# Monotonicity and inequalities for the gamma function

## Abstract

In this paper, by using the monotonicity rule for the ratio of two Laplace transforms, we prove that the function

$$x\mapsto \frac{1}{24x ( \ln \Gamma ( x+1/2 ) -x\ln x+x- \ln \sqrt{2\pi } ) +1}-\frac{120}{7}x^{2}$$

is strictly increasing from $$( 0,\infty )$$ onto $$( 1,1860/343 )$$. This not only yields some known and new inequalities for the gamma function, but also gives some completely monotonic functions related to the gamma function.

## Introduction

Stirling’s formula

$$n!\thicksim \sqrt{2\pi n}n^{n}e^{-n}$$
(1.1)

has important applications in statistical physics, probability theory and number theory. Due to its practical importance, it has attracted much interest of many mathematicians and have motivated a large number of research papers concerning various generalizations and improvements.

Burnside’s formula 

$$n!\thicksim \sqrt{2\pi } \biggl( \frac{n+1/2}{e} \biggr) ^{n+1/2}:=b _{n}$$
(1.2)

slightly improves (1.1). Gosper  replaced $$\sqrt{2\pi n}$$ by $$\sqrt{2\pi ( n+1/6 ) }$$ in (1.1) to get

$$n!\thicksim \sqrt{2\pi \biggl( n+\frac{1}{6} \biggr) } \biggl( \frac{n}{e} \biggr) ^{n}:=g_{n},$$
(1.3)

which is better than (1.1) and (1.2). Batir  obtained an asymptotic formula similar to (1.3):

$$n!\thicksim \frac{n^{n+1}e^{-n}\sqrt{2\pi }}{\sqrt{n-1/6}}:=b_{n} ^{\prime },$$
(1.4)

which is stronger than (1.1) and (1.2). A more accurate approximation for the factorial function

$$n!\thicksim \sqrt{2\pi } \biggl( \frac{n^{2}+n+1/6}{e^{2}} \biggr) ^{n/2+1/4}:=m_{n}$$
(1.5)

was presented in  by Mortici.

The gamma function $$\Gamma ( x ) =\int_{0}^{\infty }t^{x-1}e ^{-t}\,dt$$ for $$x>0$$ is closely related to Stirling’s formula since $$\Gamma (n+1)=n!$$ for all $$n\in \mathbb{N}$$. This inspires some authors to also pay attention to finding various better approximations for the gamma function. Here we list some more accurate approximations:

1. (i)

Ramanujan’s [5, p. 339] approximation formula as $$x\rightarrow \infty$$

$$\Gamma ( x+1 ) \thicksim \sqrt{\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 8x^{3}+4x^{2}+x+\frac{1}{30} \biggr) ^{1/6}:=R ( x ) ;$$
(1.6)
2. (ii)

Windschitl’s (see [6, Eq. (42)]) approximation formula

$$\Gamma ( x+1 ) \thicksim \sqrt{2\pi x} \biggl( \frac{x}{e} \biggr) ^{x} \biggl( x\sinh \frac{1}{x} \biggr) ^{x/2}:=W ( x ) ;$$
(1.7)
3. (iii)

Smith’s [6, Eq. (42)] approximation formula

$$\Gamma \biggl( x+\frac{1}{2} \biggr) \thicksim \sqrt{2\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2}:=S ( x ) ;$$
(1.8)
4. (iv)

Nemes’ formula ([7, Corollary 4.1]) states that

$$\Gamma ( x+1 ) \thicksim \sqrt{2\pi x} \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 1+\frac{1}{12x^{2}-1/10} \biggr) ^{x}:=N ( x ) ;$$
(1.9)
5. (v)

Chen’s  presented a new approximation

$$\Gamma (x+1)\thicksim \sqrt{2\pi x} \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 1+\frac{1}{12x^{3}+24x/7-1/2} \biggr) ^{x^{2}+53/210}=C ( x ) .$$
(1.10)

### Remark 1

Let $$A ( x )$$ be an approximation for $$\Gamma ( x+1 )$$ as $$x\rightarrow \infty$$. If there is $$m>0$$ such that

$$\lim_{x\rightarrow \infty }\frac{\ln \Gamma ( x+1 ) -\ln A ( x ) }{x^{-m}}=c\neq 0,\pm \infty ,$$
(1.11)

then we say that the rate of $$A ( x )$$ converging to $$\Gamma ( x+1 )$$ is like $$x^{-m}$$ as $$x\rightarrow \infty$$. Evidently, the larger m is, the higher the accuracy of $$A ( x )$$ approximating for $$\Gamma ( x+1 )$$ is. Since $$( x-1 ) /\ln x\rightarrow 1$$ as $$x\rightarrow 1$$, the limit relation can be equivalently written as

$$\lim_{x\rightarrow \infty }\frac{\Gamma ( x+1 ) /A ( x ) -1}{x^{-m}}=c\neq 0,\pm \infty ,$$

or

$$\frac{\Gamma ( x+1 ) }{A ( x ) }=1+O \bigl( x^{-m} \bigr) \quad \mbox{as }x \rightarrow\infty .$$

### Remark 2

It is easy to check that as $$n\rightarrow \infty$$ or $$x\rightarrow \infty$$,

\begin{aligned}& n! = \frac{n^{n+1}e^{-n}\sqrt{2\pi }}{\sqrt{n-1/6}} \bigl( 1+O \bigl( n^{-2} \bigr) \bigr) , \\& n! = \sqrt{2\pi } \biggl( \frac{n^{2}+n+1/6}{e^{2}} \biggr) ^{n/2+1/4} \bigl( 1+O \bigl( n^{-3} \bigr) \bigr) , \\& \Gamma \biggl( x+\frac{1}{2} \biggr) = \sqrt{2\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2} \bigl( 1+O \bigl( x^{-5} \bigr) \bigr) . \end{aligned}

These together with those shown in [8, (3.5)-(3.10)] indicate that Chen’s one $$C ( x )$$ is the best among approximation formulas listed above.

More results involving the approximation formulas for the factorial or gamma function can be found in  and the references cited therein.

