aligned}$$
(2.3)

In [8] (see also [9, 10]) Yang has proved that \(b_{2}=\frac{1}{24}\), \(b_{3}=\frac{1}{48}\).

Hence
$$\begin{aligned}& \int_{0}^{1}h(s)\,ds=\frac{e}{24}, \end{aligned}$$
(2.4)
$$\begin{aligned}& \int_{0}^{1}sh(s)\,ds=\frac{e}{48}. \end{aligned}$$
(2.5)
Now, we establish an integral expression of \(u_{n}(c)\). Equation (2.1) implies the following results:
$$\begin{aligned}& \biggl(1+\frac{1}{n+c} \biggr)^{n+c} =e \Biggl(1-\sum _{j=1}^{\infty}\frac{b_{j}}{(1+n+c)^{j}} \Biggr), \end{aligned}$$
(2.6)
$$\begin{aligned}& \biggl(1+\frac{1}{n+c-1} \biggr)^{n+c-1} =e \Biggl(1-\sum _{j=1}^{\infty}\frac{b_{j}}{(n+c)^{j}} \Biggr). \end{aligned}$$
(2.7)
Hence by (2.2), (2.3), (2.6), and (2.7), we have
$$\begin{aligned} u_{n}(c) =&\frac{e}{2} \biggl(\frac{n}{n+c} - \frac{n+1}{1+n+c} \biggr)+ \int_{0}^{1}h(s)\sum_{j=2}^{\infty} \frac {ns^{j-2}}{(n+c)^{j}}\,ds \\ &{}- \int_{0}^{1}h(s)\sum_{j=2}^{\infty} \frac{(n+1)s^{j-2}}{(1+n+c)^{j}}\,ds. \end{aligned}$$
(2.8)
Note that
$$\begin{aligned}& \frac{e}{2}= \int_{0}^{1}12h(s)\,ds, \end{aligned}$$
(2.9)
$$\begin{aligned}& \sum_{j=2}^{\infty} \frac{ns^{j-2}}{(n+c)^{j}}=\frac{1}{(n+c)(n+c-s)}, \end{aligned}$$
(2.10)
$$\begin{aligned}& \sum_{j=2}^{\infty} \frac{(n+1)s^{j-2}}{(1+n+c)^{j}}=\frac{1}{(1+n+c)(1+n+c-s)}. \end{aligned}$$
(2.11)
Therefore, from (2.8)-(2.11), we obtain the desired result:
$$\begin{aligned} u_{n}(c)= \int_{0}^{1}h(s)\frac{(1-12c)n^{2} +(1-12c+24cs-24c^{2})n+K}{(n+c)(1+n+c)(n+c-s)(1+n+c-s)}\,ds, \end{aligned}$$
(2.12)
where
$$K=s^{2}-(1+2c)s+c+c^{2}. $$
From (2.12), we get immediately
$$\begin{aligned}& \lim_{n\rightarrow\infty}u_{n}(c)=0, \\& \lim_{n\rightarrow\infty}n^{2}u_{n}(c)= \int_{0}^{1}(1-12c)h(s)\,ds \\& \hphantom{\lim_{n\rightarrow\infty}n^{2}u_{n}(c)}=\frac{e}{24}(1-12c), \\& \lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)= \int_{0}^{1}\biggl(2s-\frac{1}{6} \biggr)h(s)\,ds \\& \hphantom{\lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)}=\frac{e}{24}-\frac{1}{6}\times\frac{e}{24} \\& \hphantom{\lim_{n\rightarrow\infty}n^{3}u_{n}\biggl( \frac{1}{12}\biggr)}=\frac{5e}{144}. \end{aligned}$$

3 Conclusions

We have established an integral expression of \(u_{n}(c)\), which provides a direct proof of Theorem 1 in [1] and tell us that Theorem 1 is a natural result. We believe that the expression will lead to a significant contribution toward the study of Keller’s limit.

Declarations

Acknowledgements

The work was supported by the National Natural Science Foundation of China, No. 11471103.

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)
School of Mathematics and Information Science, Henan Polytechnic University
(2)
Department of Mathematics, Valahia University of Târgovişte

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Copyright

© Hu and Mortici 2016