Some New Properties of the Gamma Function Based on Ramanujan’s Formula and Nemes’ Formula

1. Introduction, Definitions and Motivation

In his lost notebook, Ramanujan (see [36] and [5]) claimed that

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)=\sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{{\theta }_x}{30}}\right)}^{1/6},\end{array}$$

(1)

where ${\theta }_x\to 1$ as $x\to \infty$ and ${\displaystyle \frac{3}{10}}<{\theta }_x<1$. Thus, clearly, we can rewrite (1) as follows:

$$\mathsf{\Gamma }(x+1)\approx \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}(x\to \infty )$$

(2)

and

$$\begin{array}{cc}\hfill & \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{100}}\right)}^{1/6}\hfill \\ & <\mathsf{\Gamma }(x+1)<\sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}(x\geqq 0).\hfill \end{array}$$

Ramanujan’s claim has been the subject of intense investigations and is reviewed in [6]. It has also motivated a large number of research papers (see, for example, [4,7,8,9,10,11,17,18,19,23,24,25,26,27,28,30]). In particular, Karatsuba [19] proved that the function $h\left(x\right)$ given by

$$\begin{array}{c}\hfill h\left(x\right):={\left[{\left({\displaystyle \frac{e}{x}}\right)}^x{\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{\pi }}}\right]}^6-(8x^3+4x^2+x)={\displaystyle \frac{{\theta }_x}{30}}\end{array}$$

(3)

is monotonically increasing from $[1,\infty )$ onto $\left[h\right(1),h(\infty \left)\right)$ with

$$h\left(1\right)={\displaystyle \frac{e^6}{{\pi }^3}}-13=0.0111976\dots \mathrm{and}h(\infty )={\displaystyle \frac{1}{30}}=0.0333\dots .$$

The following asymptotic representation of the gamma function was also established by Karatsuba [19]:

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}-{\displaystyle \frac{11}{240x}}+{\displaystyle \frac{79}{3360x^2}}+{\displaystyle \frac{3539}{201600x^3}}\hfill \\ \hfill & -{\displaystyle \frac{9511}{403200x^4}}-{\displaystyle \frac{10051}{716800x^5}}+{\displaystyle \frac{233934691}{6386688000x^6}}+\dots {)}^{1/6}(x\to \infty ),\hfill \end{array}$$

(4)

in which we have corrected the term involving $x^{-6}$ (see [19] for a formula for successively determining the coefficients).

Alzer [4] proved that, in the interval $(0,1]$, the constant term ${\displaystyle \frac{1}{100}}$ can be replaced by the best possible constant given by

$$\underset{0.6\leqq x\leqq 0.7}{min}{\theta }_x=0.0100450\dots$$

and that the improved two-sided inequality for ${\theta }_x$ holds true for $0\leqq x<\infty$.

Mortici [25] presented the following analogous result to (2):

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\approx \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(16x^4+{\displaystyle \frac{32}{3}}{x}^3+{\displaystyle \frac{32}{9}}{x}^2+{\displaystyle \frac{176}{405}}x-{\displaystyle \frac{128}{1215}}\right)}^{1/8}(x\to \infty ).\end{array}$$

(5)

Chen [8] unified the formulas (4) and (5), and developed (5) into a complete asymptotic expansion. In fact, Chen [8] developed the approximation formula (2) to deduce the followin complete asymptotic expansion:

$$\begin{array}{cc}\hfill & \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\hfill \\ \hfill & ·\left(1-{\displaystyle \frac{\frac{11}{11520}}{x^4+\frac{78}{77}{x}^3+\frac{365579}{355740}{x}^2+\frac{11084441}{27391980}x-\frac{308425057271}{1349876774400}-\frac{2328181507}{3654158611950x}+\dots }}\right).\hfill \end{array}$$

(6)

Mortici [27] obtained, but without a formula for the general term, the following asymptotic series associated with the Ramanujan formula:

$$\begin{array}{cc}\hfill & \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\hfill \\ \hfill & ·exp\left(-{\displaystyle \frac{11}{11520x^4}}+{\displaystyle \frac{13}{13440x^5}}+{\displaystyle \frac{1}{691200x^6}}-{\displaystyle \frac{421}{691200x^7}}+{\displaystyle \frac{121}{22118400x^8}}+\dots \right).\hfill \end{array}$$

(7)

By using the Maple software, we find, as $x\to \infty$, that

$$\begin{array}{cc}\hfill & \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\hfill \\ \hfill & ·exp\left(-{\displaystyle \frac{\frac{11}{11520}}{{(x+\frac{39}{154})}^4}}+{\displaystyle \frac{\frac{228689}{372556800}}{{\left(x+\frac{26370613}{105654318}\right)}^6}}-{\displaystyle \frac{\frac{4573069844726921}{6927753293166182400}}{{\left(x+\frac{1810818106768144787791987}{7247468634164831017423170}\right)}^8}}+\dots \right),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\hfill \\ \hfill & ·\left(1-{\displaystyle \frac{\frac{11}{11520}}{{\left(x+\frac{39}{154}\right)}^4}}+{\displaystyle \frac{\frac{228689}{372556800}}{{\left(x+\frac{26370613}{105654318}\right)}^6}}-{\displaystyle \frac{\frac{18279646485804007}{27711013172664729600}}{{\left(x+\frac{7238202225594897335677933}{28969853741080785618783390}\right)}^8}}+\dots \right)\hfill \end{array}$$

(9)

and

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}-{\displaystyle \frac{\frac{11}{240}}{x+\frac{79}{154}}}+{\displaystyle \frac{\frac{459733}{15523200}}{{(x+\frac{71181889}{212396646})}^3}}\hfill \\ \hfill & -{\displaystyle \frac{\frac{125134498502528329}{4061997157910630400}}{{\left(x+\frac{597217044207994777948097107}{1993361083562177969968840050}\right)}^5}}+\dots {)}^{1/6},\hfill \end{array}$$

(10)

which led us to pose the following problem.

Problem. Find the constants $a_{\ell }$, $b_{\ell }$, ${\alpha }_{\ell }$, ${\beta }_{\ell }$, ${\lambda }_{\ell }$ and ${\mu }_{\ell }$ such that

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}exp\left(\sum _{\ell =2}^{\infty }{\displaystyle \frac{a_{\ell }}{{(x+b_{\ell })}^{2\ell }}}\right),\end{array}$$

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\left(1+\sum _{\ell =2}^{\infty }{\displaystyle \frac{{\alpha }_{\ell }}{{(x+{\beta }_{\ell })}^{2\ell }}}\right)\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}+\sum _{\ell =1}^{\infty }{\displaystyle \frac{{\lambda }_{\ell }}{{(x+{\mu }_{\ell })}^{2\ell -1}}}\right)}^{1/6}\end{array}$$

as $x\to \infty$.

Our first aim in this paper is to solve the above problem and our solution is stated in Theorem 1, Theorem 2 and Theomrem 3.

We now recall that a function g is said to be subadditive on an interval I if

$$g(x+y)\leqq g\left(x\right)+g\left(y\right)$$

for $x,y\in I$ and $x+y\in I$. If the inequality is reversed, then g is called superadditive. Subadditive and superadditive functions have applications in, for example, the theory of differential equations, in semigroup theory, and in the theory of convex bodies (see, for details, [3] and the references therein).

Subadditivity problems are also discussed in number theory. For example, Hardy and Littlewood proposed the following well-known (still unsettled) conjecture:

$$\pi (x+y)\leqq \pi \left(x\right)+\pi \left(y\right)$$

for all integers $x,y\geqq 2$. Here $\pi \left(x\right)$ denotes the number of primes not exceeding x (see [33] and [34]).

Motivated by (7), we define $\vartheta \left(x\right)$ by the following equality:

$$\mathsf{\Gamma }(x+1)=\sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}{e}^{\vartheta \left(x\right)}.$$

(11)

We call $\vartheta \left(x\right)$ the remainder of Ramanujan’s formula. Recently, Chen [10] presented some properties of $\vartheta \left(x\right)$ including, for example, monotonicity properties, inequalities and asymptotic expansions.