It is worth mentioning that Yang and Chu  proposed a new approach to construct asymptotic formulas by bivariate means. As applications, they offered in [9, Propositions 4 and 5] two asymptotic formulas: as $$x\rightarrow \infty$$,

\begin{aligned}& \Gamma (x+1) \thicksim \sqrt{2\pi } \biggl( \frac{x+1/2}{e} \biggr) ^{x+1/2}\exp \biggl( -\frac{1}{24}\frac{x+1/2}{x^{2}+x+37/120} \biggr) :=Y_{1} ( x ) , \\& \Gamma (x+1) \thicksim \sqrt{2\pi } \biggl( \frac{x+1/2}{e} \biggr) ^{x+1/2}\exp \biggl( -\frac{1517}{44\text{,}640}\frac{1}{x+1/2}- \frac{343}{44 \text{,}640}\frac{x+1/2}{x^{2}+x+111/196} \biggr) \\& \hphantom{\Gamma (x+1) }{}:=Y_{2} ( x ) , \end{aligned}

which satisfy

$$\Gamma ( x+1 ) =Y_{1} ( x ) \bigl( 1+O \bigl( x^{-5} \bigr) \bigr) \quad \mbox{and}\quad \Gamma ( x+1 ) =Y_{2} ( x ) \bigl( 1+O \bigl( x^{-7} \bigr) \bigr) ,$$

and proved that the functions (replace x by $$x-1/2$$)

\begin{aligned}& f_{4} \biggl(x-\frac{1}{2} \biggr) = \ln \Gamma \biggl(x+ \frac{1}{2} \biggr)-\frac{1}{2} \ln 2\pi -x\ln x+x+ \frac{1}{24} \frac{x}{x^{2}+7/120}, \\& f_{5} \biggl(x-\frac{1}{2} \biggr) = \ln \Gamma \biggl(x+ \frac{1}{2} \biggr)-\frac{1}{2} \ln 2\pi -x\ln x+\frac{1}{1440} \frac{5880x^{2}+1517}{x ( 98x^{2}+31 ) } \end{aligned}

are increasingly concave and decreasingly convex on $$( 0,\infty )$$, respectively. Clearly, both $$Y_{1} ( x )$$ and $$Y_{2} ( x )$$ are accurate and simpler approximation formulas for the gamma function.

According to these inequalities given in [9, Corollary 7], it is natural to ask: What are the best α and β such that the double inequality

$$\exp \biggl[ -\frac{1}{24x}\frac{120x^{2}+7 ( \alpha -1 ) }{120x ^{2}+7\alpha } \biggr] < \frac{\Gamma ( x+1/2 ) }{\sqrt{2 \pi } ( x/e ) ^{x}}< \exp \biggl[ -\frac{1}{24x}\frac{120x^{2}+7 ( \beta -1 ) }{120x^{2}+7\beta } \biggr]$$
(1.12)

holds for all $$x>0$$? This problem is equivalent to determining the monotonicity of the function

$$f ( x ) =\frac{1}{24x ( \ln \Gamma ( x+1/2 ) -x \ln x+x-\ln \sqrt{2\pi } ) +1}-\frac{120}{7}x^{2}$$
(1.13)

on $$( 0,\infty )$$.

The aim of this paper is to answer this problem. Our main result is the following theorem.

### Theorem 1

The function f defined by (1.13) is strictly increasing from $$( 0,\infty )$$ onto $$( 1,1860/343 )$$.

As a consequence of the above theorem, the following corollary is immediate.

### Corollary 1

For $$x>x_{0}\geq 0$$, the double inequality (1.12) holds if and only if $$\alpha \geq f ( \infty ) =1860/343$$ and $$1\leq \beta \leq f ( x_{0} )$$. In particular, we have

$$\exp \biggl[ -\frac{1}{1440}\frac{5880x^{2}+1517}{x ( 98x^{2}+31 ) } \biggr] < \frac{\Gamma ( x+1/2 ) }{\sqrt{2 \pi } ( x/e ) ^{x}}< \exp \biggl( -\frac{5x}{120x^{2}+7} \biggr)$$

holds for $$x>0$$.

Replacing x by $$n+1/2$$, then putting $$x_{0}=1$$ in Corollary 1, and noting that

$$\beta_{1}:=f \biggl( \frac{3}{2} \biggr) =\frac{1}{36\ln 2-54\ln 3-18 \ln \pi +55}- \frac{270}{7}\approx 4.7243,$$
(1.14)

we deduce the following statement.

### Corollary 2

The double inequality

\begin{aligned}& \exp \biggl[ -\frac{1}{24 ( n+1/2 ) }\frac{120 ( n+1/2 ) ^{2}+7 ( \alpha -1 ) }{120 ( n+1/2 ) ^{2}+7\alpha } \biggr] \\& \quad < \frac{n!}{\sqrt{2\pi } ( ( n+1/2 ) /e ) ^{n+1/2}} \\& \quad < \exp \biggl[ -\frac{1}{24 ( n+1/2 ) }\frac{120 ( n+1/2 ) ^{2}+7 ( \beta_{1}-1 ) }{120 ( n+1/2 ) ^{2}+7\beta_{1}} \biggr] \end{aligned}

holds with the best constants $$\beta_{1}\approx 4.7243$$ given by (1.14) and $$\alpha =1860/343\approx 5.4227$$.

## Tools

To prove our main result, we need some lemmas as tools. The first lemma is the convolution formula of the Laplace transform.

### Lemma 1

()

Let $$f_{i} ( t )$$ for $$i=1,2$$ be piecewise continuous in arbitrary finite intervals included on $$( 0,\infty )$$. If there exist some constants $$M_{i}>0$$ and $$c_{i}\geq 0$$ such that $$\vert f_{i} ( t ) \vert \leq M_{i}e^{c_{i}t}$$ for $$i=1,2$$, then

$$\int_{0}^{\infty }f_{1} ( u ) e^{-su} \,du \int_{0}^{\infty }f _{2} ( v ) e^{-sv} \,dv= \int_{0}^{\infty } \biggl( \int_{0}^{t}f _{1} ( u ) f_{2} ( t-u )\,du \biggr) e^{-st}\,dt.$$
(2.1)

The second one is a special monotonicity rule for the ratio of two power series, which first appeared in [30, Lemma 6.4] and was proved in , also see .

### Lemma 2

([31, Corollary 2.3])

Let $$A ( t ) =\sum_{k=0}^{\infty }a_{k}t^{k}$$ and $$B ( t ) =\sum_{k=0}^{\infty }b_{k}t^{k}$$ be two real power series converging on $$\mathbb{R}$$ with $$b_{k}>0$$ for all k. If, for certain $$m\in \mathbb{N}$$, the sequence $$\{a_{k}/b_{k}\}$$ is increasing (decreasing) for $$0\leq k\leq m$$ and decreasing (increasing) for $$k\geq m$$, then there is a unique $$t_{0}\in ( 0,\infty )$$ such that the function $$A/B$$ is increasing (decreasing) on $$( 0,t _{0} )$$ and decreasing (increasing) on $$( t_{0},\infty )$$.

