# Sharp Smith’s bounds for the gamma function

## Abstract

Among various approximation formulas for the gamma function, Smith showed that

$$\Gamma \biggl( x+\frac{1}{2} \biggr) \thicksim S ( x ) =\sqrt{2 \pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2}, \quad x\rightarrow \infty ,$$

which is a little-known but accurate and simple one. In this note, we prove that the function $$x\mapsto \ln \Gamma ( x+1/2 ) - \ln S ( x )$$ is strictly increasing and concave on $$( 0,\infty )$$, which shows that Smith’s approximation is just an upper one.

## 1 Introduction

The Stirling formula

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

has many important applications in statistical physics, probability theory and number theory. Due to its practical importance, it has attracted much interest of many mathematicians and has also motivated a large number of research papers concerning various generalizations and improvements; see for example, Burnside’s [1], Gosper [2], Batir [3], Mortici [4].

The gamma function $$\Gamma ( x ) =\int_{0}^{\infty }t^{x-1}e ^{-t}\,dt$$ for $$x>0$$ is closely related to the Stirling formula, since $$\Gamma (n+1)=n!$$ for all $$n\in \mathbb{N}$$. This inspired some authors to also pay attention to find various better approximations for the gamma function; see, for instance, Ramanujan [5, p. 339], Windschitl (see Nemes [6, Corollary 4.1]), Yang and Chu [7], Chen [8].

More results involving the approximation formulas for the factorial or gamma function can be found in [923] and the references cited therein.

In this note, we are interested in Smith’s approximation formula (see [24, equation (42)]):

$$\Gamma \biggl( x+\frac{1}{2} \biggr) \thicksim \sqrt{2\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2}:=S ( x ), \quad \text{as }x\rightarrow \infty .$$
(1.2)

It is easy to check that

$$\Gamma \biggl( x+\frac{1}{2} \biggr) =\sqrt{2\pi } \biggl( \frac{x}{e} \biggr) ^{x} \biggl( 2x\tanh \frac{1}{2x} \biggr) ^{x/2} \biggl( 1+O \biggl( \frac{1}{x^{5}} \biggr) \biggr) ,$$

which shows that the rate of $$S ( x )$$ converging to $$\Gamma ( x+1/2 )$$ as $$x\rightarrow \infty$$ is like $$x^{-5}$$. According to the comment in [8, (3.5)–(3.10)], it is well known that Smith’s approximation is an accurate but simple one for gamma function.

The aim of this short note is to further prove the Smith approximation $$S ( x )$$ is an upper one. Our main result is stated as follows.

### Theorem 1

The function

$$f ( x ) =\ln \Gamma \biggl( x+\frac{1}{2} \biggr) -\ln \sqrt{2 \pi }-x\ln x+x-\frac{x}{2}\ln \biggl( 2x\tanh \frac{1}{2x} \biggr)$$

is strictly increasing and concave from $$( 0,\infty )$$ onto $$( -\ln \sqrt{2},0 )$$.

## 2 Proof of Theorem 1

To prove Theorem 1 we need the following two lemmas.

### Lemma 1

The inequality

$$\psi^{\prime } \biggl( x+\frac{1}{2} \biggr) < \frac{4}{3} \frac{15x^{2}+4}{x ( 20x^{2}+7 ) }$$
(2.1)

holds for all $$x>0$$.

### Proof

Let

$$f_{1} ( x ) =\psi^{\prime } \biggl( x+\frac{1}{2} \biggr) - \frac{4}{3}\frac{15x^{2}+4}{x ( 20x^{2}+7 ) }.$$
(2.2)

Using the recurrence formula [25, pp. 258–260]:

$$\psi^{ ( n ) }(x+1)-\psi^{ ( n ) }(x)=\frac{ ( -1 ) ^{n}n!}{x^{n+1}},$$

we have

\begin{aligned} f_{1} ( x+1 ) -f_{1} ( x ) =&\psi^{\prime } \biggl( x+ \frac{3}{2} \biggr) -\frac{4}{3 ( x+1 ) }\frac{15x^{2}+30x+19}{20x ^{2}+40x+27} \\ &{}-\psi^{\prime } \biggl( x+\frac{1}{2} \biggr) +\frac{4}{3} \frac{15x ^{2}+4}{x ( 20x^{2}+7 ) } \\ &{}-\frac{1}{ ( x+1/2 ) ^{2}}-\frac{4}{3 ( x+1 ) }\frac{15x ^{2}+30x+19}{20x^{2}+40x+27}+\frac{4}{3} \frac{15x^{2}+4}{x ( 20x ^{2}+7 ) } \\ =&\frac{144}{x ( x+1 ) ( 2x+1 ) ^{2} ( 20x ^{2}+7 ) ( 20x^{2}+40x+27 ) }>0. \end{aligned}

It then follows that

$$f_{1} ( x ) < f_{1} ( x+1 ) < \cdots < \lim _{n\rightarrow \infty }f_{1} ( x+n ) =0,$$

which proves the desired inequality (2.1). □

### Lemma 2

The inequality

$$\frac{\sinh^{2}t}{\cosh t}>\frac{t^{2} ( 21t^{2}+60 ) }{31t ^{2}+60}$$
(2.3)

holds for all $$t>0$$.