Nemes [32] presented the following approximation formula for the gamma function:

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{2\pi x}{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(1+{\displaystyle \frac{1}{12x^2-\frac{1}{10}}}\right)}^x(x\to \infty ).\end{array}$$

(12)

The formula (12) is stronger than the formula (2). In recent years, several inequalities and asymptotic expansions, which are associated with the Nemes formula (12), were investigated in [8,10,23,29].

In light of (12), we define $\Theta \left(x\right)$ by the equality

$$\mathsf{\Gamma }(x+1)=\sqrt{2\pi x}{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(1+{\displaystyle \frac{1}{12x^2-\frac{1}{10}}}\right)}^x{e}^{\Theta \left(x\right)}.$$

(13)

We call $\Theta \left(x\right)$ the remainder of Nemes’ formula. Chen [12] proved some properties of $\Theta \left(x\right)$ including, for example, monotonicity properties, asymptotic expansions and inequalities.

Chen [13] obtained the following approximation formula for the gamma function:

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{2\pi x}{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(1+{\displaystyle \frac{1}{12x^3+\frac{24}{7}x-\frac{1}{2}}}\right)}^{x^2+{\displaystyle \frac{53}{210}}}(x\to \infty ),\end{array}$$

(14)

which is more accurate than the Ramanujan formula and the Nemes formulas. We now define $\rho \left(x\right)$ as follows:

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)=\sqrt{2\pi x}{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(1+{\displaystyle \frac{1}{12x^3+\frac{24}{7}x-\frac{1}{2}}}\right)}^{x^2+{\displaystyle \frac{53}{210}}}{e}^{\rho \left(x\right)}.\end{array}$$

(15)

We call $\rho \left(x\right)$ the remainder of the formula (15).

The second aim of the present paper is to consider Subadditive properties for ${\theta }_x$ and $\Theta \left(x\right)$ (see Theorem 4 and Theorem 5). And, finally, we consider superadditive properties for $\vartheta \left(x\right)$ and $\rho \left(x\right)$ (see Theorem 6 and Theorem 7).

For several developments based upon the monotonicity properties, inequalities, limit formulas and so on, which are associated with the gamma and related functions, the interested reader may refer to [14,16,22,31,35].

The numerical values given in this paper have been calculated via the computer program Maple 13.

2. A Set of Lemmas

Chen [10] showed that

$$\begin{array}{c}\hfill ln\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}\right)\sim \sum _{j=1}^{\infty }{c}_j{x}^{-j}(x\to \infty )\end{array}$$

(16)

with the coefficients $c_j$ given by

$$c_j=a_j-{\displaystyle \frac{1}{j}}\sum _{k=1}^{j-1}k{c}_k{a}_{j-k}(j\geqq 1),$$

(17)

where

$$\begin{array}{c}\hfill a_1={\displaystyle \frac{1}{2}},a_2={\displaystyle \frac{1}{8}},a_3={\displaystyle \frac{1}{240}}\mathrm{and}{a}_j=0(j\geqq 4).\end{array}$$

(18)

The first few coefficients $c_j$ are given by

$$\begin{array}{cc}\hfill & c_1={\displaystyle \frac{1}{2}},c_2=0,c_3=-{\displaystyle \frac{1}{60}},c_4={\displaystyle \frac{11}{1920}},c_5=-{\displaystyle \frac{1}{960}},\hfill \\ \hfill & c_6=-{\displaystyle \frac{1}{115200}},c_7={\displaystyle \frac{67}{806400}},c_8=-{\displaystyle \frac{121}{3686400}},c_9={\displaystyle \frac{271}{41472000}}.\hfill \end{array}$$

Remark 1. 

We can give explicit representation of the coefficients $c_j$ in (16) as follows:

$$\begin{array}{c}\hfill c_j=\sum _{\begin{array}{c}{k}_1+2k_2+\dots +j{k}_j=j\\ k_1+k_2+\dots +k_j=k\\ 1\leqq k\leqq j\end{array}}{(-1)}^{k-1}(k-1)!{\displaystyle \frac{a_1^{k_1}{a}_2^{k_2}\dots a_j^{k_j}}{k_1!k_2!\dots k_j!}},\end{array}$$

where $a_j$ are given in (18) and the summation is taken over all nonnegative integer solutions $(k_1,k_2,\dots ,k_j)$ of the following equations:

$$k_1+2k_2+\dots +j{k}_j=j\mathrm{and}{k}_1+k_2+\dots +k_j=k(k=1,2,\dots ,j).$$

The representation using recursive algorithm is better for numerical evaluations. Furthermore, based on the coefficients $c_j$ in (16), Chen [10] established the asymptotic expansions for $\vartheta \left(x\right)$.

Lemma 1 

(see [10]). (i)The remainder $\vartheta \left(x\right)$ of Ramanujan’s formula has the following asymptotic expansion:

$$\begin{array}{c}\hfill \vartheta \left(x\right)\sim \sum _{j=1}^{\infty }{d}_j{x}^{-j}(x\to \infty )\end{array}$$

(19)

with the coefficients $d_j$ given by

$$\begin{array}{c}\hfill d_j={\displaystyle \frac{B_{j+1}}{j(j+1)}}-{\displaystyle \frac{c_j}{6}}(j\in \mathbb{N}),\end{array}$$

(20)

where $c_j$ are given in (17) and $B_n$ are the Bernoulli numbers defined by the following generating function:

$${\displaystyle \frac{z}{e^z-1}}=\sum _{n=0}^{\infty }{B}_n{\displaystyle \frac{z^n}{n!}}\left(\right|z|<2\pi ).$$

(21)

(ii)The function $exp\left(\vartheta \right(x\left)\right)$ has the following asymptotic expansion:

$$\begin{array}{c}\hfill exp\left(\vartheta \left(x\right)\right)\sim \sum _{j=0}^{\infty }{q}_j{x}^{-j}(x\to \infty ),\end{array}$$

(22)

with the coefficients $q_j$ given by

$$\begin{array}{c}\hfill q_0=1,q_j={\displaystyle \frac{1}{j}}\sum _{k=1}^{j}k{d}_k{q}_{j-k}(j\geqq 1),\end{array}$$

(23)

where $d_j$ are given by (20).

Remark 2. 

By using (20), as many coefficients as we please in the right-hand side of (7) can be obtained:

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\hfill \\ \hfill & ·exp(-{\displaystyle \frac{11}{11520x^4}}+{\displaystyle \frac{13}{13440x^5}}+{\displaystyle \frac{1}{691200x^6}}-{\displaystyle \frac{421}{691200x^7}}\hfill \\ \hfill & +{\displaystyle \frac{121}{22118400x^8}}+{\displaystyle \frac{2301019}{2737152000x^9}}-\dots ),x\to \infty .\hfill \end{array}$$

(24)

Moreover, from (11) and (22), we obtain the following asymptotic expansion of the gamma function:

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\left(\sum _{j=0}^{\infty }{q}_j{x}^{-j}\right)(x\to \infty ),\hfill \end{array}$$

(25)

where $q_j$ are given in (23), namely,

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\hfill \\ \hfill & ·(1-{\displaystyle \frac{11}{11520x^4}}+{\displaystyle \frac{13}{13440x^5}}+{\displaystyle \frac{1}{691200x^6}}-{\displaystyle \frac{421}{691200x^7}}\hfill \\ \hfill & +{\displaystyle \frac{1573}{265420800x^8}}+{\displaystyle \frac{64357747}{76640256000x^9}}+\dots )(x\to \infty ).\hfill \end{array}$$

(26)

Lemma 2 

(see [8]). Let $r\ne 0$ be a given real number. Then the gamma function has the following asymptotic expansion:

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(2^r{x}^r+\sum _{j=1}^{\infty }{p}_j{x}^{r-j}\right)}^{1/\left(2r\right)}(x\to \infty )\end{array}$$

(27)

with the coefficients $p_j\equiv p_j\left(r\right)$ given by

$$\begin{array}{c}\hfill p_j=2^r{b}_j(j\geqq 1),\end{array}$$

(28)

where $b_j\equiv b_j\left(r\right)$ are given by the recursive relation:

$$\begin{array}{c}\hfill b_0=1,b_j={\displaystyle \frac{1}{j}}\sum _{k=1}^j{\displaystyle \frac{2r{B}_{k+1}}{k+1}}{b}_{j-k}(j\geqq 1),\end{array}$$

(29)

and $B_n$ are the Bernoulli numbers generated by (21).

