1 Introduction
The constant e is the base in the natural logarithm. e can be defined by the limit
With the possible exception of π, e is the most important constant in mathematics since it appears in myriad mathematical contexts involving limits and derivatives. It is noted (see, e.g., [1]; see also [2], pp.26-27) that approximations to e were first discovered in the 1600s. For the various inequalities and approximations of the constants e, the reader may be referred to several recent works (see, e.g., [3‐9]).
$$\begin{aligned} e=\lim_{n \to\infty} \biggl(1+\frac{1}{n} \biggr)^{n} = 2.7182818284590452353602875\ldots. \end{aligned}$$
Batir and Cancan (see [3], Theorem 2.5 and Theorem 2.6) proved that for \(n\in\mathbb{N}:=\{1, 2, \ldots\}\),
and
with the best possible constants
and
$$ \exp \biggl(1-\frac{n}{2(n+c)^{2}} \biggr)\leq \biggl(1+ \frac{1}{n} \biggr)^{n}< \exp \biggl(1-\frac{n}{2(n+d)^{2}} \biggr) $$
(1.1)
$$ \exp \biggl(1+\frac{1}{2(n+\alpha)} \biggr)< \biggl(1+ \frac{1}{n} \biggr)^{n+1}\leq\exp \biggl(1+\frac{1}{2(n+\beta)} \biggr) $$
(1.2)
$$ c = \frac{1}{\sqrt{2-\ln4}}-1=0.27649\ldots,\qquad d=\frac{1}{3}, $$
$$ \alpha=\frac{1}{3}, \qquad\beta=\frac{1}{4\ln2-2}-1=0.2943497\ldots. $$
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Inspired by (1.1) and (1.2), Hu and Mortici (see [6], Theorem 1 and Theorem 2) established the following inequalities:
and
for \(x\geq1\).
$$ \exp \biggl(1-\frac{x}{2(x+\frac{1}{3}-\frac{1}{12x})^{2}} \biggr)< \biggl(1+ \frac{1}{x} \biggr)^{x}<\exp \biggl(1-\frac{x}{2(x+\frac{1}{3}-\frac {1}{12x}+\frac{23}{540x^{2}})^{2}} \biggr) $$
(1.3)
$$ \exp \biggl(1+\frac{x}{2(x+\frac{1}{6}-\frac{1}{24x}+\frac {43}{2{,}160x^{2}})^{2}} \biggr)< \biggl(1+ \frac{1}{x} \biggr)^{x+1}<\exp \biggl(1+\frac{x}{2(x+\frac{1}{6}-\frac{1}{24x})^{2}} \biggr) $$
(1.4)
Also in [6], the authors proved
(see [6], Theorem 6), and they proposed as an open problem the following approximation formula:
for \(a\in[0, 1]\setminus\{\frac{1}{2}\}\).
$$ \exp \biggl(1+\frac{x}{12(x+\frac{1}{3})^{3}} \biggr)< \biggl(1+ \frac {1}{x} \biggr)^{x+\frac{1}{2}}<\exp \biggl(1+\frac{x}{12(x+\frac {1}{3}-\frac{7}{90x})^{3}} \biggr),\quad x\geq1 $$
(1.5)
$$ \biggl(1+\frac{1}{n} \biggr)^{n+a}\approx\exp \biggl(1+\frac{(a-\frac {1}{2})n}{ (n+\frac{3a-2}{6(2a-1)} )^{2}} \biggr) $$
(1.6)
For
where \(\theta=0.82462 \ldots\) is the unique zero of \(675a^{3}-1{,}098a^{2}+558a-92\), Hu and Mortici (see [6], Theorems 3 to 5) considered the inequalities for \((1+1/x)^{x+a}\). In fact, the authors obtained the following approximation formula:
$$ a\in \biggl[0, \frac{4-\sqrt{2}}{7} \biggr)\cup \biggl(\frac{1}{2}, \frac{4+\sqrt{2}}{7} \biggr)\cup(\theta,1], $$
$$ \biggl(1+\frac{1}{x} \biggr)^{x+a}\approx\exp \biggl(1 +\frac{(a-\frac{1}{2}) x}{ (x + \frac{3a-2}{6(2a-1)}+\frac{-7a^{2}+8a-2}{24(2a-1)^{2}x} )^{2}} \biggr),\quad x \to\infty. $$
(1.7)
In the case \(a = \frac{1}{2}\), Hu and Mortici [6] presented the following asymptotic representation:
Moreover, the authors showed that the coefficients \(a_{j}\) in (1.8) can be inductively obtained by equating the following relation:
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}\sim\exp \biggl(1+ \frac{\frac {1}{12}x}{ (x+\sum_{j=0}^{\infty}a_{j}x^{-j} )^{3}} \biggr),\quad x\to\infty. $$