### Proof

It is obvious that the inequality what we consider is equivalent to

$$f_{2} ( t ) = \bigl( 31t^{2}+60 \bigr) ( \sinh t ) ^{2}-t^{2} \bigl( 21t^{2}+60 \bigr) \cosh t>0.$$

Simplifying and expanding it in power series lead us to

\begin{aligned} 2f_{2} ( t ) =&60\cosh 2t+31t^{2}\cosh 2t-120t^{2}\cosh t-42t ^{4}\cosh t-31t^{2}-60 \\ =&60\sum_{n=0}^{\infty }\frac{2^{2n}}{ ( 2n ) !}t^{2n}+31 \sum_{n=1}^{\infty }\frac{2^{2n-2}}{ ( 2n-2 ) !}t^{2n} \\ &{}-120\sum_{n=1}^{\infty }\frac{1}{ ( 2n-2 ) !}t^{2n}-42 \sum_{n=2}^{\infty }\frac{1}{ ( 2n-4 ) !}t^{2n}-31t^{2}-60 \\ :=&\sum_{n=2}^{\infty }\frac{a_{n}}{ ( 2n ) !}t^{2n}, \end{aligned}

where

$$a_{n}= \bigl( 62n^{2}-31n+120 \bigr) 2^{2n-1}-24n ( 2n-1 ) \bigl( 14n^{2}-35n+31 \bigr) .$$

It is easy to check that $$a_{2}=a_{3}=0$$ and $$a_{4}=49\,184>0$$. It remains to prove $$a_{n}>0$$ for $$n\geq 5$$.

To this end, it suffices to prove $$b_{n}=2^{2n-1}-6n ( 2n-1 ) >0$$ for $$n\geq 5$$, because the inequality

$$\bigl( 62n^{2}-31n+120 \bigr) >4 \bigl( 14n^{2}-35n+31 \bigr)$$

is clearly valid for $$n\geq 5$$. We easily obtain

$$b_{n+1}-4b_{n}=6 \bigl( 6n^{2}-7n-1 \bigr) >0$$

for $$n\geq 5$$, which in combination with $$b_{5}=242>0$$ yields $$b_{n}>0$$ for $$n\geq 5$$. This completes the proof. □

Now we are in a position to prove Theorem 1.

### Theorem 1

Differentiating and simplifying yields

\begin{aligned}& f^{\prime } ( x ) = \psi \biggl( x+\frac{1}{2} \biggr) - \ln x- \frac{1}{2}\ln \biggl( 2x\tanh \frac{1}{2x} \biggr) + \frac{1}{2x \sinh ( 1/x ) }-\frac{1}{2}, \\& f^{\prime \prime } ( x ) = \psi^{\prime } \biggl( x+ \frac{1}{2} \biggr) +\frac{1}{2x^{3}}\frac{\cosh ( 1/x ) }{ \sinh^{2} ( 1/x ) }-\frac{3}{2x}. \end{aligned}

As an application of inequalities (2.1) and (2.3) it gives

\begin{aligned}& \begin{aligned} f^{\prime \prime } ( x ) &< \frac{4}{3}\frac{15x^{2}+4}{x ( 20x^{2}+7 ) }+\frac{1}{2x^{3}} \frac{\cosh ( 1/x ) }{\sinh^{2} ( 1/x ) }-\frac{3}{2x} \\ &=\frac{1}{2x^{3}}\frac{\cosh ( 1/x ) }{\sinh^{2} ( 1/x ) }-\frac{1}{6}\frac{60x^{2}+31}{x ( 20x^{2}+7 ) } \\ & \mathop{=}^{x=1/t}\frac{t^{3}}{2} \biggl( \frac{\cosh t}{ \sinh^{2}t}- \frac{31t^{2}+60}{t^{2} ( 21t^{2}+60 ) } \biggr) < 0. \end{aligned} \end{aligned}

Then it is deduced that

$$f^{\prime } ( x ) >\lim_{x\rightarrow \infty }f^{\prime } ( x ) =0,$$

which in turn implies that

$$-\frac{1}{2}\ln 2=\lim_{x\rightarrow 0^{+}}f ( x ) < f ( x ) < \lim _{x\rightarrow \infty }f ( x ) =0.$$

This completes the proof. □

## 3 Corollaries and remarks

Using the increasing property of $$f ( x+1/2 )$$ given in Theorem 1 and noting that

$$f \biggl( \frac{1}{2} \biggr) =\ln \frac{\sqrt{e}}{\sqrt{\pi } ( \tanh 1 ) ^{1/4}}\quad \text{and}\quad f \biggl( \frac{3}{2} \biggr) = \ln \biggl( \frac{2e\sqrt{e}3^{3/4}}{27\sqrt{\pi }\tanh^{3/4} ( 1/3 ) } \biggr) ,$$

we have the corollaries.