Lemma 3 

([2]). Let $n\geqq 0$ be an integer. The functions

$$F_n\left(x\right)=ln\mathsf{\Gamma }\left(x\right)-\left(x-{\displaystyle \frac{1}{2}}\right)lnx+x-{\displaystyle \frac{1}{2}}ln\left(2\pi \right)-\sum _{j=1}^{2n}{\displaystyle \frac{B_{2j}}{2j(2j-1)x^{2j-1}}}$$

and

$$G_n\left(x\right)=-ln\mathsf{\Gamma }\left(x\right)+\left(x-{\displaystyle \frac{1}{2}}\right)lnx-x+{\displaystyle \frac{1}{2}}ln\left(2\pi \right)+\sum _{j=1}^{2n+1}{\displaystyle \frac{B_{2j}}{2j(2j-1)x^{2j-1}}}$$

are strictly completely monotonic on $(0,\infty ),$$B_n$ being the Bernoulli numbers generated by (21)

Remark 3. 

Lemma 3 states that, for every $m\in {\mathbb{N}}_0,$ the function

$$R_m\left(x\right)={(-1)}^m\left[ln\mathsf{\Gamma }\left(x\right)-\left(x-{\displaystyle \frac{1}{2}}\right)lnx+x-ln\sqrt{2\pi }-\sum _{j=1}^m{\displaystyle \frac{B_{2j}}{2j(2j-1)x^{2j-1}}}\right]$$

is completely monotonic on $(0,\infty )$. As a matter of fact, in the year 2006, Koumandos [20] presented a simpler proof of the complete monotonicity property of $R_m\left(x\right),$ whereas Koumandos and Pedersen [21] strengthened this result.

From Lemma 3, we obtain, for $x>0$ and $m\in {\mathbb{N}}_0$, that

$$\sum _{j=1}^{2n}{\displaystyle \frac{B_{2j}}{2j(2j-1)x^{2j-1}}}<ln\mathsf{\Gamma }\left(x\right)-\left(x-{\displaystyle \frac{1}{2}}\right)lnx+x-{\displaystyle \frac{1}{2}}ln\left(2\pi \right)<\sum _{j=1}^{2n+1}{\displaystyle \frac{B_{2j}}{2j(2j-1)x^{2j-1}}}$$

and

$$\sum _{j=1}^{2m}{\displaystyle \frac{B_{2j}}{2j{x}^{2j}}}<lnx-\psi \left(x\right)-{\displaystyle \frac{1}{2x}}<\sum _{j=1}^{2m+1}{\displaystyle \frac{B_{2j}}{2j{x}^{2j}}}.$$

(30)

In particular, for $x>0$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{12x}}-{\displaystyle \frac{1}{360x^3}}+{\displaystyle \frac{1}{1260x^5}}-{\displaystyle \frac{1}{1680x^7}}+{\displaystyle \frac{1}{1188x^9}}-{\displaystyle \frac{691}{360360x^{11}}}\hfill \\ \hfill & <ln\left({\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{2\pi x}}}{\left({\displaystyle \frac{e}{x}}\right)}^x\right)<{\displaystyle \frac{1}{12x}}-{\displaystyle \frac{1}{360x^3}}+{\displaystyle \frac{1}{1260x^5}}-{\displaystyle \frac{1}{1680x^7}}+{\displaystyle \frac{1}{1188x^9}}\hfill \end{array}$$

(31)

and

$$\begin{array}{c}\hfill \psi \left(x\right)-lnx+{\displaystyle \frac{1}{2x}}<-{\displaystyle \frac{1}{12x^2}}+{\displaystyle \frac{1}{120x^4}}-{\displaystyle \frac{1}{252x^6}}+{\displaystyle \frac{1}{240x^8}}.\end{array}$$

(32)

Lemma 4. 

Let

$$\begin{array}{c}\hfill G\left(x\right)={\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{2\pi x}}}{\left({\displaystyle \frac{e}{x}}\right)}^x.\end{array}$$

(33)

Then, for $x\geqq 2,$

$$\begin{array}{c}\hfill {\left[G\left(x\right)\right]}^6<1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}-{\displaystyle \frac{11}{1920x^4}}+{\displaystyle \frac{79}{26880x^5}}+{\displaystyle \frac{3539}{1612800x^6}}.\end{array}$$

(34)

Proof. 

Using the second inequality in (31), we have

$$\begin{array}{cc}\hfill & lnG\left(x\right)-{\displaystyle \frac{1}{6}}ln\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}-{\displaystyle \frac{11}{1920x^4}}+{\displaystyle \frac{79}{26880x^5}}+{\displaystyle \frac{3539}{1612800x^6}}\right)\hfill \\ \hfill & <{\displaystyle \frac{1}{12x}}-{\displaystyle \frac{1}{360x^3}}+{\displaystyle \frac{1}{1260x^5}}-{\displaystyle \frac{1}{1680x^7}}+{\displaystyle \frac{1}{1188x^9}}\hfill \\ \hfill & -{\displaystyle \frac{1}{6}}ln\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}-{\displaystyle \frac{11}{1920x^4}}+{\displaystyle \frac{79}{26880x^5}}+{\displaystyle \frac{3539}{1612800x^6}}\right)=:F\left(x\right).\hfill \end{array}$$

In order to prove (34), it suffices to show that $F\left(x\right)<0$ for $x\geqq 2$. Indeed, upon differentiation with respect to x, we obtain

$$\begin{array}{c}\hfill F^{\prime }\left(x\right)={\displaystyle \frac{P_8(x-2)}{55440x^{10}(1612800x^6+806400x^5+201600x^4+6720x^3-9240x^2+4740x+3539)}},\end{array}$$

where

$$\begin{array}{cc}\hfill P_8\left(x\right)& =50316218728+248467730132x+497175600429x^2+541336258940x^3\hfill \\ \hfill & +355755510460x^4+145641719976x^5+36440322618x^6\hfill \\ \hfill & +5108361720x^7+307585740x^8.\hfill \end{array}$$

We then see that $F^{\prime }\left(x\right)>0$ for $x\geqq 2$. Hence, clearly, the function $F\left(x\right)$ is strictly increasing for $x\geqq 2$, and we have

$$\begin{array}{c}\hfill F\left(x\right)<\underset{t\to \infty }{lim}F\left(t\right)=0(x\geqq 2).\end{array}$$

The proof of Lemma 4 is thus completed. □

3. Asymptotic Expansions

In this section, we first state and prove the following result.

Theorem 1. 

The gamma function has the following asymptotic expansion:

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}exp\left(\sum _{\ell =2}^{\infty }{\displaystyle \frac{a_{\ell }}{{(x+b_{\ell })}^{2\ell }}}\right),\hfill \end{array}$$

(35)

where $a_{\ell }$ and $b_{\ell }$ are given by a pair of recurrence relations as follows:

$$a_{\ell }=-{\displaystyle \frac{c_{2\ell }}{6}}-\sum _{k=2}^{\ell -1}{a}_k{b}_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k}}\right)(\ell \geqq 3)$$

(36)

and

$$b_{\ell }={\displaystyle \frac{1}{2\ell a_{\ell }}}\left\{{\displaystyle \frac{c_{2\ell +1}}{6}}-\sum _{k=2}^{\ell -1}{a}_k{b}_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell }{2\ell -2k+1}}\right)-{\displaystyle \frac{B_{2\ell +2}}{(2\ell +1)(2\ell +2)}}\right\}(\ell \geqq 3),$$

(37)

with

$$a_2=-{\displaystyle \frac{11}{11520}}\mathrm{and}{b}_2={\displaystyle \frac{39}{154}}.$$

Here $c_j$ are given in (17) and $B_n$ are the Bernoulli numbers generated by (21).

Proof. 