(1.8)
$$ \Biggl(\sum_{j=3}^{\infty}(-1)^{j-1} \frac{6(j-2)}{j(j-1)x^{j}} \Biggr) \Biggl(x+\sum_{j=0}^{\infty} \frac{a_{j}}{x^{j}} \Biggr)^{3}=1. $$
(1.9)
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The first few coefficients \(a_{j}\) are
We then obtain the following explicit asymptotic expansion:
$$ a_{0}=\frac{1}{3},\qquad a_{1}=- \frac{7}{90},\qquad a_{2}=\frac{16}{405},\qquad a_{3}=- \frac{2{,}141}{85{,}050}. $$
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}\sim\exp \biggl(1+\frac{\frac {1}{12}x}{ (x+\frac{1}{3}-\frac{7}{90x}+\frac{16}{405x^{2}}-\frac {2{,}141}{85{,}050x^{3}}+\cdots )^{3}} \biggr),\quad x\to\infty. $$
(1.10)
In this paper, our tasks are as follows:
(i)
We consider the above open problem of Hu and Mortici and develop the approximation formula (1.6) to produce a complete asymptotic expansion. More precisely, we give a formula for determining the coefficients \(b_{j}\equiv b_{j}(a)\) such that
$$ \biggl(1+\frac{1}{x} \biggr)^{x+a}\sim\exp \biggl(1+ \frac{(a-\frac {1}{2})x}{ (x+\sum_{j=0}^{\infty}b_{j}x^{-j} )^{2}} \biggr) \quad \biggl(x \to\infty; a\neq\frac{1}{2} \biggr). $$
(ii)
We provide a recurrence relation for determining the coefficients \(a_{j}\) in (1.8). The representation using recursive algorithm is better for numerical evaluations.
(iii)
We present new inequalities and asymptotic expansions for \((1+1/x)^{x+a}\).
The following lemma is required in the sequel.
Lemma 1.1
(see [10])
Let
\(g(x)\)
be a function with an asymptotic expansion (\(q_{0}=1\))
Then for all real
r
we have
where
$$ g(x)\sim\sum_{j=0}^{\infty}q_{j}x^{-j}, \quad x\to\infty. $$
$$ \bigl[g(x) \bigr]^{r}\sim\sum _{j=0}^{\infty}P_{j}(r)x^{-j},\quad x \to\infty, $$
$$ P_{0}(r)=1,\qquad P_{j}(r)=\frac{1}{j} \sum_{k=1}^{j} \bigl[k(1+r)-j \bigr]q_{k}P_{j-k}(r),\quad j\in \mathbb{N}. $$
(1.11)
2 Asymptotic formulas
Theorem 2.1
Let
\(a\neq\frac{1}{2}\)
be a given real number. The following asymptotic expansion holds:
with the coefficients
\(b_{j}\equiv b_{j}(a)\)
given by
where the
\(P_{j}(-\frac{1}{2})\)
are given by the recurrence relation
$$ \biggl(1+\frac{1}{x} \biggr)^{x+a}\sim\exp \biggl(1+ \frac{(a-\frac {1}{2})x}{ (x+\sum_{j=0}^{\infty}b_{j}x^{-j} )^{2}} \biggr),\quad x \to\infty, $$
(2.1)
$$\begin{aligned} b_{j}=P_{j+1} \biggl(-\frac{1}{2} \biggr),\quad j \in\mathbb{N}_{0}:=\mathbb{N}\cup\{0\}, \end{aligned}$$
(2.2)
$$ \begin{aligned} &P_{0} \biggl(-\frac{1}{2} \biggr)=1,\\ &P_{j} \biggl(-\frac{1}{2} \biggr)=\frac{1}{j}\sum _{k=1}^{j}(-1)^{k} \biggl( \frac{k}{2}-j \biggr)\frac{2 ((a-1)k+2a-1 )}{(2a-1)(k+1)(k+2)}P_{j-k} \biggl(- \frac{1}{2} \biggr),\quad j\in\mathbb{N}. \end{aligned} $$
(2.3)
Proof
To determine the coefficients \(b_{j}\) in (2.1), we write (2.1) as
where
$$ \bigl(g(x) \bigr)^{-1/2}\sim x + \sum _{j=0}^{\infty}b_{j}x^{-j}, $$
(2.4)
$$ g(x)=\frac{(x+a)\ln(1+\frac{1}{x})-1}{(a-\frac{1}{2})x}. $$
Using the Maclaurin expansion of \(\ln(1+t)\) with \(t = x^{-1}\) yields
with
$$ g(x)\sim x^{-2}\sum_{k=0}^{\infty}q_{k}x^{-k},\quad x\to\infty $$
$$ q_{k}=(-1)^{k}\frac{2 ((a-1)k+2a-1 )}{(2a-1)(k+1)(k+2)}. $$
(2.5)
By Lemma 1.1, we have
with the coefficients \(P_{j}(-\frac{1}{2})\) given by
where the \(q_{k}\) are given by (2.5).