### Corollary 1

The double inequality

$$\alpha_{1}< \frac{e^{x+1/2}\Gamma ( x+1 ) }{\sqrt{2\pi } ( x+1/2 ) ^{x+1/2} [ ( 2x+1 ) \tanh ( 1/ ( 2x+1 ) ) ] ^{ ( 2x+1 ) /4}}< 1$$

holds for all $$x>0$$ with the best constants 1 and $$\alpha_{1}=\sqrt{e/ \pi }/ ( \tanh 1 ) ^{1/4}\approx 0.99573$$.

### Corollary 2

The double inequality

$$\alpha_{2}< \frac{n!}{\sqrt{2\pi } ( ( n+1/2 ) /e ) ^{n+1/2} [ ( 2n+1 ) \tanh ( 1/ ( 2n+1 ) ) ] ^{ ( 2n+1 ) /4}}< 1$$

holds for all $$n\in \mathbb{N}$$ with the best constants 1 and

$$\alpha_{2}=\frac{2e\sqrt{e}3^{3/4}}{27\sqrt{\pi }\tanh^{3/4} ( 1/3 ) }\approx 0.99994.$$

By the decreasing property of $$f^{\prime } ( x+1/2 )$$ given in Theorem 1 and the facts that

\begin{aligned}& f^{\prime } \biggl( \frac{1}{2} \biggr) =\frac{1}{\sinh 2}- \frac{1}{2} \ln ( \tanh 1 ) +\ln 2-\frac{1}{2}-\gamma \approx 0.027823, \\& f^{\prime } \biggl( \frac{3}{2} \biggr) =\frac{1}{3\sinh ( 2/3 ) }- \frac{1}{2}\ln \biggl( 3\tanh \frac{1}{3} \biggr) -\ln \frac{3}{2}+\psi ( 1 ) +\frac{1}{2}\approx 0.00016946, \end{aligned}

the following corollaries are immediate.

### Corollary 3

For $$x>0$$, the inequalities

\begin{aligned}& \frac{1}{2}+\frac{1}{2}\ln \biggl( ( 2x+1 ) \tanh \frac{1}{2x+1} \biggr) -\frac{1}{ ( 2x+1 ) \sinh ( 2/ ( 2x+1 ) ) } \\& \quad < \psi ( x+1 ) -\ln \biggl( x+\frac{1}{2} \biggr) \\& \quad < \beta_{1}+\frac{1}{2}\ln \biggl( ( 2x+1 ) \tanh \frac{1}{2x+1} \biggr) -\frac{1}{ ( 2x+1 ) \sinh ( 2/ ( 2x+1 ) ) } \end{aligned}

hold, where the constants $$1/2$$ and

$$\beta_{1}=\ln 2-\frac{1}{2}\ln ( \tanh 1 ) +\frac{1}{\sinh 2}- \gamma \approx 0.52782$$

are the best possible.

### Corollary 4

Let $$H_{n}=\sum_{k=1}^{n}$$ for $$n\in \mathbb{N}$$. The inequalities

\begin{aligned}& \biggl( \frac{1}{2}+\gamma \biggr) +\frac{1}{2}\ln \biggl( ( 2n+1 ) \tanh \frac{1}{2n+1} \biggr) -\frac{1}{ ( 2n+1 ) \sinh ( 2/ ( 2x+1 ) ) } \\& \quad < H_{n}-\ln \biggl( n+\frac{1}{2} \biggr) \\& \quad < \beta_{2}+\frac{1}{2}\ln \biggl( ( 2n+1 ) \tanh \frac{1}{2n+1} \biggr) -\frac{1}{ ( 2n+1 ) \sinh ( 2/ ( 2n+1 ) ) } \end{aligned}

hold, where $$1/2+\gamma \approx 1.0772$$ and

$$\beta_{2}=\frac{1}{3\sinh ( 2/3 ) }-\frac{1}{2}\ln \biggl( 3 \tanh \frac{1}{3} \biggr) -\ln \frac{3}{2}+1\approx 1.0774$$

are the best possible constants.

Finally, as a by-product of Lemma 1, we draw the following conclusion.