The third lemma is called L’Hospital piecewise monotonicity rule .

### Lemma 3

([33, Theorem 8])

Let $$-\infty \leq a< b\leq \infty$$. Suppose that (i) f and g are differentiable functions on $$(a,b)$$; (ii) $$g^{\prime }\neq 0$$ on $$(a,b)$$; (iii) $$f(a^{+})=g(a^{+})=0$$; (iv) there is $$c\in ( a,b )$$ such that $$f^{\prime }/g^{\prime }$$ is increasing (decreasing) on $$( a,c )$$ and decreasing (increasing) on $$( c,b )$$. Then

1. (i)

when $$\operatorname{sgn}g^{\prime }\operatorname{sgn}H_{f,g} ( b^{-} ) \geq ( \leq ) 0$$, $$f/g$$ is increasing (decreasing) on $$( a,b )$$, where $$H_{f,g}= ( f^{\prime }/g^{\prime } ) g-f$$;

2. (ii)

when $$\operatorname{sgn}g^{\prime }\operatorname{sgn}H_{f,g} ( b^{-} ) < ( > ) 0$$, there is a unique number $$x_{a}\in ( a,b )$$ such that $$f/g$$ is increasing (decreasing) on $$( a,x _{a} )$$ and decreasing (increasing) on $$( x_{a},b )$$.

The last one gives a monotonicity rule for the ratio of two Laplace transforms, which is crucial to proving our main result (see [34, Remark 3]).

### Lemma 4

Let the functions A, B be defined on $$( 0,\infty )$$ such that their Laplace transforms exist with $$B ( t ) \neq 0$$ for all $$t>0$$. Then the function

$$x\mapsto U ( x ) =\frac{\int_{0}^{\infty }A ( t ) e ^{-xt}\,dt}{\int_{0}^{\infty }B ( t ) e^{-xt}\,dt}$$

is decreasing (increasing) on $$( 0,\infty )$$ if $$A/B$$ is increasing (decreasing) on $$( 0,\infty )$$.

### Proof

Differentiation yields

\begin{aligned}& \biggl( \int_{0}^{\infty }B ( t ) e^{-xt}\,dt \biggr) ^{2}U ^{\prime } ( x ) \\& \quad = - \int_{0}^{\infty }A ( t ) e^{-xt}\,dt \int_{0}^{\infty }B ( t ) e^{-xt}\,dt+ \int_{0}^{\infty }A ( t ) e^{-xt}\,dt \int_{0}^{\infty }tB ( t ) e^{-xt}\,dt \\& \quad = \int_{0}^{\infty } \int_{0}^{\infty }t \biggl[ \frac{A ( s ) }{B ( s ) }- \frac{A ( t ) }{B ( t ) } \biggr] B ( t ) B ( s ) e^{-xt-xs}\,ds\,dt:=D. \end{aligned}

Exchanging the integral variables s and t, we have

$$D= \int_{0}^{\infty } \int_{0}^{\infty }s \biggl[ \frac{A ( t ) }{B ( t ) }- \frac{A ( s ) }{B ( s ) } \biggr] B ( s ) B ( t ) e^{-xs-xt}\,dt\,ds,$$

$$2D=- \int_{0}^{\infty } \int_{0}^{\infty } [ t-s ] \biggl[ \frac{A ( t ) }{B ( t ) }- \frac{A ( s ) }{B ( s ) } \biggr] B ( s ) B ( t ) e^{-xt-xs}\,dt\,ds.$$

By the assumptions, the desired assertions follow. □

## Proof of Theorem 1

Before proving Theorem 1, we also need several concrete lemmas.

### Lemma 5

([28, Lemma 4])

Let $$g_{0}$$ be defined on $$( 0,\infty )$$ by

$$g_{0} ( x ) =\ln \Gamma \biggl( x+\frac{1}{2} \biggr) -x\ln x+x- \frac{1}{2}\ln ( 2\pi ) .$$
(3.1)

Then $$g_{0} ( x )$$ has the following integral representation:

$$g_{0} ( x ) =- \int_{0}^{\infty }h ( t ) e^{-xt}\,dt,$$
(3.2)

where

$$h ( t ) =\frac{1}{t^{2}}- \frac{1}{2t\sinh ( t/2 ) }.$$
(3.3)

### Lemma 6

Let $$h ( t )$$ be defined on $$( 0,\infty )$$ by (3.3). Then we have

\begin{aligned}& x \int_{0}^{\infty }h ( t ) e^{-xt}\,dt = \frac{1}{24}+ \int _{0}^{\infty }h^{\prime } ( t ) e^{-xt} \,dt, \end{aligned}
(3.4)
\begin{aligned}& x \int_{0}^{\infty }h^{\prime } ( t ) e^{-xt} \,dt = \int_{0} ^{\infty }h^{\prime \prime } ( t ) e^{-xt} \,dt, \end{aligned}
(3.5)
\begin{aligned}& x \int_{0}^{\infty }h^{\prime \prime } ( t ) e^{-xt} \,dt = - \frac{7}{2880}+ \int_{0}^{\infty }h^{\prime \prime \prime } ( t ) e^{-xt} \,dt. \end{aligned}
(3.6)

### Proof

Integration by parts yields

$$x \int_{0}^{\infty }h ( t ) e^{-xt}\,dt=- \int_{0}^{\infty }h ( t )\,de^{-xt}=- \bigl[ h ( t ) e^{-xt} \bigr] _{t=0} ^{t=\infty }+ \int_{0}^{\infty }h^{\prime } ( t ) e^{-xt} \,dt,$$

which, by a simple computation,

\begin{aligned}& \lim_{t\rightarrow 0}h ( t ) e^{-xt} = \lim_{t\rightarrow 0} \biggl( \frac{1}{t^{2}}-\frac{1}{2t\sinh ( t/2 ) } \biggr) e ^{-xt}= \frac{1}{24}, \\& \lim_{t\rightarrow \infty }h ( t ) e^{-xt} = \lim_{t\rightarrow \infty } \biggl( \frac{1}{t^{2}}-\frac{1}{2t\sinh ( t/2 ) } \biggr) e^{-xt}=0, \end{aligned}

gives (3.4).