In view of (8), we let

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}exp\left(\sum _{\ell =2}^{\infty }{\displaystyle \frac{a_{\ell }}{{(x+b_{\ell })}^{2\ell }}}\right),\hfill \end{array}$$

where $a_{\ell }$ and $b_{\ell }$ are real numbers to be determined. This can be written by (16) as follows:

$$ln{\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{2\pi x}{(x/e)}^x}}\sim \sum _{j=1}^{\infty }{\displaystyle \frac{c_j}{6x^j}}+\sum _{j=2}^{\infty }{\displaystyle \frac{a_j}{x^{2j}}}{\left(1+{\displaystyle \frac{b_j}{x}}\right)}^{-2j},$$

(38)

where $c_j$ are given in (17). Direct computation yields

$$\begin{array}{cc}\hfill \sum _{j=2}^{\infty }{\displaystyle \frac{a_j}{x^{2j}}}{\left(1+{\displaystyle \frac{b_j}{x}}\right)}^{-2j}& =\sum _{j=2}^{\infty }{\displaystyle \frac{a_j}{x^{2j}}}\sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{-2j}{k}}\right){\displaystyle \frac{b_j^k}{x^k}}\hfill \\ \hfill & =\sum _{j=2}^{\infty }{\displaystyle \frac{a_j}{x^{2j}}}\sum _{k=0}^{\infty }{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k+2j-1}{k}}\right){\displaystyle \frac{b_j^k}{x^k}}\hfill \\ \hfill & =\sum _{j=4}^{\infty }\sum _{k=0}^{j-4}{a}_{k+2}{b}_{k+2}^{j-k-4}{(-1)}^{j-k}\left({\displaystyle \genfrac{}{}{0pt}{}{j+k-1}{j-k-4}}\right){\displaystyle \frac{1}{x^{j+k}}},\hfill \end{array}$$

which can be written as follows:

$$\begin{array}{c}\hfill \sum _{j=2}^{\infty }{\displaystyle \frac{a_j}{x^{2j}}}{\left(1+{\displaystyle \frac{b_j}{x}}\right)}^{-2j}\sim \sum _{j=4}^{\infty }\left\{\sum _{k=2}^{\lfloor j/2\rfloor }{a}_k{b}_k^{j-2k}{(-1)}^{j-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k}}\right)\right\}{\displaystyle \frac{1}{x^j}}.\end{array}$$

(39)

Now, upon substituting from (39) into (38), we have

$$ln{\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{2\pi x}{(x/e)}^x}}\sim {\displaystyle \frac{1}{12x}}-{\displaystyle \frac{1}{360x^3}}+\sum _{j=4}^{\infty }\left\{{\displaystyle \frac{c_j}{6}}+\sum _{k=2}^{\lfloor j/2\rfloor }{a}_k{b}_k^{j-2k}{(-1)}^{j-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k}}\right)\right\}{\displaystyle \frac{1}{x^j}}.$$

(40)

It is well known that (see [1])

$$ln\left({\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{2\pi x}{(x/e)}^x}}\right)\sim \sum _{j=1}^{\infty }{\displaystyle \frac{B_{j+1}}{j(j+1)x^j}}.$$

(41)

Equating coefficients of the term $x^{-j}$ on the right-hand sides of (40) and (41), we obtain

$${\displaystyle \frac{c_j}{6}}+\sum _{k=2}^{\lfloor j/2\rfloor }{a}_k{b}_k^{j-2k}{(-1)}^{j-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k}}\right)={\displaystyle \frac{B_{j+1}}{j(j+1)}}(j\geqq 4).$$

(42)

If we set $j=2\ell$ and $j=2\ell +1$ in (42), we obtain

$${\displaystyle \frac{c_{2\ell }}{6}}+\sum _{k=2}^{\ell }{a}_k{b}_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k}}\right)=0$$

(43)

and

$${\displaystyle \frac{c_{2\ell +1}}{6}}-\sum _{k=2}^{\ell }{a}_k{b}_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell }{2\ell -2k+1}}\right)={\displaystyle \frac{B_{2\ell +2}}{(2\ell +1)(2\ell +2)}},$$

(44)

respectively. For $\ell =2$, from (43) and (44), we get

$$\begin{array}{c}\hfill a_2=-{\displaystyle \frac{11}{11520}}\mathrm{and}{b}_2={\displaystyle \frac{39}{154}}.\end{array}$$

Also, for $\ell \geqq 3$, we have

$${\displaystyle \frac{c_{2\ell }}{6}}+\sum _{k=2}^{\ell -1}{a}_k{b}_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k}}\right)+a_{\ell }=0$$

and

$${\displaystyle \frac{c_{2\ell +1}}{6}}-\sum _{k=2}^{\ell -1}{a}_k{b}_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell }{2\ell -2k+1}}\right)-2\ell a_{\ell }{b}_{\ell }={\displaystyle \frac{B_{2\ell +2}}{(2\ell +1)(2\ell +2)}}.$$

We are led easily to the recurrence relations (36) and (37). This evidently completes proof of Theorem 1. □

Here we give explicit numerical values of some first terms of $a_{\ell }$ and $b_{\ell }$ by using the formulas (36) and (37). This shows how easily we can determine the constants $a_{\ell }$ and $b_{\ell }$ in (35).

$$\begin{array}{cc}\hfill a_2& =-{\displaystyle \frac{11}{11520}},b_2={\displaystyle \frac{39}{154}},\hfill \\ \hfill a_3& =-{\displaystyle \frac{1}{6}}{c}_6-10a_2{b}_2^2=-{\displaystyle \frac{1}{6}}·\left(-{\displaystyle \frac{1}{115200}}\right)-10·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^2={\displaystyle \frac{228689}{372556800}},\hfill \\ \hfill b_3& =-{\displaystyle \frac{-280c_7+33600a_2{b}_2^3-1}{10080a_3}}\hfill \\ \hfill & =-{\displaystyle \frac{-280·\left({\displaystyle \frac{67}{806400}}\right)+33600·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^3-1}{10080·\left({\displaystyle \frac{228689}{372556800}}\right)}}={\displaystyle \frac{26370613}{105654318}},\hfill \\ \hfill a_4& =-{\displaystyle \frac{1}{6}}{c}_8-35a_2{b}_2^4-21a_3{b}_3^2\hfill \\ \hfill & =-{\displaystyle \frac{1}{6}}·\left(-{\displaystyle \frac{121}{3686400}}\right)-35·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^4-21·\left({\displaystyle \frac{228689}{372556800}}\right)·{\left({\displaystyle \frac{26370613}{105654318}}\right)}^2\hfill \\ \hfill & =-{\displaystyle \frac{4573069844726921}{6927753293166182400}},\hfill \\ \hfill b_4& =-{\displaystyle \frac{-198c_9+66528a_2{b}_2^5+66528a_3{b}_3^3+1}{9504a_4}}\hfill \\ \hfill & =-{\displaystyle \frac{-198·\left(\frac{271}{41472000}\right)+66528·\left(-\frac{11}{11520}\right)·{\left(\frac{39}{154}\right)}^5+66528·\left(\frac{228689}{372556800}\right)·{\left(\frac{26370613}{105654318}\right)}^3+1}{9504·\left(-\frac{4573069844726921}{6927753293166182400}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1810818106768144787791987}{7247468634164831017423170}}.\hfill \end{array}$$

We note that the values of $a_{\ell }$ and $b_{\ell }$ (for $\ell =2,3,4$) are equal to the constants of the term ${\displaystyle \frac{a_{\ell }}{{(x+b_{\ell })}^{2\ell }}}$ (for $\ell =2,3,4$) in (8), respectively. We thus obtain an alternating even-type asymptotic series for $\mathsf{\Gamma }(x+1)$. From a computational viewpoint, the formula (8) improves the formula the formula (24).

Theorem 2. 

The gamma function has the following asymptotic expansion:

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\left(1+\sum _{\ell =2}^{\infty }{\displaystyle \frac{{\alpha }_{\ell }}{{(x+{\beta }_{\ell })}^{2\ell }}}\right),\hfill \end{array}$$

(45)

where ${\alpha }_{\ell }$ and ${\beta }_{\ell }$ are given by a pair of recurrence relations as follows:

$${\alpha }_{\ell }=q_{2\ell }-\sum _{k=2}^{\ell -1}{\alpha }_k{\beta }_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k}}\right)(\ell \geqq 3)$$

(46)

and

$${\beta }_{\ell }=-{\displaystyle \frac{1}{2\ell {\alpha }_{\ell }}}\left[q_{2\ell +1}+\sum _{k=2}^{\ell -1}{\alpha }_k{\beta }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell }{2\ell -2k+1}}\right)\right](\ell \geqq 3)$$

(47)

with

$${\alpha }_2=-{\displaystyle \frac{11}{11520}}\mathrm{and}{\beta }_2={\displaystyle \frac{39}{154}}.$$

Here $q_j$ are given in (23).