### Theorem 2

Let g be defined on $$( 0,\infty )$$ by

$$g ( x ) =\ln \Gamma \biggl( x+\frac{1}{2} \biggr) - \biggl[ \frac{1}{2}\ln 2\pi +\frac{16}{21}x\ln x+\frac{5x}{42}\ln \biggl( x ^{2}+\frac{7}{20} \biggr) -x-\frac{\sqrt{35}}{42} \operatorname{arccot} \biggl( \sqrt{\frac{20}{7}}x \biggr) \biggr] .$$

Then g is strictly increasing and concave on $$( 0,\infty )$$.

### Proof

Differentiation yields

\begin{aligned}& g^{\prime } ( x ) =\psi \biggl( x+\frac{1}{2} \biggr) - \biggl[ \frac{5}{42}\ln \biggl( x^{2}+\frac{7}{20} \biggr) + \frac{16}{21}\ln x \biggr] , \\& g^{\prime \prime } ( x ) =\psi^{\prime } \biggl( x+ \frac{1}{2} \biggr) -\frac{4}{3}\frac{15x^{2}+4}{x ( 20x^{2}+7 ) }=f_{1} ( x ) < 0, \end{aligned}

where the inequality holds due to Lemma 1. This completes the proof. □

### Remark 1

Theorem 2 gives a new approximation for the gamma function

$$\Gamma \biggl( x+\frac{1}{2} \biggr) \thicksim \sqrt{2\pi }x^{26x/21} \biggl( x^{2}+\frac{7}{20} \biggr) ^{5x/42}\exp \biggl[ -x-\frac{ \sqrt{35}}{42}\operatorname{arccot} \biggl( \sqrt{\frac{20}{7}}x \biggr) \biggr] ,$$

as $$x\rightarrow \infty$$, which satisfies

$$\Gamma \biggl( x+\frac{1}{2} \biggr) =\sqrt{2\pi }x^{26x/21} \biggl( x ^{2}+\frac{7}{20} \biggr) ^{5x/42}\exp \biggl[ -x- \frac{\sqrt{35}}{42}\operatorname{arccot} \biggl( \sqrt{\frac{20}{7}}x \biggr) \biggr] \bigl( 1+O \bigl( x^{-5} \bigr) \bigr) .$$

### Remark 2

Theorem 2 also offers an asymptotic formula for the psi function

$$\psi \biggl( x+\frac{1}{2} \biggr) \thicksim \frac{5}{42}\ln \biggl( x ^{2}+\frac{7}{20} \biggr) +\frac{16}{21}\ln x\quad \text{as }x \rightarrow \infty .$$

Furthermore, by replacing x with $$x+1/2$$, we have the following sharp inequalities:

\begin{aligned}& \frac{5}{42}\ln \biggl( x^{2}+x+\frac{3}{5} \biggr) + \frac{16}{21} \ln \biggl( x+\frac{1}{2} \biggr) \\& \quad < \psi ( x+1 ) < \lambda_{0}+\frac{5}{42}\ln \biggl( x^{2}+x+ \frac{3}{5} \biggr) +\frac{16}{21}\ln \biggl( x+ \frac{1}{2} \biggr) \end{aligned}
(3.1)

for $$x>0$$ with the best constant

\begin{aligned}& \lambda_{0}=\frac{16}{21}\ln 2-\frac{5}{42}\ln \frac{3}{5}-\gamma \approx 0.011709; \\& \gamma +\frac{5}{42}\ln \biggl( n^{2}+n+\frac{3}{5} \biggr) + \frac{16}{21}\ln \biggl( n+\frac{1}{2} \biggr) \\& \quad < H_{n}< \lambda_{0}+\gamma +\frac{5}{42}\ln \biggl( n^{2}+n+ \frac{3}{5} \biggr) +\frac{16}{21}\ln \biggl( n+ \frac{1}{2} \biggr) \end{aligned}

for $$n\in \mathbb{N}$$ with the best constant

$$\lambda_{1}=1-\frac{16}{21}\ln \frac{3}{2}- \frac{5}{42}\ln \frac{13}{5}-\gamma \approx 0.00010718.$$

Inequalities (3.1) first appeared in [26, Corollary 3.4].

## 4 Conclusions

In this note, we mainly presented an upper bound of Smith’s approximation in accordance with the fact that the function $$x\mapsto \ln \Gamma ( x+1/2 ) - \ln S ( x )$$ is strictly increasing and concave on $$( 0,\infty )$$. As a consequence, we get some new sharp estimates to various classical inequalities concerning the gamma function and hyperbolic functions.

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## Acknowledgements

This paper is supported by the National Science Foundation of China grant No. 11371050.

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All authors contributed to each part of this work equally, and they all read and approved the final manuscript.

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Correspondence to Shen-Zhou Zheng.

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Li, XQ., Liu, ZM., Yang, ZH. et al. Sharp Smith’s bounds for the gamma function. J Inequal Appl 2018, 27 (2018). https://doi.org/10.1186/s13660-018-1620-3