Similarly, integration by parts and limit relations $$\lim_{t\rightarrow 0}h^{\prime } ( t ) e^{-xt}=0$$ and $$\lim_{t\rightarrow \infty }h^{\prime } ( t ) e^{-xt}=0$$ yield (3.5). Integration by parts in combination with $$\lim_{t\rightarrow 0}h^{ \prime \prime } ( t ) e^{-xt}=-7/2880$$ and $$\lim_{t\rightarrow \infty }h^{\prime \prime } ( t ) e^{-xt}=0$$ gives (3.6). □

### Lemma 7

Let $$h ( t )$$ be defined by (3.3). Then (i) $$h^{\prime } ( t ) <0$$ for $$t>0$$; (ii) there is $$t_{0}>0$$ such that the function $$h^{\prime \prime \prime }/h^{\prime }$$ is increasing on $$( 0,t_{0} )$$ and decreasing on $$( t_{0},\infty )$$. Therefore, we have

$$-\frac{31}{98}< \frac{h^{\prime \prime \prime } ( t ) }{h^{ \prime } ( t ) }< \lambda_{0}\approx 0.051704,$$

where $$\lambda_{0}=h^{\prime \prime \prime } ( t_{0} ) /h ^{\prime } ( t_{0} )$$, here $$t_{0}$$ is the unique solution of the equation $$[ h^{\prime \prime \prime } ( t ) /h^{ \prime } ( t ) ] ^{\prime }=0$$ on $$(0,\infty )$$.

### Proof

Differentiation yields

\begin{aligned}& h^{\prime } ( t ) = \frac{1}{4}\frac{2\sinh ( t/2 ) +t\cosh ( t/2 ) }{t^{2}\sinh^{2} ( t/2 ) }- \frac{2}{t ^{3}}, \\& h^{\prime \prime } ( t ) = \frac{6}{t^{4}}-\frac{1}{16} \frac{t ^{2}\cosh t+8\cosh t+4t\sinh t+3t^{2}-8}{t^{3}\sinh^{3} ( t/2 ) }, \\& h^{\prime \prime \prime } ( t ) = -\frac{24}{t^{5}}+\frac{1}{64t ^{4}\sinh^{4} ( t/2 ) } \biggl( 6t^{2}\sinh \frac{3t}{2}+48 \sinh \frac{3t}{2}+t^{3} \cosh \frac{3t}{2}+24t\cosh \frac{3t}{2} \\& \hphantom{h^{\prime \prime \prime } ( t ) = }{}+23t^{3}\cosh \frac{t}{2}-24t\cosh \frac{t}{2}+30t^{2}\sinh \frac{t}{2}-144\sinh \frac{t}{2} \biggr) . \end{aligned}

Simplifying and expanding in power series yield

\begin{aligned} - \biggl( 4t^{3}\sinh^{2}\frac{t}{2} \biggr) h^{\prime } ( t ) =&4 \cosh t-t^{2}\cosh \frac{t}{2}-2t \sinh \frac{t}{2}-4 \\ =&\sum_{n=3}^{\infty }\frac{2^{2n-2}-n^{2}}{2^{2n-4} ( 2n ) !}t^{2n}>0, \end{aligned}

which proves $$h^{\prime } ( t ) <0$$ for $$t>0$$.

Then $$h^{\prime \prime \prime } ( t ) /h^{\prime } ( t )$$ can be expressed as

\begin{aligned}& \frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) } = \frac{96\sinh^{4}s-3s^{3}\sinh 3s-6s\sinh 3s-s^{4} \cosh 3s-6s^{2}\cosh 3s}{16s^{2} ( 2\sinh^{2}s-s\sinh s-s^{2} \cosh s ) \sinh^{2}s} \\& \hphantom{\frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) } =}{}+\frac{-15s^{3}\sinh s+18s\sinh s-23s^{4}\cosh s+6s^{2}\cosh s}{16s ^{2} ( 2\sinh^{2}s-s\sinh s-s^{2}\cosh s ) \sinh^{2}s}:=\frac{h _{1} ( s ) }{h_{2} ( s ) }, \end{aligned}

where $$s=2t$$. Using ‘product into sum’ formula for hyperbolic functions and expanding in power series give

\begin{aligned} h_{1} ( s ) &:=12\cosh 4s-48\cosh 2s-3s^{3}\sinh 3s-6s \sinh 3s-s^{4}\cosh 3s-6s^{2}\cosh 3s \\ &\quad {}-15s^{3}\sinh s+18s\sinh s-23s^{4}\cosh s+6s^{2}\cosh s+36 \\ &=12\sum_{n=0}^{\infty }\frac{4^{2n}}{ ( 2n ) !}s^{2n}-48 \sum_{n=0}^{\infty }\frac{2^{2n}}{ ( 2n ) !}s^{2n}-3 \sum_{n=2}^{\infty }\frac{3^{2n-3}}{ ( 2n-3 ) !}s^{2n} \\ &\quad {}-6\sum_{n=1}^{\infty } \frac{3^{2n-1}}{ ( 2n-1 ) !}s ^{2n}-\sum_{n=2}^{\infty } \frac{3^{2n-4}}{ ( 2n-4 ) !}s^{2n}-6\sum_{n=1}^{ \infty } \frac{3^{2n-2}}{ ( 2n-2 ) !}s^{2n}-15\sum_{n=2} ^{\infty }\frac{1}{ ( 2n-3 ) !}s^{2n} \\ &\quad {}+18\sum_{n=1}^{\infty } \frac{1}{ ( 2n-1 ) !}s^{2n}-23 \sum_{n=2}^{\infty } \frac{1}{ ( 2n-4 ) !}s^{2n}+6 \sum_{n=1}^{\infty } \frac{1}{ ( 2n-2 ) !}s^{2n} \\ &:=4 \sum_{n=2}^{\infty }\frac{a_{n}}{ ( 2n ) !}s^{2n}, \end{aligned}

where

\begin{aligned}& a_{n} = 3\times 4^{2n}-12\times 2^{2n}-2n \bigl( 2n^{3}+3n^{2}+19n+30 \bigr) 3^{2n-4} \\& \hphantom{a_{n} =}{}-2n ( 2n-3 ) \bigl( 23n^{2}-27n+10 \bigr) , \\& h_{2} ( s ) : =16s^{2} \bigl( 2\sinh^{2}s-s\sinh s-s^{2} \cosh s \bigr) \sinh^{2}s \\& \hphantom{h_{2} ( s ) }{}= 4s^{2} \bigl( \cosh 4s-4\cosh 2s-s\sinh 3s-s^{2}\cosh 3s+s^{2} \cosh s+3s\sinh s+3 \bigr) \\& \hphantom{h_{2} ( s )}=4s^{2} \Biggl( \sum_{n=1}^{\infty } \frac{4^{2n}}{ ( 2n ) !}s^{2n}-4\sum_{n=1}^{\infty } \frac{2^{2n}}{ ( 2n ) !}s^{2n}-\sum_{n=1}^{\infty } \frac{3^{2n-1}}{ ( 2n-1 ) !}s^{2n}-\sum_{n=1}^{\infty } \frac{3^{2n-2}}{ ( 2n-2 ) !}s^{2n} \\& \hphantom{h_{2} ( s ) : =}{}+\sum_{n=1}^{\infty } \frac{1}{ ( 2n-2 ) !}s ^{2n}+3\sum_{n=1}^{\infty } \frac{1}{ ( 2n-1 ) !}s^{2n}+3 \Biggr) :=4\sum_{n=2}^{\infty } \frac{b_{n}}{ ( 2n ) !}s^{2n}, \end{aligned}

where

$$b_{n}=2n ( 2n-1 ) \bigl( 4^{2n-2}-2^{2n}-4n ( n-1 ) 3^{2n-4}+4n ( n-1 ) \bigr) .$$