Proof. 

In view of (9), we put

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}\right)}^{1/6}\left(1+\sum _{\ell =2}^{\infty }{\displaystyle \frac{{\alpha }_{\ell }}{{(x+{\beta }_{\ell })}^{2\ell }}}\right),\end{array}$$

where ${\alpha }_{\ell }$ and ${\beta }_{\ell }$ are real numbers to be determined. This can be written by (39) as follows:

$$\begin{array}{cc}\hfill \mathsf{\Gamma }(x+1)& \sim \sqrt{2\pi x}{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}\right)}^{1/6}\hfill \\ \hfill & ·\left(1+\sum _{j=4}^{\infty }\left\{\sum _{k=2}^{\lfloor j/2\rfloor }{\alpha }_k{\beta }_k^{j-2k}{(-1)}^{j-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k}}\right)\right\}{\displaystyle \frac{1}{x^j}}\right).\hfill \end{array}$$

(48)

From (25) and (48), we have

$$\sum _{j=0}^{\infty }{q}_j{x}^{-j}=1+\sum _{j=4}^{\infty }\left\{\sum _{k=2}^{\lfloor j/2\rfloor }{\alpha }_k{\beta }_k^{j-2k}{(-1)}^{j-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k}}\right)\right\}{\displaystyle \frac{1}{x^j}},$$

(49)

where $q_j$ are given in (23). Equating the coefficients of the term $x^{-j}$ on both sides of (49), we obtain

$$q_j=\sum _{k=2}^{\lfloor j/2\rfloor }{\alpha }_k{\beta }_k^{j-2k}{(-1)}^{j-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k}}\right)(j\geqq 4).$$

(50)

Upon setting $j=2\ell$ and $j=2\ell +1$ in (50), we get

$$q_{2\ell }=\sum _{k=2}^{\ell }{\alpha }_k{\beta }_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k}}\right)$$

(51)

and

$$q_{2\ell +1}=-\sum _{k=2}^{\ell }{\alpha }_k{\beta }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell }{2\ell -2k+1}}\right).$$

(52)

respectively. For $\ell =2$, from (51) and (52) we obtain

$$\begin{array}{c}\hfill {\alpha }_2=-{\displaystyle \frac{11}{11520}}\mathrm{and}{\beta }_2={\displaystyle \frac{39}{154}}.\end{array}$$

Also, for $\ell \geqq 3$, we have

$$q_{2\ell }=\sum _{k=2}^{\ell -1}{\alpha }_k{\beta }_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k}}\right)+{\alpha }_{\ell }$$

and

$$q_{2\ell +1}=-\sum _{k=2}^{\ell -1}{\alpha }_k{\beta }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell }{2\ell -2k+1}}\right)-2\ell {\alpha }_{\ell }{\beta }_{\ell }.$$

We then obtain the recurrence relations (46) and (47). The proof of Theorem 2 is thus completed. □

Here we give explicit numerical values of the first few terms of $a_{\ell }$ and $b_{\ell }$ by using the formulas (46) and (47). This shows how easily we can determine the constants $a_{\ell }$ and $b_{\ell }$ in (45).

$$\begin{array}{cc}\hfill {\alpha }_2& =-{\displaystyle \frac{11}{11520}},{\beta }_2={\displaystyle \frac{39}{154}},\hfill \\ \hfill {\alpha }_3& =q_6-10{\alpha }_2{\beta }_2^2={\displaystyle \frac{1}{691200}}-10·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^2={\displaystyle \frac{228689}{372556800}},\hfill \\ \hfill {\beta }_3& =-{\displaystyle \frac{q_7+20{\alpha }_2{\beta }_2^3}{6{\alpha }_3}}=-{\displaystyle \frac{-{\displaystyle \frac{421}{691200}}+20·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^3}{6·\left({\displaystyle \frac{228689}{372556800}}\right)}}={\displaystyle \frac{26370613}{105654318}},\hfill \\ \hfill {\alpha }_4& =q_8-35{\alpha }_2{\beta }_2^4-21{\alpha }_3{\beta }_3^2\hfill \\ \hfill & ={\displaystyle \frac{1573}{265420800}}-35·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^4-21·\left({\displaystyle \frac{228689}{372556800}}\right)·{\left({\displaystyle \frac{26370613}{105654318}}\right)}^2\hfill \\ \hfill & =-{\displaystyle \frac{18279646485804007}{27711013172664729600}},\hfill \\ \hfill {\beta }_4& =-{\displaystyle \frac{q_9+56{\alpha }_2{\beta }_2^5+56{\alpha }_3{\beta }_3^3}{8{\alpha }_4}}\hfill \\ \hfill & =-{\displaystyle \frac{{\displaystyle \frac{64357747}{76640256000}}+56·\left(-{\displaystyle \frac{11}{11520}}\right)·{\left({\displaystyle \frac{39}{154}}\right)}^5+56·\left({\displaystyle \frac{228689}{372556800}}\right){\left({\displaystyle \frac{26370613}{105654318}}\right)}^3}{8·\left(-{\displaystyle \frac{18279646485804007}{27711013172664729600}}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{7238202225594897335677933}{28969853741080785618783390}}.\hfill \end{array}$$

We note that the values of ${\alpha }_{\ell }$ and ${\beta }_{\ell }$ (for $\ell =2,3,4$) are equal to the constants of the term ${\displaystyle \frac{{\alpha }_{\ell }}{{(x+{\beta }_{\ell })}^{2\ell }}}$ (for $\ell =2,3,4$) in (9), respectively. We can obtain an alternating even-type asymptotic series for $\mathsf{\Gamma }(x+1)$. From a computational viewpoint, the formula (9) improves the formula (26).

Theorem 3. 

The gamma function has the following asymptotic expansion:

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}+\sum _{\ell =1}^{\infty }{\displaystyle \frac{{\lambda }_{\ell }}{{(x+{\mu }_{\ell })}^{2\ell -1}}}\right)}^{1/6}(x\to \infty ),\end{array}$$

(53)

where ${\lambda }_{\ell }$ and ${\mu }_{\ell }$ are given by a pair of recurrence relations as follows:

$${\lambda }_{\ell }=p_{2\ell +2}-\sum _{k=1}^{\ell -1}{\lambda }_k{\mu }_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -2}{2\ell -2k}}\right)(\ell \geqq 2)$$

(54)

and

$$\begin{array}{c}\hfill {\mu }_{\ell }=-{\displaystyle \frac{1}{(2\ell -1){\lambda }_{\ell }}}\left\{p_{2\ell +3}+\sum _{k=1}^{\ell -1}{\lambda }_k{\mu }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k+1}}\right)\right\}(\ell \geqq 2)\end{array}$$

(55)

with

$${\lambda }_1=-{\displaystyle \frac{11}{240}}\mathrm{and}{\mu }_1={\displaystyle \frac{79}{154}}.$$

Here $p_j$ are given by

$$\begin{array}{c}\hfill p_j=8b_j(j\geqq 1)\end{array}$$

(56)

and

$$\begin{array}{c}\hfill b_0=1,b_j={\displaystyle \frac{1}{j}}\sum _{k=1}^j{\displaystyle \frac{6B_{k+1}}{k+1}}{b}_{j-k}(j\geqq 1).\end{array}$$

(57)

Proof. 

