With the aid of the lemmas in Sect. 2, we can prove Theorem 1.
Proof of Theorem 1
Differentiation yields
$$\begin{aligned} &f_{0}^{\prime} ( x ) =\psi( x+1 ) -\frac{1}{2}\ln \biggl( x\sinh\frac{1}{x} \biggr) +\frac{1}{2x}\coth \frac{1}{x} \\ &\phantom{f_{0}^{\prime} ( x )=}{}-\ln x-\frac{1}{2x}-\frac{1}{2}+\frac{7}{324} \frac{175x^{2}+99}{x^{4} ( 35x^{2}+33 ) ^{2}}, \\ &f_{0}^{\prime\prime} ( x ) =\psi^{\prime} ( x+1 ) +\frac{1}{2x^{3}}\frac{1}{\sinh^{2} ( 1/x ) } \\ &\phantom{f_{0}^{\prime\prime} ( x ) =}{}-\frac{3}{2x}+\frac{1}{2x^{2}}-\frac{7}{54}\frac {6125x^{4}+6545x^{2}+2178}{x^{5} ( 35x^{2}+33 ) ^{3}}. \end{aligned}$$
Since \(\lim_{x\rightarrow\infty}f_{0} ( x ) =\lim_{x\rightarrow \infty}f_{0}^{\prime} ( x ) =0\), it suffices to prove \(f_{0}^{\prime\prime} ( x ) >0\) for \(x\geq1\). Replacing x by \(( x+1/2 ) \) in Lemma 1 leads to
$$ \psi^{\prime} ( x+1 ) >\frac{7}{30}\frac{ ( 2x+1 ) ( 165x^{4}+330x^{3}+815x^{2}+650x+417 ) }{77x^{6}+231x^{5}+560x^{4}+735x^{3}+623x^{2}+294x+60}, $$
which indicates that
$$\begin{aligned} f_{0}^{\prime\prime} ( x ) >{}&\frac{7}{30}\frac{ ( 2x+1 ) ( 165x^{4}+330x^{3}+815x^{2}+650x+417 ) }{77x^{6}+231x^{5}+560x^{4}+735x^{3}+623x^{2}+294x+60}+ \frac{1}{2x^{3}}\frac{1}{\sinh^{2} ( 1/x ) } \\ &{}-\frac{3}{2x}+\frac{1}{2x^{2}}-\frac{7}{54}\frac {6125x^{4}+6545x^{2}+2178}{x^{5} ( 35x^{2}+33 ) ^{3}}:=f_{01} \biggl( \frac{1}{x} \biggr). \end{aligned}$$
Arranging gives
$$\begin{aligned} f_{01} ( t ) ={}&\frac{t}{2} \biggl( \frac{t}{\sinh t} \biggr) ^{2}+\frac{7}{30}\frac{t ( t+2 ) ( 417t^{4}+650t^{3}+815t^{2}+330t+165 ) }{60t^{6}+294t^{5}+623t^{4}+735t^{3}+560t^{2}+231t+77} \\ &{}-\frac{3}{2}t+\frac{1}{2}t^{2}-\frac{7}{54}t^{7} \frac{2178t^{4}+6545t^{2}+6125}{ ( 33t^{2}+35 ) ^{3}}, \end{aligned}$$
where \(t=1/x\in ( 0,1 ) \). Applying the first inequality of (2.1) we have
$$\begin{aligned} f_{01} ( t ) >{}&\frac{t}{2} \biggl( 1-\frac{1}{6}t^{2}+ \frac{7}{360}t^{4}-\frac{31}{15{,}120}t^{6}+ \frac{127}{604{,}800}t^{8}-\frac{73}{3{,}421{,}440}t^{10} \biggr) ^{2} \\ &{}+\frac{7}{30}\frac{t ( t+2 ) ( 417t^{4}+650t^{3}+815t^{2}+330t+165 ) }{60t^{6}+294t^{5}+623t^{4}+735t^{3}+560t^{2}+231t+77} \\ &{}-\frac{3}{2}t+\frac{1}{2}t^{2}-\frac{7}{54}t^{7} \frac{2178t^{4}+6545t^{2}+6125}{ ( 33t^{2}+35 ) ^{3}} \\ ={}&\frac{t^{11}\times p_{22} ( t ) }{ ( 33t^{2}+35 ) ^{3} ( 60t^{6}+294t^{5}+623t^{4}+735t^{3}+560t^{2}+231t+77 ) }, \end{aligned}$$