$$ \bigl(g(x) \bigr)^{-1/2}\sim x \Biggl(\sum _{j=0}^{\infty}q_{k}x^{-k} \Biggr)^{-1/2}\sim x\sum_{j=0}^{\infty}P_{j} \biggl(-\frac{1}{2} \biggr)x^{-j} $$
(2.6)
$$ P_{0} \biggl(- \frac{1}{2} \biggr)=1,\qquad P_{j} \biggl(-\frac{1}{2} \biggr)= \frac{1}{j}\sum_{k=1}^{j} \biggl( \frac{k}{2}-j \biggr)q_{k}P_{j-k} \biggl(- \frac{1}{2} \biggr),\quad j\in \mathbb{N}, $$
(2.7)
Hence, it follows from (2.4) and (2.6) that
we then obtain
where the \(P_{j}(-\frac{1}{2})\) are given in (2.7). The proof of Theorem 2.1 is complete. □
$$\begin{aligned}& x + \sum_{j=0}^{\infty}b_{j}x^{-j} \sim x\sum_{j=0}^{\infty}P_{j} \biggl(- \frac{1}{2} \biggr)x^{-j}, \\& \sum_{j=0}^{\infty}b_{j}x^{-j} \sim \sum_{j=0}^{\infty}P_{j+1} \biggl(- \frac{1}{2} \biggr)x^{-j}, \end{aligned}$$
$$\begin{aligned} b_{j}=P_{j+1} \biggl(-\frac{1}{2} \biggr),\quad j \in\mathbb{N}_{0}, \end{aligned}$$
Remark 2.1
Using (2.2) and (2.3), we now show that we easily can determine the \(b_{j}\) in (2.1). The first few coefficients \(b_{j}\equiv b_{j}(a)\) are
$$\begin{aligned}& b_{0}=P_{1} \biggl(- \frac{1}{2} \biggr)=\frac{3a-2}{6(2a-1)}P_{0} \biggl(- \frac{1}{2} \biggr)=\frac {3a-2}{6(2a-1)}, \\& b_{1}=P_{2} \biggl(-\frac{1}{2} \biggr)= \frac{(9a-6)P_{1}(-\frac{1}{2})+(3-4a)P_{0}(-\frac {1}{2})}{12(2a-1)}=-\frac{2-8a+7a^{2}}{24(2a-1)^{2}}, \\& \begin{aligned}[b] b_{2}&=P_{3} \biggl(-\frac{1}{2} \biggr)= \frac{(150a-100)P_{2}(-\frac {1}{2})+(60-80a)P_{1}(-\frac{1}{2})+(45a-36)P_{0}(-\frac {1}{2})}{108(2a-1)} \\ &=\frac{-92+558a-1{,}098a^{2}+675a^{3}}{2{,}160(2a-1)^{3}}, \end{aligned}\\& \begin{aligned}[b] b_{3}&=P_{4} \biggl(-\frac{1}{2} \biggr)\\ &=-\frac{(140-210a)P_{3}(-\frac{1}{2})+(120a-90)P_{2}(-\frac {1}{2})+(60-75a)P_{1}(-\frac{1}{2})+(48a-40)P_{0}(-\frac {1}{2})}{240(2a-1)}\\ &=-\frac{1{,}412-11{,}424a+34{,}308a^{2}-44{,}928a^{3}+21{,}249a^{4}}{51{,}840(2a-1)^{4}}. \end{aligned} \end{aligned}$$
In particular, setting \(a=0\) and \(a=1\) in (2.1), respectively, we have the following complete asymptotic expansions:
and
as \(x\to\infty\).