In view of (10), we set

$$\begin{array}{c}\hfill \mathsf{\Gamma }(x+1)\sim \sqrt{\pi }{\left({\displaystyle \frac{x}{e}}\right)}^x{\left(8x^3+4x^2+x+{\displaystyle \frac{1}{30}}+\sum _{\ell =1}^{\infty }{\displaystyle \frac{{\lambda }_{\ell }}{{(x+{\mu }_{\ell })}^{2\ell -1}}}\right)}^{1/6}(x\to \infty ),\end{array}$$

where ${\lambda }_{\ell }$ and ${\nu }_{\ell }$ are real numbers to be determined. This can be written as follows:

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{\pi }{\left(x/e\right)}^x}}\right)}^6-8x^3-4x^2-x-{\displaystyle \frac{1}{30}}\sim \sum _{j=1}^{\infty }{\displaystyle \frac{{\lambda }_j}{x^{2j-1}}}{\left(1+{\displaystyle \frac{{\mu }_j}{x}}\right)}^{-2j+1}.\end{array}$$

(58)

Direct computation yields

$$\begin{array}{cc}\hfill \sum _{j=1}^{\infty }{\displaystyle \frac{{\lambda }_j}{x^{2j-1}}}{\left(1+{\displaystyle \frac{{\mu }_j}{x}}\right)}^{-2j+1}& =\sum _{j=1}^{\infty }{\displaystyle \frac{{\lambda }_j}{x^{2j-1}}}\sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{-2j+1}{k}}\right){\displaystyle \frac{{\mu }_j^k}{x^k}}\hfill \\ \hfill & =\sum _{j=1}^{\infty }{\displaystyle \frac{{\lambda }_j}{x^{2j-1}}}\sum _{k=0}^{\infty }{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k+2j-2}{k}}\right){\displaystyle \frac{{\mu }_j^k}{x^k}}\hfill \\ \hfill & =\sum _{j=1}^{\infty }\sum _{k=0}^{j-1}{\lambda }_{k+1}{\mu }_{k+1}^{j-k-1}{(-1)}^{j-k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j+k-1}{j-k-1}}\right){\displaystyle \frac{1}{x^{j+k}}},\hfill \end{array}$$

which can be written as follows:

$$\begin{array}{c}\hfill \sum _{j=1}^{\infty }{\displaystyle \frac{{\lambda }_j}{x^{2j-1}}}{\left(1+{\displaystyle \frac{{\mu }_j}{x}}\right)}^{-2j+1}\sim \sum _{j=1}^{\infty }\left\{\sum _{k=1}^{\lfloor {\displaystyle \frac{j+2}{2}}\rfloor }{\lambda }_k{\mu }_k^{j-2k+1}{(-1)}^{j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k+1}}\right)\right\}{\displaystyle \frac{1}{x^j}}.\end{array}$$

(59)

Upon substituting from (59) into (58), we have

$${\left({\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{\pi }{\left(x/e\right)}^x}}\right)}^6-8x^3-4x^2-x-{\displaystyle \frac{1}{30}}\sim \sum _{j=1}^{\infty }\left\{\sum _{k=1}^{\lfloor {\displaystyle \frac{j+2}{2}}\rfloor }{\lambda }_k{\mu }_k^{j-2k+1}{(-1)}^{j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k+1}}\right)\right\}{\displaystyle \frac{1}{x^j}}.$$

(60)

On the other hand, it follows from Lemma 2 (with $r=3$) that

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{\mathsf{\Gamma }(x+1)}{\sqrt{\pi }{\left(x/e\right)}^x}}\right)}^6-8x^3-4x^2-x-{\displaystyle \frac{1}{30}}\sim \sum _{j=1}^{\infty }{\displaystyle \frac{p_{j+3}}{x^j}}\end{array}$$

(61)

with

$$\begin{array}{c}\hfill p_j=8b_j(j\geqq 1)\end{array}$$

and

$$\begin{array}{c}\hfill b_0=1,b_j={\displaystyle \frac{1}{j}}\sum _{k=1}^j{\displaystyle \frac{6B_{k+1}}{k+1}}{b}_{j-k}(j\geqq 1).\end{array}$$

Equating the coefficients of the term $x^{-j}$ on the right-hand sides of (60) and (61), we obtain

$$p_{j+3}=\sum _{k=1}^{\lfloor {\displaystyle \frac{j+2}{2}}\rfloor }{\lambda }_k{\mu }_k^{j-2k+1}{(-1)}^{j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{j-2k+1}}\right)(j\in \mathbb{N}).$$

(62)

If we set $j=2\ell -1$ and $j=2\ell$ in (62), we get

$$p_{2\ell +2}=\sum _{k=1}^{\ell }{\lambda }_k{\mu }_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -2}{2\ell -2k}}\right)$$

(63)

and

$$\begin{array}{cc}\hfill p_{2\ell +3}& =-\sum _{k=1}^{\ell +1}{\lambda }_k{\mu }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k+1}}\right)\hfill \\ \hfill & =-\sum _{k=1}^{\ell }{\lambda }_k{\mu }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k+1}}\right)-{\lambda }_{\ell +1}{\mu }_{\ell +1}^{-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{-1}}\right)\hfill \\ \hfill & =-\sum _{k=1}^{\ell }{\lambda }_k{\mu }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k+1}}\right),\hfill \end{array}$$

(64)

respectively. For $\ell =1$, from (63) and (64), we obtain

$$\begin{array}{c}\hfill {\lambda }_1=p_4=-{\displaystyle \frac{11}{240}}\mathrm{and}{\mu }_1=-{\displaystyle \frac{p_5}{{\lambda }_1}}={\displaystyle \frac{79}{154}}.\end{array}$$

Also, for $\ell \geqq 2$, we have

$$p_{2\ell +2}=\sum _{k=1}^{\ell -1}{\lambda }_k{\mu }_k^{2\ell -2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -2}{2\ell -2k}}\right)+{\lambda }_{\ell }$$

and

$$\begin{array}{cc}\hfill p_{2\ell +3}& =-\sum _{k=1}^{\ell -1}{\lambda }_k{\mu }_k^{2\ell -2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2\ell -1}{2\ell -2k+1}}\right)-(2\ell -1){\lambda }_{\ell }{\mu }_{\ell }.\hfill \end{array}$$

We are now led to the recurrence relations (54) and (55). The proof of Theorem 3 is thus completed. □

Here we give explicit numerical values of the first few terms of $a_{\ell }$ and $b_{\ell }$ by using the formulas (54) and (55). This shows how easily we can determine the constants $a_{\ell }$ and $b_{\ell }$ in (53).

$$\begin{array}{cc}\hfill {\lambda }_1& =-{\displaystyle \frac{11}{240}},{\mu }_1={\displaystyle \frac{79}{154}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\lambda }_2& =p_6-{\lambda }_1{\mu }_1^2={\displaystyle \frac{3539}{201600}}-\left(-{\displaystyle \frac{11}{240}}\right)·{\left({\displaystyle \frac{79}{154}}\right)}^2={\displaystyle \frac{459733}{15523200}},\hfill \\ \hfill {\mu }_2& =-{\displaystyle \frac{p_7+{\lambda }_1{\mu }_1^3}{3{\lambda }_2}}=-{\displaystyle \frac{-{\displaystyle \frac{9511}{403200}}+\left(-{\displaystyle \frac{11}{240}}\right)·{\left({\displaystyle \frac{79}{154}}\right)}^3}{3·\left({\displaystyle \frac{459733}{15523200}}\right)}}={\displaystyle \frac{71181889}{212396646}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\lambda }_3& =p_8-{\lambda }_1{\mu }_1^4-6{\lambda }_2{\mu }_2^2\hfill \\ \hfill & =-{\displaystyle \frac{10051}{716800}}-\left(-{\displaystyle \frac{11}{240}}\right)·{\left({\displaystyle \frac{79}{154}}\right)}^4-6·\left({\displaystyle \frac{459733}{15523200}}\right)·{\left({\displaystyle \frac{71181889}{212396646}}\right)}^2\hfill \\ \hfill & =-{\displaystyle \frac{125134498502528329}{4061997157910630400}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\mu }_3& =-{\displaystyle \frac{p_9+{\lambda }_1{\mu }_1^5+10{\lambda }_2{\mu }_2^3}{5{\lambda }_3}}\hfill \\ \hfill & =-{\displaystyle \frac{{\displaystyle \frac{233934691}{6386688000}}+\left(-{\displaystyle \frac{11}{240}}\right)·{\left({\displaystyle \frac{79}{154}}\right)}^5+10·\left({\displaystyle \frac{459733}{15523200}}\right)·{\left({\displaystyle \frac{71181889}{212396646}}\right)}^3}{5·\left(-{\displaystyle \frac{125134498502528329}{4061997157910630400}}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{597217044207994777948097107}{1993361083562177969968840050}}.\hfill \end{array}$$

We note that the values of ${\lambda }_{\ell }$ and ${\mu }_{\ell }$ (for $\ell =1,2,3$) are equal to the constants of the term ${\displaystyle \frac{{\lambda }_{\ell }}{{(x+{\mu }_{\ell })}^{2\ell -1}}}$ (for $\ell =1,2,3$) in (9), respectively. We thus obtain an alternating odd-type asymptotic series for $\mathsf{\Gamma }(x+1)$. From a computational viewpoint, the formula (10) improves the Ramanujan-Karatsuba formula (4).