where \(p_{22} ( t ) =\sum_{k=0}^{22}a_{k}t^{k}\) with \(a_{0}=\frac{2{,}341{,}955}{27}\), \(a_{1}=\frac{2{,}341{,}955}{9}\), \(a_{2}= \frac{4{,}592{,}761{,}525{,}177}{41{,}057{,}280}\), \(a_{3}= \frac {3{,}740{,}791{,}861{,}177}{13{,}685{,}760}\), \(a_{4}= -\frac{21{,}774{,}907{,}040{,}747}{615{,}859{,}200}\), \(a_{5}=\frac {1{,}776{,}198{,}096{,}757}{51{,}321{,}600}\), \(a_{6}=-\frac{2{,}348{,}474{,}362{,}865{,}491}{59{,}122{,}483{,}200} \), \(a_{7}=-\frac{444{,}392{,}576{,}792{,}851}{19{,}707{,}494{,}400}\), \(a_{8}= \frac {722{,}576{,}509{,}559{,}549}{344{,}881{,}152{,}000} \), \(a_{9}=\frac {734{,}284{,}235{,}570{,}623}{229{,}920{,}768{,}000}\), \(a_{10}=-\frac {27{,}685{,}269{,}148{,}007{,}477}{74{,}494{,}328{,}832{,}000}\), \(a_{11}=-\frac {13{,}202{,}571{,}814{,}150{,}457}{24{,}831{,}442{,}944{,}000}\), \(a_{12}=\frac{1{,}859{,}898{,}503{,}651{,}431}{585{,}312{,}583{,}680{,}000}\), \(a_{13}=\frac{40{,}990{,}762{,}057{,}313{,}921}{682{,}864{,}680{,}960{,}000}\), \(a_{14}=\frac{1{,}227{,}464{,}630{,}525{,}327}{573{,}606{,}332{,}006{,}400}\), \(a_{15}=-\frac{107{,}829{,}513{,}340{,}517}{19{,}510{,}419{,}456{,}000}\), \(a_{16}=-\frac{1{,}469{,}516{,}232{,}022{,}339}{4{,}780{,}052{,}766{,}720{,}000}\), \(a_{17}=\frac{224{,}320{,}158{,}179}{492{,}687{,}360{,}000}\), \(a_{18}=\frac{214{,}165{,}238{,}137}{6{,}437{,}781{,}504{,}000}\), \(a_{19}=-\frac{402{,}182{,}039}{11{,}943{,}936{,}000}\), \(a_{20}=-\frac{150{,}639{,}953}{50{,}164{,}531{,}200}\), \(a_{21}= \frac{2{,}872{,}331}{1{,}194{,}393{,}600}\), \(a_{22}=\frac{58{,}619}{119{,}439{,}360}\).
It remains to prove \(p_{22} ( t ) =\sum_{k=0}^{22}a_{k}t^{k}>0\) for \(t\in(0,1]\). Since \(a_{k}>0\) for \(k=0\), 1, 2, 3, 8, 9, 12, 13, 14, 17, 18, 21, 22 and \(a_{k}<0\) for \(k=4\), 6, 7, 10, 11, 15, 16, 19, 20, we have
$$ p_{22} ( t ) =\sum_{k=0}^{22}a_{k}t^{k}= \sum_{a_{k}>0}a_{k}t^{k}+\sum_{a_{k}< 0}a_{k}t^{k}>\sum _{k=4,6,7,10,11,15,16,19,20}a_{k}t^{k}+\sum _{k=0}^{3}a_{k}t^{k}:=p_{20} ( t ). $$
Clearly, the coefficients of the polynomial \(-p_{20} ( t ) \) satisfy the conditions in Lemma 3, and
$$ -p_{20} ( 1 ) =\sum_{k=4,6,7,10,11,15,16,19,20} ( -a_{k} ) -\sum_{k=0}^{3}a_{k}=- \frac{1{,}135{,}768{,}202{,}621{,}781{,}774{,}901}{1{,}792{,}519{,}787{,}520{,}000}< 0. $$
It then follows that \(p_{20} ( t ) >0\) for \(t\in(0,1]\), and so is \(p_{22} ( t ) \), which implies \(f_{01} ( t ) >0\) for \(t\in (0,1]\). Consequently, \(f_{0}^{\prime\prime} ( x ) >0\) for all \(x\geq1\). This completes the proof. □
As a direct consequence of Theorem 1, we immediately get the following.