$$ \biggl(1+\frac{1}{x} \biggr)^{x}\sim\exp \biggl(1- \frac{x}{2 (x+\frac {1}{3}-\frac{1}{12x}+\frac{23}{540x^{2}}-\frac{353}{12{,}960x^{3}}+\cdots )^{2}} \biggr) $$
(2.8)
$$ \biggl(1+\frac{1}{x} \biggr)^{x+1}\sim\exp \biggl(1+\frac{x}{2 (x+\frac {1}{6}-\frac{1}{24x}+\frac{43}{2{,}160x^{2}}-\frac{617}{51{,}840x^{3}}+\cdots )^{2}} \biggr) $$
(2.9)
Theorem 2.2
The coefficients
\(a_{j}\)
in (1.8) are given by
where the
\(P_{j}(-\frac{1}{3})\)
are given by the recurrence relation
$$\begin{aligned} a_{j}=P_{j+1} \biggl(-\frac{1}{3} \biggr),\quad j \in\mathbb{N}_{0}, \end{aligned}$$
(2.10)
$$ \begin{aligned} &P_{0} \biggl(-\frac{1}{3} \biggr)=1,\\ &P_{j} \biggl(-\frac{1}{3} \biggr)=\frac{1}{j}\sum _{k=1}^{j}(-1)^{k} \biggl( \frac {2k}{3}-j \biggr)\frac{6(k+1)}{(k+2)(k+3)}P_{j-k} \biggl(- \frac{1}{3} \biggr),\quad j\in\mathbb{N}. \end{aligned} $$
(2.11)
Proof
A similar argument to the proof of Theorem 2.1 will establish the result in Theorem 2.2. To determine the coefficients \(a_{j}\) in (1.8), we write (1.8) as
where
$$ \bigl(f(x) \bigr)^{-1/3}\sim x + \sum _{j=0}^{\infty}a_{j}x^{-j}, $$
(2.12)
$$ f(x)=\frac{(x+\frac{1}{2})\ln(1+\frac{1}{x})-1}{\frac{1}{12}x}. $$
Using the Maclaurin expansion of \(\ln(1+t)\) with \(t = x^{-1}\) yields
with
$$ f(x)\sim x^{-3}\sum_{k=0}^{\infty}p_{k}x^{-k},\quad x\to\infty, $$
$$ p_{k}=(-1)^{k}\frac{6(k+1)}{(k+2)(k+3)}. $$
(2.13)
By Lemma 1.1, we have
with the coefficients \(P_{j}(-\frac{1}{3})\) given by
where the \(p_{k}\) are given by (2.13).
$$ \bigl(f(x) \bigr)^{-1/3}\sim x \Biggl(\sum _{k=0}^{\infty}p_{k}x^{-k} \Biggr)^{-1/3}\sim x\sum_{j=0}^{\infty}P_{j} \biggl(-\frac{1}{3} \biggr)x^{-j} $$
(2.14)
$$ P_{0} \biggl(- \frac{1}{3} \biggr)=1,\qquad P_{j} \biggl(-\frac{1}{3} \biggr)= \frac{1}{j}\sum_{k=1}^{j} \biggl( \frac{2k}{3}-j \biggr)p_{k}P_{j-k} \biggl(- \frac{1}{3} \biggr),\quad j\in \mathbb{N}, $$
(2.15)
Hence, it follows from (2.12) and (2.14) that
we then obtain
where the \(P_{j}(-\frac{1}{3})\) are given in (2.15). The proof of Theorem 2.2 is complete. □
$$\begin{aligned}& x + \sum_{j=0}^{\infty}a_{j}x^{-j} \sim x\sum_{j=0}^{\infty}P_{j} \biggl(- \frac{1}{3} \biggr)x^{-j}, \\& \sum_{j=0}^{\infty}a_{j}x^{-j} \sim \sum_{j=0}^{\infty}P_{j+1} \biggl(- \frac{1}{3} \biggr)x^{-j}, \end{aligned}$$
$$\begin{aligned} a_{j}=P_{j+1} \biggl(-\frac{1}{3} \biggr),\quad j \in\mathbb{N}_{0}, \end{aligned}$$
Remark 2.2
Using (2.10) and (2.11), we now show how easily we can determine the \(a_{j}\) in (1.8). The first few coefficients \(a_{j}\) are
We note that the values of \(a_{j}\) (for \(j=0, 1, 2, 3\)) here are equal to the coefficients of \(1/x^{j}\) (for \(j=0, 1, 2, 3\)) in (1.10), respectively.