It follows from (8), (9) and (10) that

$$\begin{array}{c}\hfill n!\sim \sqrt{\pi }{\left({\displaystyle \frac{n}{e}}\right)}^n{\left(8n^3+4n^2+n+{\displaystyle \frac{1}{30}}\right)}^{1/6}exp\left(-{\displaystyle \frac{\frac{11}{11520}}{{(n+\frac{39}{154})}^4}}\right)=:u_n,\end{array}$$

(65)

$$\begin{array}{c}\hfill n!\sim \sqrt{\pi }{\left({\displaystyle \frac{n}{e}}\right)}^n{\left(8n^3+4n^2+n+{\displaystyle \frac{1}{30}}\right)}^{1/6}\left(1-{\displaystyle \frac{\frac{11}{11520}}{{(n+\frac{39}{154})}^4}}\right)=:v_n\end{array}$$

(66)

and

$$\begin{array}{c}\hfill n!\sim \sqrt{\pi }{\left({\displaystyle \frac{n}{e}}\right)}^n{\left(8n^3+4n^2+n+{\displaystyle \frac{1}{30}}-{\displaystyle \frac{\frac{11}{240}}{n+\frac{79}{154}}}\right)}^{1/6}=:w_n.\end{array}$$

(67)

Moreover, as $n\to \infty$, we have

$$\begin{array}{cc}& n!=u_n(1+O\left(n^{-6}\right)),n!=v_n(1+O\left(n^{-6}\right))\mathrm{and}n!=w_n(1+O\left(n^{-6}\right)).\hfill \end{array}$$

It is observed from Table 1 that, among the approximation formulas (65) to (67), for $n\in \mathbb{N}$, the formula (65) happens to be the best one.

Table 1. Comparison between the approximation formulas (65) to (67).

n	${\displaystyle \frac{n!-u_n}{n!}}$	${\displaystyle \frac{n!-v_n}{n!}}$	${\displaystyle \frac{n!-w_n}{n!}}$
1	$1.037\times {10}^{-4}$	$1.038\times {10}^{-4}$	$1.045\times {10}^{-4}$
10	$5.24108\times {10}^{-10}$	$5.24112\times {10}^{-10}$	$5.2647\times {10}^{-10}$
100	$6.04659108\times {10}^{-16}$	$6.04659153\times {10}^{-16}$	$6.0772\times {10}^{-16}$
1000	$6.129174881\times {10}^{-22}$	$6.129174886\times {10}^{-22}$	$6.1606\times {10}^{-22}$

Table 1. Comparison between the approximation formulas (65) to (67).

n	${\displaystyle \frac{n!-u_n}{n!}}$	${\displaystyle \frac{n!-v_n}{n!}}$	${\displaystyle \frac{n!-w_n}{n!}}$
1	$1.037\times {10}^{-4}$	$1.038\times {10}^{-4}$	$1.045\times {10}^{-4}$
10	$5.24108\times {10}^{-10}$	$5.24112\times {10}^{-10}$	$5.2647\times {10}^{-10}$
100	$6.04659108\times {10}^{-16}$	$6.04659153\times {10}^{-16}$	$6.0772\times {10}^{-16}$
1000	$6.129174881\times {10}^{-22}$	$6.129174886\times {10}^{-22}$	$6.1606\times {10}^{-22}$

4. Subadditive Property

Our first result in this section is contained in Theorem 4 below.

Theorem 4. 

The correction term ${\theta }_x$ of Ramanujan’s formula (1) satisfies the following inequality:

$$\begin{array}{c}\hfill {\theta }_x+{\theta }_y>{\theta }_{x+y}(x,y\geqq 2).\end{array}$$

(68)

Proof. 

We first write (3) as follows:

$$\begin{array}{c}\hfill {\displaystyle \frac{{\theta }_x}{30x}}=8x^2{\left(G\left(x\right)\right)}^6-(8x^2+4x+1),\end{array}$$

where $G\left(x\right)$ is defined in (33). Then, by using the inequalities (32) and (34), we find for $x\geqq 2$ that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{480x}}{\left({\displaystyle \frac{{\theta }_x}{x}}\right)}^{\prime }& ={\left(G\left(x\right)\right)}^6\left[1+3x\left(\psi \left(x\right)-lnx+{\displaystyle \frac{1}{2x}}\right)\right]-1-{\displaystyle \frac{1}{4x}}\hfill \\ \hfill & <\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}-{\displaystyle \frac{11}{1920x^4}}+{\displaystyle \frac{79}{26880x^5}}+{\displaystyle \frac{3539}{1612800x^6}}\right)\hfill \\ \hfill & ·\left[1+3x\left(-{\displaystyle \frac{1}{12x^2}}+{\displaystyle \frac{1}{120x^4}}-{\displaystyle \frac{1}{252x^6}}+{\displaystyle \frac{1}{240x^8}}\right)\right]-1-{\displaystyle \frac{1}{4x}}\hfill \\ \hfill & =-{\displaystyle \frac{P_{10}(x-2)}{2709504000x^{13}}},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill P_{10}\left(x\right)& =6101113+3820147324x+14712540868x^2+26124948576x^3+28131938562x^4\hfill \\ \hfill & +20222466600x^5+10011267420x^6+3386675040x^7+748591200x^8\hfill \\ \hfill & +97372800x^9+5644800x^{10}.\hfill \end{array}$$

We easily see that

$${\left({\displaystyle \frac{{\theta }_x}{x}}\right)}^{\prime }<0(x\geqq 2).$$

Therefore, the function $x\mapsto {\displaystyle \frac{{\theta }_x}{x}}$ is strictly decreasing for $x\geqq 2$, and we have

$$\begin{array}{c}\hfill {\theta }_x>{\displaystyle \frac{x}{x+y}}{\theta }_{x+y}\mathrm{and}{\theta }_y>{\displaystyle \frac{y}{x+y}}{\theta }_{x+y}(x,y\geqq 2).\end{array}$$

Adding these two expressions, we obtain (68). This completes the proof of Theorem 4. □

Remark 4. 

Some computer experiments indicate that the function $x\mapsto {\displaystyle \frac{{\theta }_x}{x}}$ is strictly decreasing for $x>0$. This implies that

$${\theta }_x+{\theta }_y>{\theta }_{x+y}(\forall x,y>0).$$

Theorem 5. 

The remainder $\Theta \left(x\right)$ of Nemes’ formula (13) satisfies the following inequality:

$$\begin{array}{c}\hfill \Theta \left(x\right)+\Theta \left(y\right)>\Theta (x+y)(x,y\geqq 1).\end{array}$$

(69)

Proof. 

