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).

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 [3–6].

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.