$$\begin{aligned}& a_{0}=P_{1} \biggl(- \frac{1}{3} \biggr)=\frac{1}{3}P_{0} \biggl(- \frac{1}{3} \biggr)=\frac{1}{3}, \\& a_{1}=P_{2} \biggl(-\frac{1}{3} \biggr)= \frac{2}{3}P_{1} \biggl(-\frac{1}{3} \biggr)- \frac {3}{10}P_{0} \biggl(-\frac{1}{3} \biggr)=- \frac{7}{90}, \\& a_{2}=P_{3} \biggl(-\frac{1}{3} \biggr)= \frac{7}{9}P_{2} \biggl(-\frac{1}{3} \biggr)- \frac {1}{2}P_{1} \biggl(-\frac{1}{3} \biggr)+ \frac{4}{15}P_{0} \biggl(-\frac{1}{3} \biggr)= \frac {16}{405}, \\& a_{3}=P_{4} \biggl(-\frac{1}{3} \biggr)= \frac{5}{6}P_{3} \biggl(-\frac{1}{3} \biggr)- \frac {3}{5}P_{2} \biggl(-\frac{1}{3} \biggr)+ \frac{2}{5}P_{1} \biggl(-\frac{1}{3} \biggr)- \frac {5}{21}P_{0} \biggl(-\frac{1}{3} \biggr)=- \frac{2{,}141}{85{,}050}. \end{aligned}$$
Theorems 2.3 and 2.4 present new asymptotic expansions for \((1+1/x)^{x+\alpha}\) (with \(\alpha\neq\frac{1}{2}\)) and \((1+1/x)^{x+1/2}\), respectively. As the proofs of Theorems 2.3 and 2.4 are similar to the proof of Theorem 2.1, we omit them.
Theorem 2.3
Let
\(\alpha\neq\frac{1}{2}\)
be a given real number. The following asymptotic expansion holds:
with the coefficients
\(\beta_{j}\equiv\beta_{j}(\alpha)\)
given by
where the
\(P_{j}(-1)\)
are given by the recurrence relation
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\alpha}\sim\exp \biggl(1+\frac{\alpha-\frac {1}{2}}{x+\sum_{j=0}^{\infty}\beta_{j}x^{-j}} \biggr),\quad x \to\infty, $$
(2.16)
$$\begin{aligned} \beta_{j}=P_{j+1}(-1),\quad j\in \mathbb{N}_{0}, \end{aligned}$$
(2.17)
$$ P_{0}(-1)=1,\qquad P_{j}(-1)=\sum _{k=1}^{j}(-1)^{k-1}\frac{2 ((\alpha-1)k+2\alpha-1 )}{(2\alpha-1)(k+1)(k+2)}P_{j-k}(-1),\quad j\in\mathbb{N}. $$
(2.18)
Remark 2.3
The first few coefficients \(\beta_{j}\equiv\beta_{j}(\alpha)\) are
$$\begin{aligned}& \beta_{0}=\frac{3\alpha-2}{3(2\alpha-1)}, \\& \beta_{1}=-\frac{6\alpha^{2}-6\alpha+1}{18(2\alpha-1)^{2}}, \\& \beta_{2}=\frac{90\alpha^{3}-132\alpha^{2}+57\alpha-8}{270(2\alpha-1)^{3}}, \\& \beta_{3}=-\frac{684\alpha^{4}-1{,}332\alpha^{3}+918\alpha^{2}-276\alpha +31}{1{,}620(2\alpha-1)^{4}}. \end{aligned}$$
In particular, setting \(\alpha=0\) and \(\alpha=1\) in (2.16), respectively, we have the following complete asymptotic expansions:
and
as \(x\to\infty\).