Thus, if we prove the sequence $$\{a_{n}/b_{n}\}_{n\geq 5}$$ is increasing then decreasing, then by Lemma 2 we deduce that there is $$t_{0}$$ such that $$h^{\prime \prime }/h$$ is increasing on $$( 0,t _{0} )$$ and decreasing on $$( t_{0},\infty )$$, and the proof is done. To this end, if $$b_{n}>0$$ for $$n\geq 5$$, then it suffices to show that there is $$n_{0}>5$$ such that $$d_{n}=a_{n}b_{n+1}-b_{n}a _{n+1}\leq 0$$ for $$5\leq n\leq n_{0}$$ and $$d_{n}\geq 0$$ for $$n\geq n_{0}$$.

Now, it is easy to check that

$$\frac{b_{n+1}}{2 ( n+1 ) ( 2n+1 ) }-16\frac{b_{n}}{2n ( 2n-1 ) }=4n ( 7n-25 ) 3^{2n-4}+12\times 2^{2n}-4n ( 15n-17 ) >0,$$

which together with $$b_{4}=0$$ yields $$b_{n}>0$$ for $$n\geq 5$$. On the other hand, by an elementary computation, we obtain

$$d_{n}=a_{n}b_{n+1}-b_{n}a_{n+1}= \sum_{k=1,2,3,4,6,8,9,12,16}p_{k} ( n ) k^{2n},$$

where

\begin{aligned}& p_{16} ( n ) = 6 ( 4n+1 ) , \\& p_{12} ( n ) = -\frac{1}{324}n \bigl( 28n^{5}+12n^{4}-1181n ^{3}+9678n^{2}+3457n+1830 \bigr) , \\& p_{9} ( n ) = \frac{64}{243}n^{2} ( n+1 ) \bigl( n ^{3}+8n^{2}+20n-2 \bigr) , \\& p_{8} ( n ) = 6 \bigl( 18n^{2}-41n-8 \bigr) , \\& p_{6} ( n ) = -\frac{4}{81}n \bigl( 20n^{5}+132n^{4}+185n ^{3}-678n^{2}-997n-822 \bigr) , \\& p_{4} ( n ) = -\frac{1}{4} \bigl( 1380n^{6}-1804n^{5}+989n ^{4}-3134n^{3}+1327n^{2}-2118n-384 \bigr) , \\& p_{3} ( n ) = \frac{128}{27}n^{2} ( n+1 ) \bigl( 32n ^{5}-32n^{4}-33n^{3}+48n^{2}-50n+8 \bigr), \\& p_{2} ( n ) = 4n \bigl( 276n^{5}-508n^{4}-295n^{3}+106n^{2}+43n-150 \bigr), \\& p_{1} ( n ) = 192n^{2} ( n+1 ) \bigl( 9n^{3}-8n ^{2}+2 \bigr) . \end{aligned}

An easy verification yields

\begin{aligned}& d_{5}=-4\text{,}007\text{,}555\text{,}481\text{,}600, \\& d_{6}=-3\text{,}910\text{,}448\text{,}396\text{,}574\text{,}720, \\& d_{7}=-1\text{,}900\text{,}746\text{,}298\text{,}639\text{,}319\text{,}040, \\& d_{8}=-630\text{,}125\text{,}315\text{,}460\text{,}849 \text{,}991 \text{,}680, \\& d_{9}=-150\text{,}180\text{,}694\text{,}294\text{,}194\text{,}463 \text{,}408\text{,}128, \\& d_{10}=-20\text{,}155\text{,}436 \text{,}802\text{,}005\text{,}011 \text{,}207\text{,}151\text{,}616, \end{aligned}

and $$d_{11}=3\text{,}463\text{,}285\text{,}943\text{,}229\text{,}784\text{,}738\text{,}339\text{,}553\text{,}280>0$$. It remains to show $$d_{n}>0$$ for $$n\geq 11$$. To this end, we write $$d_{n}$$ as

\begin{aligned} d_{n} &= \bigl[ p_{16} ( n ) \times 16^{2n}+p_{12} ( n ) \times 12^{2n} \bigr] \\ &\quad {}+ \bigl[ p_{9} ( n ) \times 9^{2n}+p_{6} ( n ) \times 6^{2n} \bigr] + \bigl[ p_{8} ( n ) \times 8^{2n}+p_{4} ( n ) \times 4^{2n} \bigr] \\ &\quad {}+ \bigl[ p_{3} ( n ) \times 3^{2n}+p_{2} ( n ) \times 2^{2n}+p_{1} ( n ) \bigr] , \end{aligned}

and denote the expressions in the square brackets by $$d_{n}^{\prime }$$, $$d_{n}^{\prime \prime }$$, $$d_{n}^{\prime \prime \prime }$$ and $$d_{n}^{\prime \prime \prime \prime }$$, respectively. We easily get the recurrence relation of $$d_{n}^{\prime }$$

\begin{aligned} &\frac{p_{16} ( n ) d_{n+1}^{\prime }-16^{2}p_{16} ( n+1 ) d_{n}^{\prime }}{12^{2n}} \\ &\quad =144p_{16} ( n ) p_{12} ( n+1 ) -16^{2}p_{16} ( n+1 ) p_{12} ( n ) \\ &\quad =\frac{8}{27} \bigl( 784n^{7}-3724n^{6}-51 \text{,}008n^{5}+328\text{,}397n^{4}+10 \text{,}762n^{3} \\ &\quad \quad {}-1\text{,}037 \text{,}977n^{2}-650\text{,}802n-124\text{,}416 \bigr) \\ &\quad =\frac{8}{27} \bigl( 784m^{7}+23\text{,}716m^{6}+248\text{,}872m^{5}+1\text{,}086\text{,}697m ^{4}+1\text{,}666 \text{,}702m^{3} \\ &\quad \quad {}+1\text{,}160\text{,}503m^{2}+10\text{,}500\text{,}078m+20 \text{,}928\text{,}024 \bigr) >0, \end{aligned}

where $$m=n-5\geq 6$$. This in combination with $$p_{16} ( n ) >0$$ and $$d_{11}^{\prime }=2^{45}\times 71\text{,}481\text{,}197\text{,}516\text{,}733>0$$ leads us to $$d_{n}^{\prime }>0$$ for $$n\geq 11$$.

