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.

1 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 [1]

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

(1.2)

slightly improves (1.1). Gosper [2] 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 [3] 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 [4] 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:

• (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)

• (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)

• (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)

• (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)

• (v) Chen’s [8] 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 [9–28] and the references cited therein.

It is worth mentioning that Yang and Chu [9] 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 $\alpha =1860/343\approx 5.4227$.

2 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

([29])

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 [31], also see [32].

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 [33].

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

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

• (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

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

then adding gives

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

By the assumptions, the desired assertions follow. □

3 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. □

4 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 $x\geq 1/2$.