# On the Keller limit and generalization

## 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).

## 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 . Such a limit is very useful in many mathematical contexts and contributes as a tool for establishing some interesting inequalities .

In the recent paper , 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  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.

## Main results

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

### 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  (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}

## Conclusions

We have established an integral expression of $$u_{n}(c)$$, which provides a direct proof of Theorem 1 in  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.

## References

1. Mortici, C, Jang, X: Estimates of $$(1+ 1/x)^{x}$$ involved in Carleman’s inequality and Keller’s limit. Filomat 7, 1535-1539 (2015)

2. Sandor, J, Debnath, L: On certain inequalities involving the constant e and their applications. J. Math. Anal. Appl. 249(2), 569-582 (2000)

3. Polya, G, Szego, G: Problems and Theorems in Analysis, vol. I. Springer, New York (1972)

4. Hardy, GH, Littlewood, JE, Polya, G: Inequalities. Cambridge University Press, London (1952)

5. Mortici, C, Hu, Y: On some convergences to the constant e and improvements of Carleman’s inequality. Carpath. J. Math. 31, 249-254 (2015)

6. Mortici, C, Hu, Y: On an infinite series for $$(1+ 1/x)^{x}$$. arXiv:1406.7814 [math.CA]

7. Hu, Y, Mortici, C: On the coefficients of an expansion of $$(1+\frac{1}{x})^{x}$$ related to Carleman’s inequality. arXiv:1401.2236 [math.CA]

8. Yang, X: Approximations for constant e and their applications. J. Math. Anal. Appl. 262, 651-659 (2001)

9. Gyllenberg, M, Yan, P: On a conjecture by Yang. J. Math. Anal. Appl. 264, 687-690 (2001)

10. Chen, H: On an infinite series for $$(1+\frac{1}{x})^{x}$$ and its application. Int. J. Math. Math. Sci. 11, 675-680 (2002)

## Acknowledgements

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

## Author information

Authors

### Corresponding author

Correspondence to Yue Hu.

### 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 also read and approved the final manuscript.

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