Similarly, we have

\begin{aligned} &\frac{p_{9} ( n ) d_{n+1}^{\prime \prime }-81p_{9} ( n+1 ) d_{n}^{\prime \prime }}{6^{2n}} \\ &\quad =36p_{9} ( n ) p _{6} ( n+1 ) -81p_{9} ( n+1 ) p_{6} ( n ) \\ &\quad =\frac{256}{2187}n ( n+1 ) ^{2} \bigl( 100n^{9}+1960n^{8}+15 \text{,}413n^{7}+55\text{,}819n^{6}+53\text{,}414n^{5} \\ &\quad\quad {}-273\text{,}428n^{4}-1\text{,}024\text{,}655n^{3}-1\text{,}559\text{,}511n^{2}-1\text{,}278\text{,}612n-399\text{,}492 \bigr) >0 \end{aligned}

for $$n\geq 3$$. This together with $$p_{9} ( n ) >0$$ and $$d_{3}^{\prime \prime }=717\text{,}610\text{,}752>0$$ yields $$d_{n}^{\prime \prime }>0$$ for $$n\geq 3$$.

Also, we get

\begin{aligned} &\frac{p_{8} ( n ) d_{n+1}^{\prime \prime \prime }-64p_{8} ( n+1 ) d_{n}^{\prime \prime \prime }}{4^{2n}} \\ &\quad =16p_{8} ( n ) p_{4} ( n+1 ) -64p_{8} ( n+1 ) p_{4} ( n ) \\ &\quad = 1\text{,}788\text{,}480n^{8}-5\text{,}219\text{,}424n^{7}-367 \text{,}632n^{6}+8\text{,}703 \text{,}096n^{5}+13\text{,}278 \text{,}240n^{4} \\ &\quad \quad {}+9\text{,}974\text{,}760n^{3}-7\text{,}438\text{,}608n^{2}+1\text{,}718 \text{,}592n+423\text{,}936, \end{aligned}

which can be rewritten as

\begin{aligned}& 1\text{,}788\text{,}480m^{8}+23\text{,}396\text{,}256m^{7}+126 \text{,}870\text{,}192m^{6}+367\text{,}098\text{,}936m ^{5}+619\text{,}910\text{,}160m^{4} \\& \quad {}+687\text{,}582\text{,}120m^{3}+676\text{,}606 \text{,}944m^{2}+635\text{,}328\text{,}864m+311\text{,}091\text{,}840>0, \end{aligned}

where $$m=n-2\geq 9$$. This in combination with $$p_{8} ( n ) >0$$ for $$n\geq 3$$ and $$d_{7}^{\prime \prime \prime }=2^{30}\times 6\text{,}089\text{,}535>0$$ indicates that $$d_{n}^{\prime \prime \prime }>0$$ for $$n\geq 7$$.

As far as $$d_{n}^{\prime \prime \prime \prime }>0$$ for $$n\geq 11$$, it is clear, since

\begin{aligned} \frac{27}{128n^{2} ( n+1 ) }p_{3} ( n ) &= \bigl( 32n ^{5}-32n^{4}-33n^{3}+48n^{2}-50n+8 \bigr) \\ &=32m^{5}+288m^{4}+991m^{3}+1642m^{2}+1282m+348>0, \end{aligned}

where $$m=n-2>0$$,

\begin{aligned} \frac{p_{2} ( n ) }{4n} =& \bigl( 276n^{5}-508n^{4}-295n^{3}+106n ^{2}+43n-150 \bigr) \\ =&276m^{5}+3632m^{4}+18\text{,}449m^{3}+44 \text{,}539m^{2}+49\text{,}630m+18\text{,}888>0 \end{aligned}

for $$m=n-3>0$$, $$p_{1} ( n ) =192n^{2} ( n+1 ) ( 9n^{3}-8n^{2}+2 ) >0$$ for $$n\geq 1$$. This proves the piecewise monotonicity of $$h^{\prime \prime \prime }/h^{\prime }$$ on $$( 0, \infty )$$.

It is easy to verify that

$$\lim_{t\rightarrow 0}\frac{h^{\prime \prime \prime } ( t ) }{h ^{\prime } ( t ) }=-\frac{31}{98}\quad \mbox{and} \quad \lim_{t\rightarrow \infty }\frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) }=0.$$

Solving the equation $$[ h^{\prime \prime \prime } ( t ) /h^{\prime } ( t ) ] ^{\prime }=0$$ yields $$t=t_{0} \approx 10.96011$$, which gives $$\lambda_{0}=h^{\prime \prime \prime } ( t_{0} ) / h^{\prime } ( t_{0} ) \approx 0.051704$$.

By the piecewise monotonicity of $$h^{\prime \prime \prime }/h^{\prime }$$ on $$( 0,\infty )$$, we conclude that

$$-\frac{7}{120}=\min \biggl( \lim_{t\rightarrow 0} \frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) }, \lim_{t\rightarrow \infty }\frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) } \biggr) < \frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) }< \frac{h ^{\prime \prime \prime } ( t_{0} ) }{h^{\prime } ( t_{0} ) }=\lambda_{0}\approx 0.051704,$$

which completes the proof. □

We now are in a position to prove Theorem 1.