$$ \biggl(1+\frac{1}{x} \biggr)^{x}\sim\exp \biggl(1- \frac{\frac{1}{2}}{x+\frac {2}{3}-\frac{1}{18x}+\frac{4}{135x^{2}}-\frac{31}{1{,}620x^{3}}+\cdots} \biggr) $$
(2.19)
$$ \biggl(1+\frac{1}{x} \biggr)^{x+1}\sim\exp \biggl(1+\frac{\frac {1}{2}}{x+\frac{1}{3}-\frac{1}{18x}+\frac{7}{270x^{2}}-\frac {5}{324x^{3}}+\cdots} \biggr) $$
(2.20)
Theorem 2.4
The following asymptotic expansion holds:
with the coefficients
\(c_{j}\)
given by
where the
\(P_{j}\)
are given by the recurrence relation
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}\sim\exp \biggl(1+\frac{\frac {1}{12}}{x^{2}+x+\frac{1}{10}+\sum_{j=1}^{\infty}c_{j}x^{-j}} \biggr),\quad x \to\infty, $$
(2.21)
$$\begin{aligned} c_{j}=P_{j+2},\quad j\in\mathbb{N}, \end{aligned}$$
(2.22)
$$ P_{0}=1,\qquad P_{j}=\sum _{k=1}^{j}(-1)^{k-1} \frac{6(k+1)}{(k+2)(k+3)}P_{j-k}, \quad j\in\mathbb{N}. $$
(2.23)
Remark 2.4
Here, from (2.21), we give the following explicit asymptotic expansion:
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}\sim\exp \biggl(1+\frac{\frac {1}{12}}{x^{2}+x+\frac{1}{10}-\frac{3}{700x^{2}}+\frac{3}{700x^{3}}-\frac {79}{21{,}000x^{4}}+\cdots} \biggr),\quad x \to\infty. $$
(2.24)
Here, we show the superiority of our new approximation formulas over Hu and Mortici’s approximation formulas.
It follows from (2.8) and (2.19) that
and
$$ \biggl(1+\frac{1}{x} \biggr)^{x}\approx\exp \biggl(1-\frac{\frac{1}{2}x}{ (x+\frac{1}{3}-\frac{1}{12x}+\frac{23}{540x^{2}}-\frac{353}{12{,}960x^{3}} )^{2}} \biggr):=f_{1}(x) $$
(2.25)
$$ \biggl(1+\frac{1}{x} \biggr)^{x}\approx\exp \biggl(1-\frac{\frac {1}{2}}{x+\frac{2}{3}-\frac{1}{18x}+\frac{4}{135x^{2}}-\frac {31}{1{,}620x^{3}}} \biggr):=g_{1}(x). $$
(2.26)
Moreover, we find by the Maple software that, as \(x \rightarrow\infty\),
$$ \biggl(1+\frac{1}{x} \biggr)^{x}=f_{1}(x)+O \bigl(x^{-6} \bigr) \quad\mbox{and} \quad\biggl(1+\frac{1}{x} \biggr)^{x}=g_{1}(x)+O \bigl(x^{-6} \bigr). $$
We now prove that
$$\begin{aligned} f_{1}(x)< g_{1}(x)< \biggl(1+ \frac{1}{x} \biggr)^{x},\quad x\geq1. \end{aligned}$$
(2.27)
Elementary calculation shows that
with
Hence, we have \(f_{1}(x)< g_{1}(x)\) for \(x\geq1\).
$$\begin{aligned} \frac{f_{1}(x)}{g_{1}(x)}=\exp \biggl(-\frac{810x^{3}(4{,}242{,}240x^{3}-1{,}067{,}184x^{2}+389{,}712x-124{,}609)}{A(x)} \biggr) \end{aligned}$$
$$ \begin{aligned}[b] A(x)={}& \bigl(12{,}960x^{4}+4{,}320x^{3}-1{,}080x^{2}+552x-353 \bigr)^{2} \\ &{}\times\bigl(1{,}620x^{4}+1{,}080x^{3}-90x^{2}+48x-31 \bigr). \end{aligned} $$
The second inequality in (2.27) is obtained by considering the function \(F(x)\) defined for \(x\geq1\) by
Differentiation yields
Hence, \(F(x)\) is strictly decreasing for \(x\geq1\), and we have
This means that the second inequality in (2.27) holds for \(x\geq1\).