Corollary 1
For
\(n\in\mathbb{N}\), the double inequality
$$ \exp\frac{7}{324n^{3} ( 35n^{2}+33 ) }< \frac{n!}{\sqrt{2\pi n}( n/e ) ^{n} ( n\sinh n^{-1} ) ^{n/2}}< \lambda\exp\frac{7}{324n^{3} ( 35n^{2}+33 ) } $$
holds with the best constant
$$ \lambda=\exp f_{0} ( 1 ) =\frac{1}{\sqrt{2\pi\sinh1}}\exp\frac{22{,}025}{22{,}032} \approx1.000024067. $$
Set
$$ D_{0} ( y ) =y-\ln( 1+y ),\quad y=\frac{7}{324x^{3} ( 35x^{2}+33 ) }. $$
Then it is easy to check that, for \(x>1\),
$$\begin{aligned} &\frac{dD_{0} ( y ) }{dx}=-\frac{49}{324}\frac{175x^{2}+99}{x^{4} ( 35x^{2}+33 ) ^{2} ( 11{,}340x^{5}+10{,}692x^{3}+7 ) }< 0, \\ &\frac{d^{2}D_{0} ( y ) }{dx^{2}}=\frac{343}{54}\frac{ ( 18{,}191{,}250x^{9}+37{,}110{,}150x^{7}+24{,}992{,}550x^{5}+6125x^{4}+5{,}821 {,}794x^{3}+6545x^{2}+2178 ) }{x^{5} ( 35x^{2}+33 ) ^{3} ( 11{,}340x^{5}+10{,}692x^{3}+7 ) ^{2}}\\ &\phantom{\frac{d^{2}D_{0} ( y ) }{dx^{2}}}>0. \end{aligned}$$
That is to say, \(x\mapsto D_{0} ( y ) \) is decreasing and convex on \(( 1,\infty ) \), and so is the function \(f_{0}^{\ast} ( x ):=f_{0} ( x ) +D_{0} ( y ) \) by Theorem 1.
Corollary 2
The function
$$\begin{aligned} f_{0}^{\ast} ( x ) ={}&\ln\Gamma( x+1 ) -\ln\sqrt{2\pi }- \biggl( x+\frac{1}{2} \biggr) \ln x+x-\frac{x}{2}\ln \biggl( x\sinh\frac{1}{x}\biggr) \\ &{}-\ln\biggl( 1+ \frac{7}{324x^{3} ( 35x^{2}+33 ) } \biggr) \end{aligned}$$
is strictly decreasing and convex from
\(( 1,\infty ) \)
onto
\(( 0,f_{0}^{\ast} ( 1 ) ) \), where
$$ f_{0}^{\ast} ( 1 ) =1-\ln\frac{22{,}039}{22{,}032}-\ln\sqrt{2\pi \sinh1}\approx0.00002412. $$
Remark 1
Corollary 2 offers another approximation formula
$$ \Gamma( x+1 ) \thicksim\sqrt{2\pi x} \biggl( \frac{x}{e} \biggr) ^{x} \biggl( x\sinh\frac{1}{x} \biggr) ^{x/2} \biggl( 1+ \frac{7}{324}\frac{1}{x^{3} ( 35x^{2}+33 ) } \biggr) =W_{2}^{\ast} ( x ) . $$
(3.1)
Also, for \(n\in\mathbb{N}\),
$$\begin{aligned} 1+\frac{7}{324n^{3} ( 35n^{2}+33 ) }< \frac{n!}{\sqrt{2\pi n} ( n/e ) ^{n} ( n\sinh n^{-1} ) ^{n/2}}< \lambda^{\ast} \biggl( 1+\frac{7}{324n^{3} ( 35n^{2}+33 ) } \biggr) \end{aligned}$$
with the best constant
$$ \lambda^{\ast}=\exp f_{0}^{\ast} ( 1 ) = \frac{22{,}032}{22{,}039}\frac{e}{\sqrt{2\pi\sinh1}}\approx1.000024117. $$