### Proof of Theorem 1

We first prove that

$$f ( x ) =-\frac{1}{168}\frac{\int_{0}^{\infty } ( 7+2880h ^{\prime \prime } ( t ) ) e^{-xt}\,dt}{\int_{0}^{\infty } ( \int_{0}^{t}h^{\prime } ( s )\,ds ) e^{-xt}\,dt}:=- \frac{1}{168} \frac{\int_{0}^{\infty }A ( t ) e^{-xt}\,dt}{\int _{0}^{\infty }B ( t ) e^{-xt}\,dt},$$
(3.7)

where

$$A ( t ) =7+2880h^{\prime \prime } ( t ) \quad \mbox{and} \quad B ( t ) = \int_{0}^{t}h^{\prime } ( s )\,ds.$$

In fact, by Lemma 5 and identities (3.4) and (3.5), $$f ( x )$$ can be expressed as

\begin{aligned} f ( x ) =&-\frac{1}{24\int_{0}^{\infty }h^{\prime } ( t ) e^{-xt}\,dt}-\frac{120}{7}x^{2} \\ =&-\frac{7+2880x^{2}\int_{0}^{\infty }h^{\prime } ( t ) e ^{-xt}\,dt}{168\int_{0}^{\infty }h^{\prime } ( t ) e^{-xt}\,dt}=- \frac{1}{168}\frac{7/x+2880\int_{0}^{\infty }h^{\prime \prime } ( t ) e^{-xt}\,dt}{ ( 1/x ) \int_{0}^{\infty }h^{\prime } ( t ) e^{-xt}\,dt}. \end{aligned}

Application of the identity

$$\frac{1}{x^{n}}=\frac{1}{\Gamma ( n ) } \int_{0}^{\infty }t ^{n-1}e^{-xt}\,dt \quad \mbox{for }n>0$$

and Lemma 1 give (3.7).

Now, to prove f is strictly increasing on $$( 0,\infty )$$, it suffices to prove $$t\mapsto A ( t ) /B ( t )$$ is increasing on $$( 0,\infty )$$ by Lemma 4. Similar to the proof of Theorem 1, we easily see that

\begin{aligned}& \lim_{t\rightarrow 0}A ( t ) = \lim_{t\rightarrow 0} \bigl( 7+2880h ^{\prime \prime } ( t ) \bigr) =\lim_{t\rightarrow 0} \biggl[ 7+2880 \biggl( \frac{1}{t^{2}}-\frac{1}{2t\sinh ( t/2 ) } \biggr) ^{\prime \prime } \biggr] =0, \\& \lim_{t\rightarrow 0}B ( t ) = \lim_{t\rightarrow 0} \biggl( \int_{0}^{t}h^{\prime } ( s )\,ds \biggr) =0, \end{aligned}

and the function $$A^{\prime }/B^{\prime }=2880h^{\prime \prime \prime }/h^{\prime }$$ is increasing on $$( 0,t_{0} )$$ and decreasing on $$( t_{0},\infty )$$ by Lemma 7. Then by Lemma 3 it is enough to check that $$\operatorname{sgn}B^{\prime } ( t ) \operatorname{sgn} H_{A,B} ( \infty ) >0$$. In fact, $$B^{\prime } ( t ) =h^{\prime } ( t ) <0$$ for $$t>0$$ in view of Lemma 7, and

\begin{aligned}& \lim_{t\rightarrow \infty }\frac{A^{\prime } ( t ) }{B^{ \prime } ( t ) } = \lim_{t\rightarrow \infty } \frac{2800h ^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) }=0, \\& \lim_{t\rightarrow \infty }B ( t ) = \lim_{t\rightarrow \infty } \int_{0}^{t}h^{\prime } ( s )\,ds=h ( \infty ) -h \bigl( 0^{+} \bigr) =-\frac{1}{24}, \\& \lim_{t\rightarrow \infty }A ( t ) = \lim_{t\rightarrow \infty } \bigl( 7+2880h^{\prime \prime } ( t ) \bigr) =7, \end{aligned}

which imply that

$$H_{A,B} ( t ) =\frac{A^{\prime } ( t ) }{B^{\prime } ( t ) }B ( t ) -A ( t ) \rightarrow -7 \quad \mbox{as }t\rightarrow \infty .$$

This indicates $$\operatorname{sgn}B^{\prime } ( t ) \operatorname{sgn}H _{A,B} ( \infty ) >0$$.

Using the asymptotic formula [35, p. 32, (5)]

$$\ln \Gamma \biggl( x+\frac{1}{2} \biggr) =x\ln x-x+\frac{1}{2}\ln ( 2\pi ) -\sum_{k=1}^{\infty }\frac{ ( 1-2^{1-2k} ) B _{2k}}{2k ( 2k-1 ) } \frac{1}{x^{2k-1}}$$
(3.8)

as $$x\rightarrow \infty$$, we find that

\begin{aligned} f ( x ) \thicksim &\frac{1}{24x ( -\frac{1}{24x}+\frac{7}{2880x ^{3}}-\frac{31}{40\text{,}320x^{5}} ) +1}-\frac{120}{7}x^{2} \\ =&\frac{3720}{7}\frac{x^{2}}{98x^{2}-31}\rightarrow \frac{1860}{343}\quad \mbox{as }x\rightarrow \infty . \end{aligned}

While $$f ( 0^{+} ) =1$$ is clear. This completes the proof. □

## Concluding remarks

### Remark 3

In this paper, we investigate the monotonicity of the function $$f ( x )$$. In general, it is difficult to deal with such monotonicity since the gamma function Γ occurs in denominator. However, by the aid of Lemma 5, $$f ( x )$$ is equivalently changed into the ratio of two Laplace transformations of $$A ( x )$$ and $$B ( x )$$. While Lemma 4 provides exactly an approach to confirm the monotonicity of such ratio. Undoubtedly, it is a novel idea.

Moreover, it is known that Laplace transformation is related to the completely monotonic function. A function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and satisfies

$$(-1)^{n}f^{(n)}(x)\geq 0\quad \mbox{for all }x\in I \mbox{ and }n=0,1,2,\ldots.$$
(4.1)

If inequality (4.1) is strict, then f is said to be strictly completely monotonic on I. The classical Bernstein’s theorem [36, 37] states that a function f is completely monotonic on $$(0,\infty )$$ if and only if it is a Laplace transform of some nonnegative measure μ, that is,

$$f ( x ) = \int_{0}^{\infty }e^{-xt}\,d\mu ( t ) ,$$

where $$\mu ( t )$$ is non-decreasing and the integral converges for $$0< x<\infty$$.

### Remark 4

Let $$\alpha >\beta$$. If $$B ( t ) >0$$ for $$t>0$$ and

$$\beta < \frac{\int_{0}^{\infty }A ( t ) e^{-xt}\,dt}{\int_{0} ^{\infty }B ( t ) e^{-xt}\,dt}< \alpha ,$$

then, by Bernstein’s theorem, both the functions

$$x\mapsto \int_{0}^{\infty } \bigl[ A ( t ) -\beta B ( t ) \bigr] e^{-xt}\,dt\quad \mbox{and} \quad x\mapsto \int_{0}^{\infty } \bigl[ \alpha B ( t ) -A ( t ) \bigr] e^{-xt}\,dt$$

are completely monotonic on $$( 0,\infty )$$. And then, by Theorem 1, we immediately get the following.