$$\begin{aligned} F(x)&=\ln \biggl(1+\frac{1}{x} \biggr)-\frac{1}{x}\ln g_{1}(x) \\ &=\ln \biggl(1+\frac{1}{x} \biggr)-\frac{1}{x} \biggl(1- \frac{\frac {1}{2}}{x+\frac{2}{3}-\frac{1}{18x}+\frac{4}{135x^{2}}-\frac {31}{1{,}620x^{3}}} \biggr). \end{aligned}$$
$$\begin{aligned} F'(x)=-\frac{119{,}951+364{,}668(x-1)+369{,}576(x-1)^{2}+125{,}820(x-1)^{3}}{x^{2}(x+1)(1{,}620x^{4}+1{,}080x^{3}-90x^{2}+48x-31)^{2}}< 0, \quad x\geq1. \end{aligned}$$
$$\begin{aligned} F(x)>\lim_{t\to\infty}F(t)=0,\quad x\geq1. \end{aligned}$$
It follows from (2.9) and (2.20) that
and
$$ \biggl(1+\frac{1}{x} \biggr)^{x+1}\approx\exp \biggl(1+\frac{\frac {1}{2}x}{ (x+\frac{1}{6}-\frac{1}{24x}+\frac{43}{2{,}160x^{2}}-\frac {617}{51{,}840x^{3}} )^{2}} \biggr):=f_{2}(x) $$
(2.28)
$$ \biggl(1+\frac{1}{x} \biggr)^{x+1}\approx\exp \biggl(1+\frac{\frac {1}{2}}{x+\frac{1}{3}-\frac{1}{18x}+\frac{7}{270x^{2}}-\frac {5}{324x^{3}}} \biggr):=g_{2}(x). $$
(2.29)
Moreover, we find by the Maple software that, as \(x \rightarrow\infty\),
$$ \begin{aligned} \biggl(1+\frac{1}{x} \biggr)^{x+1}=f_{2}(x)+O \bigl(x^{-6} \bigr) \quad\mbox{and}\quad \biggl(1+\frac{1}{x} \biggr)^{x+1}=g_{2}(x)+O \bigl(x^{-6} \bigr). \end{aligned} $$
In fact, the following double inequality holds:
As the proofs of the inequalities (2.30) are similar to the proofs of the inequalities (2.27), we omit them. The double inequality (2.30) shows that (2.29) is better than (2.28).
$$\begin{aligned} \biggl(1+\frac{1}{x} \biggr)^{x+1}< g_{2}(x)<f_{2}(x),\quad x\geq1. \end{aligned}$$
(2.30)
It follows from (1.10) and (2.24) that
and
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}\approx\exp \biggl(1+\frac{\frac {1}{12}x}{ (x+\frac{1}{3}-\frac{7}{90x}+\frac{16}{405x^{2}}-\frac {2{,}141}{85{,}050x^{3}} )^{3}} \biggr):=f_{3}(x) $$
(2.31)
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}\approx\exp \biggl(1+\frac{\frac {1}{12}}{x^{2}+x+\frac{1}{10}-\frac{3}{700x^{2}}} \biggr):=g_{3}(x). $$
(2.32)
Moreover, we find by the Maple software that, as \(x \rightarrow\infty\),
$$ \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}=f_{3}(x)+O \bigl(x^{-7} \bigr) \quad\mbox{and}\quad \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}=g_{3}(x)+O \bigl(x^{-7} \bigr). $$
In fact, the following double inequality holds:
As the proofs of the inequalities (2.33) are similar to the proofs of the inequalities (2.27), we omit them. The double inequality (2.33) shows that (2.32) is better than (2.31).
$$\begin{aligned} \biggl(1+\frac{1}{x} \biggr)^{x+\frac{1}{2}}< g_{3}(x)<f_{3}(x),\quad x\geq1. \end{aligned}$$
(2.33)
3 Inequalities
Theorem 3.1
(i) The following inequality holds:
The first inequality in (3.1) is valid provided
\(x\geq1\), while the second inequality in (3.1) holds for
\(x>0\).
$$ \exp \biggl(1-\frac{\frac{1}{2}}{x+\frac{2}{3}-\frac{1}{18x}} \biggr)< \biggl(1+ \frac{1}{x} \biggr)^{x}<\exp \biggl(1-\frac{\frac{1}{2}}{x+\frac {2}{3}-\frac{1}{18x}+\frac{4}{135x^{2}}} \biggr). $$
(3.1)
(ii) The following inequality holds:
The first inequality in (3.2) holds for
\(x>0\), while the second inequality in (3.2) is valid provided
\(x\geq1\).
$$ \exp \biggl(1+\frac{\frac{1}{2}}{x+\frac{1}{3}-\frac{1}{18x}+\frac {7}{270x^{2}}} \biggr)< \biggl(1+ \frac{1}{x} \biggr)^{x+1}<\exp \biggl(1+\frac {\frac{1}{2}}{x+\frac{1}{3}-\frac{1}{18x}} \biggr). $$
(3.2)
(iii) The following inequality holds:
The first inequality in (3.3) holds for
\(x>0\), while the second inequality in (3.3) is valid provided
\(x\geq1\).
$$ \exp \biggl(1+\frac{\frac{1}{12}}{x^{2}+x+\frac{1}{10}} \biggr) < \biggl(1+ \frac {1}{x} \biggr)^{x+\frac{1}{2}}<\exp \biggl(1+\frac{\frac {1}{12}}{x^{2}+x+\frac{1}{10}-\frac{3}{700x^{2}}} \biggr). $$
(3.3)
Proof
In order to prove the inequalities (3.1), (3.2) and (3.3), it suffices to show that
and
Differentiation yields
We then obtain
The proof of Theorem 3.1 is complete. □
$$\begin{aligned}& L_{1}(x):=\ln \biggl(1+\frac{1}{x} \biggr)-\frac{1}{x} \biggl(1-\frac{\frac {1}{2}}{x+\frac{2}{3}-\frac{1}{18x}} \biggr)>0,\quad x\geq1, \\& U_{1}(x):=\frac{1}{x} \biggl(1-\frac{\frac{1}{2}}{x+\frac{2}{3}-\frac {1}{18x}+\frac{4}{135x^{2}}} \biggr)-\ln \biggl(1+\frac{1}{x} \biggr)>0,\quad x>0, \\& L_{2}(x):=\ln \biggl(1+\frac{1}{x} \biggr)-\frac{1}{x+1} \biggl(1+\frac{\frac {1}{2}}{x+\frac{1}{3}-\frac{1}{18x}+\frac{7}{270x^{2}}} \biggr)>0,\quad x>0, \\& U_{2}(x):=\frac{1}{x+1} \biggl(1+\frac{\frac{1}{2}}{x+\frac{1}{3}-\frac {1}{18x}} \biggr)-\ln \biggl(1+\frac{1}{x} \biggr)>0,\quad x\geq1, \\& L_{3}(x):=\ln \biggl(1+\frac{1}{x} \biggr)- \biggl(x+ \frac{1}{2} \biggr)^{-1} \biggl(1+\frac{\frac{1}{12}}{x^{2}+x+\frac{1}{10}} \biggr)>0,\quad x>0, \end{aligned}$$
$$ U_{3}(x):= \biggl(x+\frac{1}{2} \biggr)^{-1} \biggl(1+ \frac{\frac {1}{12}}{x^{2}+x+\frac{1}{10}-\frac{3}{700x^{2}}} \biggr)-\ln \biggl(1+\frac {1}{x} \biggr)>0,\quad x\geq1. $$
$$\begin{aligned}& L'_{1}(x)=-\frac{24x-1}{x^{2}(x+1)(18x^{2}+12x-1)^{2}}< 0,\quad x\geq1, \\& U'_{1}(x)=-\frac{4{,}185 (x-\frac{8}{279} )^{2}+\frac {5{,}632}{93}}{x^{2}(x+1)(270x^{3}+180x^{2}-15x+8)^{2}}<0,\quad x>0, \\& L'_{2}(x)=-\frac{3{,}375 (x-\frac{7}{225} )^{2}+\frac {686}{15}}{x(x+1)^{2}(270x^{3}+90x^{2}-15x+7)^{2}}<0,\quad x>0, \\& U'_{2}(x)=-\frac{21x-1}{x(x+1)^{2}(18x^{2}+6x-1)^{2}}<0,\quad x\geq1, \\& L'_{3}(x)=-\frac{1}{x(x+1)(10x^{2}+10x+1)^{2}(2x+1)^{2}}<0, \quad x>0, \\& U'_{3}(x)=-\frac {6{,}711+19{,}040(x-1)+17{,}920(x-1)^{2}+5{,}600(x-1)^{3}}{x(700x^{4}+700x^{3}+70x^{2}-3)^{2}(2x+1)^{2}(x+1)}<0, \quad x\geq1. \end{aligned}$$
$$\begin{aligned}& L_{1}(x)>\lim_{t\to\infty}L_{1}(t)=0,\quad x\geq1,\qquad U_{1}(x)>\lim_{t\to\infty}U_{1}(t)=0,\quad x>0, \\& L_{2}(x)>\lim_{t\to\infty}L_{2}(t)=0,\quad x>0,\qquad U_{2}(x)>\lim_{t\to\infty}U_{2}(t)=0,\quad x\geq1, \\& L_{3}(x)>\lim_{t\to\infty}L_{3}(t)=0,\quad x>0,\qquad U_{3}(x)>\lim_{t\to\infty}U_{3}(t)=0,\quad x\geq1. \end{aligned}$$
Acknowledgements
The authors thank the referees for helpful comments.
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Authors’ contributions
The authors have made equal contributions to each part of this paper. All authors have read and approved the final manuscript.