### Proposition 1

Both the functions

\begin{aligned}& g_{1} ( x ) = 2880 \biggl( x^{2}+\frac{31}{98} \biggr) \biggl[ \ln \Gamma ( x+1/2 ) -x\ln x+x-\frac{1}{2}\ln ( 2 \pi ) \biggr] +120x+\frac{1517}{49x}, \\& g_{2} ( x ) = - \biggl( x^{2}+\frac{7}{120} \biggr) \biggl[ \ln \Gamma ( x+1/2 ) -x\ln x+x-\frac{1}{2}\ln ( 2 \pi ) \biggr] - \frac{1}{24}x \end{aligned}

are completely monotonic on $$( 0,\infty )$$.

Furthermore, by Bernstein’s theorem and Lemma 7, Proposition 1 can be improved as follows.

### Theorem 2

The function

$$g ( x ) =24x \biggl( \frac{120}{7}x^{2}+a \biggr) \bigl[ \ln \Gamma ( x+1/2 ) -x\ln x+x-\ln \sqrt{2\pi } \bigr] + \frac{120}{7}x^{2}+a-1$$

is completely monotonic on $$( 0,\infty )$$ if and only if $$a\geq 1860/343$$, and so is $$-g ( x )$$ on $$( 0,\infty )$$ if and only if $$a\leq -120\lambda_{0}/7\approx -0.88635$$, where $$\lambda_{0}$$ is defined in Lemma 7.

### Proof

By Lemma 5 and identities (3.4), (3.5) and (3.6), $$g ( x )$$ can be written as

\begin{aligned} g ( x ) =&-24 \biggl( \frac{120}{7}x^{2}+a \biggr) \biggl( \frac{1}{24}+ \int_{0}^{\infty }h^{\prime } ( t ) e^{-xt} \,dt \biggr) +\frac{120}{7}x^{2}+a-1 \\ =&-1-\frac{2880}{7}x^{2} \int_{0}^{\infty }h^{\prime } ( t ) e^{-xt} \,dt-24a \int_{0}^{\infty }h^{\prime } ( t ) e^{-xt} \,dt \\ =&-24 \int_{0}^{\infty } \biggl[ a- \biggl( -\frac{120}{7} \frac{h^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) } \biggr) \biggr] h^{\prime } ( t ) e^{-xt}\,dt. \end{aligned}

Since $$h^{\prime } ( t ) <0$$ for $$t>0$$, by Bernstein’s theorem and Lemma 7, g is completely monotonic on $$( 0, \infty )$$ if and only if

$$a\geq \frac{120}{7}\sup_{t\in ( 0,\infty ) } \biggl( -\frac{h ^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) } \biggr) =-\frac{120}{7}\inf_{t\in ( 0,\infty ) }\frac{h ^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) }=- \frac{120}{7} \biggl( -\frac{31}{98} \biggr) =\frac{1860}{343},$$

and so is −g on $$( 0,\infty )$$ if and only if

$$a\leq \frac{120}{7}\inf_{t\in ( 0,\infty ) } \biggl( -\frac{h ^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) } \biggr) =-\frac{120}{7}\sup_{t\in ( 0,\infty ) }\frac{h ^{\prime \prime \prime } ( t ) }{h^{\prime } ( t ) }=- \frac{120}{7}\lambda_{0}\approx -0.88635.$$

This ends the proof. □

### Remark 5

The expression of $$f ( x )$$ reminds us to consider the asymptotic expansion of

$$\frac{1}{24x ( \ln \Gamma ( x+1/2 ) -x\ln x+x-\ln \sqrt{2 \pi } ) +1}:=x^{2}\sum_{n=0}^{\infty } \frac{c_{n}}{x^{2n}}.$$

Using asymptotic expansion (3.8), we have

$$- \Biggl( 24\sum_{n=2}^{\infty } \frac{ ( 1-2^{1-2n} ) B_{2n}}{2n ( 2n-1 ) }\frac{1}{x^{2n-2}} \Biggr) \Biggl( x^{2}\sum _{n=0} ^{\infty }c_{n}x^{-2n} \Biggr) =1,$$

that is,

$$\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{ ( 1-2^{-2k-3} ) B_{2k+4}}{2 ( k+2 ) ( 2k+3 ) }c_{n-k} \Biggr) \frac{1}{x ^{2n}}=-\frac{1}{24}.$$

Comparing coefficients gives

\begin{aligned}& -\frac{7}{2880}c_{0} = -\frac{1}{24}, \\& \sum_{k=0}^{n}\frac{ ( 1-2^{-2k-3} ) B_{2k+4}}{2 ( k+2 ) ( 2k+3 ) }c_{n-k} = 0 \quad \mbox{for }n\geq 1, \end{aligned}

which show that $$c_{n}$$ has the recurrence formula

$$c_{n}=\frac{2880}{7}\sum_{k=1}^{n} \frac{ ( 1-2^{-2k-3} ) B _{2k+4}}{2 ( k+2 ) ( 2k+3 ) }c_{n-k}\quad \mbox{and} \quad c_{0}= \frac{120}{7},$$

from which we obtain a new asymptotic expansion for the gamma function:

$$\frac{\Gamma ( x+1/2 ) }{\sqrt{2\pi } ( x/e ) ^{x}}\thicksim \exp \biggl( -\frac{1}{24x}+ \frac{1}{24x^{3}}\frac{1}{ \frac{120}{7}+\frac{1860}{343}x^{-2}+\cdots+c_{n}x^{-2n}+\cdots} \biggr) \quad \mbox{as }x\rightarrow \infty.$$

Moreover, it is easy to prove the inequalities

\begin{aligned} \exp \biggl( -\frac{1}{1440}\frac{5880x^{2}+1517}{x ( 98x^{2}+31 ) } \biggr) &=\exp \biggl( - \frac{1}{24x}+\frac{1}{24x ^{3}}\frac{1}{\frac{120}{7}+\frac{1860}{343}x^{-2}} \biggr) \\ &< \frac{\Gamma ( x+1/2 ) }{\sqrt{2\pi } ( x/e ) ^{x}}< \exp \biggl( -\frac{1}{24x}+\frac{1}{24x^{3}} \frac{1}{ \frac{120}{7}} \biggr) \\ &=\exp \biggl( -\frac{120x^{2}-7}{2880x^{3}} \biggr) \end{aligned}

hold for $$x\geq 1/2$$.

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## Acknowledgements

The authors would like to express their sincere thanks to the editors and reviewers for their great efforts to improve this paper. This work was supported by the Fundamental Research Funds for the Central Universities (No. 2015ZD29) and the Higher School Science Research Funds of Hebei Province of China (No. Z2015137).

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Jing-Feng Tian.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests. 