It follows from (13) that

$$\begin{array}{c}\hfill {\displaystyle \frac{\Theta \left(x\right)}{x}}={\displaystyle \frac{1}{x}}ln\mathsf{\Gamma }\left(x\right)-\left(1-{\displaystyle \frac{1}{2x}}\right)lnx+1-{\displaystyle \frac{ln\left(\sqrt{2\pi }\right)}{x}}-ln\left(1+{\displaystyle \frac{1}{12x^2-\frac{1}{10}}}\right),\end{array}$$

which, upon differentiation with respect to x, yields

$$\begin{array}{cc}\hfill x^2{\left({\displaystyle \frac{\Theta \left(x\right)}{x}}\right)}^{\prime }& =x\psi \left(x\right)-ln\mathsf{\Gamma }\left(x\right)-{\displaystyle \frac{1}{2}}lnx+ln\sqrt{2\pi }\hfill \\ \hfill & -{\displaystyle \frac{9600x^5-4800x^4-960x^3-320x^2-6x+3}{2(40x^2+3)(120x^2-1)}}=:g\left(x\right)\hfill \end{array}$$

and

$$\begin{array}{c}\hfill g^{\prime }\left(x\right)=x{\psi }^{\prime }\left(x\right)-{\displaystyle \frac{h\left(x\right)}{2x{(40x^2+3)}^2{(120x^2-1)}^2}}\end{array}$$

with

$$\begin{array}{cc}\hfill h\left(x\right)& =46080000x^9+23040000x^8+13824000x^7+3072000x^6-364800x^5\hfill \\ \hfill & +73600x^4+10560x^3-1920x^2+18x+9.\hfill \end{array}$$

Now, by using the asymptotic formulas for $ln\mathsf{\Gamma }\left(x\right)$ and $\psi \left(x\right)$ (see [1]), we have

$$\begin{array}{c}\hfill g\left(x\right)=-{\displaystyle \frac{2369}{604800x^5}}+O\left({\displaystyle \frac{1}{x^7}}\right)(x\to \infty ),\end{array}$$

which implies that

$$\begin{array}{c}\hfill \underset{x\to \infty }{lim}g\left(x\right)=0.\end{array}$$

Using the first inequality in (71), we find for $x\geqq 1$ that

$$\begin{array}{cc}\hfill g^{\prime }\left(x\right)& >x\left({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{2x^2}}+{\displaystyle \frac{7(15x^2+22)}{30x(21x^4+35x^2+4)}}\right)-{\displaystyle \frac{h\left(x\right)}{2x{(40x^2+3)}^2{(120x^2-1)}^2}}\hfill \\ \hfill & ={\displaystyle \frac{284280000x^6+34292800x^4-1158735x^2+1386}{30(21x^4+35x^2+4){(40x^2+3)}^2{(120x^2-1)}^2}}>0.\hfill \end{array}$$

Hence, clearly, $g\left(x\right)$ is strictly increasing for $x\geqq 1$, and we have

$$\begin{array}{c}\hfill g\left(x\right)<\underset{t\to \infty }{lim}f\left(t\right)=0\mathrm{and}{\left({\displaystyle \frac{\Theta \left(x\right)}{x}}\right)}^{\prime }<0(x\geqq 1).\end{array}$$

Therefore, the function $x\mapsto {\displaystyle \frac{\Theta \left(x\right)}{x}}$ is strictly decreasing for $x\geqq 1$. So, we have

$$\begin{array}{c}\hfill \Theta \left(x\right)>{\displaystyle \frac{x}{x+y}}\Theta (x+y)\mathrm{and}\Theta \left(y\right)>{\displaystyle \frac{y}{x+y}}\Theta (x+y)(x,y\geqq 1).\end{array}$$

Adding these two expressions, we are led to (69). This completes the proof of Theorem 5. □

Remark 5. 

Let $x_j\geqq 1$ ($j=1,2,\dots ,n$). Since the function $x\mapsto {\displaystyle \frac{\Theta \left(x\right)}{x}}$ is strictly decreasing for $x\geqq 1$, we have

$$\begin{array}{cc}\hfill \Theta \left(x_1\right)& >{\displaystyle \frac{x_1}{x_1+x_2+\dots +x_n}}\Theta (x_1+x_2+\dots +x_n),\hfill \\ \hfill \Theta \left(x_2\right)& >{\displaystyle \frac{x_2}{x_1+x_2+\dots +x_n}}\Theta (x_1+x_2+\dots +x_n),\hfill \\ \hfill & ⋮\hfill \\ \hfill \Theta \left(x_n\right)& >{\displaystyle \frac{x_n}{x_1+x_2+\dots +x_n}}\Theta (x_1+x_2+\dots +x_n).\hfill \end{array}$$

Adding these expressions, we obtain

$$\begin{array}{c}\hfill \sum _{j=1}^n\Theta \left(x_j\right)>\Theta \left(\sum _{j=1}^n{x}_j\right).\end{array}$$

5. Superadditive Property

We first state and prove the following result.

Theorem 6. 

The remainder $\vartheta \left(x\right)$ of Ramanujan’s formula (11) satisfies the following inequality:

$$\begin{array}{c}\hfill \vartheta \left(x\right)+\vartheta \left(y\right)<\vartheta (x+y)(x,y\geqq 1).\end{array}$$

(70)

Proof. 

It follows from (11) that

$$\begin{array}{c}\hfill {\displaystyle \frac{\vartheta \left(x\right)}{x}}={\displaystyle \frac{1}{x}}ln\mathsf{\Gamma }\left(x\right)-\left(1-{\displaystyle \frac{1}{2x}}\right)lnx+1-{\displaystyle \frac{ln\left(\sqrt{2\pi }\right)}{x}}-{\displaystyle \frac{1}{6x}}ln\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}\right),\end{array}$$

which, upon differentiation with respect to x, yields

$$\begin{array}{cc}\hfill x^2{\left({\displaystyle \frac{\vartheta \left(x\right)}{x}}\right)}^{\prime }& =x\psi \left(x\right)-ln\mathsf{\Gamma }\left(x\right)-{\displaystyle \frac{1}{2}}lnx+ln\sqrt{2\pi }-{\displaystyle \frac{240x^4-50x^2-24x-1}{240x^3+120x^2+30x+1}}\hfill \\ \hfill & +{\displaystyle \frac{1}{6}}ln\left(1+{\displaystyle \frac{1}{2x}}+{\displaystyle \frac{1}{8x^2}}+{\displaystyle \frac{1}{240x^3}}\right)=:f\left(x\right)\hfill \end{array}$$

and

$$\begin{array}{c}\hfill f^{\prime }\left(x\right)=x{\psi }^{\prime }\left(x\right)-{\displaystyle \frac{57600x^7+86400x^6+67200x^5+31680x^4+7860x^3+1090x^2+61x+1}{x{(240x^3+120x^2+30x+1)}^2}}.\end{array}$$

Now, by using the asymptotic formulas for $ln\mathsf{\Gamma }\left(x\right)$ and $\psi \left(x\right)$ (see [1]), we get

$$\begin{array}{c}\hfill f\left(x\right)={\displaystyle \frac{11}{2304x^4}}+O\left({\displaystyle \frac{1}{x^5}}\right)(x\to \infty ),\end{array}$$

which implies that

$$\begin{array}{c}\hfill \underset{x\to \infty }{lim}f\left(x\right)=0.\end{array}$$

From the following well-known continued fraction for ${\psi }^{\prime }$ (see [15]):

$$\begin{array}{c}\hfill {\psi }^{\prime }\left(z\right)={\displaystyle \frac{1}{z}}+{\displaystyle \frac{1}{2z^2}}+{\displaystyle \frac{2\pi }{z}}\left(\begin{array}{cccc}{a}_1^{\left(1\right)}& & a_2^{\left(1\right)}& \\ 1-13-35-5z^2& +& 1& +\end{array}\begin{array}{cccc}{a}_3^{\left(1\right)}& & a_4^{\left(1\right)}& \\ 1-13-35-5z^2& +& 1& +\dots \end{array}\right)\left(|arg\left(z\right)|<{\displaystyle \frac{\pi }{2}}\right),\end{array}$$

where

$$\begin{array}{c}\hfill a_1^{\left(1\right)}={\displaystyle \frac{1}{12\pi }}\mathrm{and}{a}_m^{\left(1\right)}={\displaystyle \frac{m^2(m^2-1)}{4(4m^2-1)}}(m\geqq 2),\end{array}$$

we find for $x>0$ that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2\pi }{x}}\left(\begin{array}{cccc}{a}_1^{\left(1\right)}& & a_2^{\left(1\right)}& \\ 1-13-35-5x^2& +& 1& +\end{array}\begin{array}{cccc}{a}_3^{\left(1\right)}& & a_4^{\left(1\right)}& \\ 1-13-35-5x^2& +& 1\end{array}\right)<{\psi }^{\prime }\left(x\right)-{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{2x^2}}\hfill \\ \hfill & <{\displaystyle \frac{2\pi }{x}}\left(\begin{array}{cccc}{a}_1^{\left(1\right)}& & a_2^{\left(1\right)}& \\ 1-13-35-5x^2& +& 1& +\end{array}\begin{array}{cc}{a}_3^{\left(1\right)}& \\ 1-13-3x^2\end{array}\right),\hfill \end{array}$$

that is,

$$\begin{array}{c}\hfill {\displaystyle \frac{7(15x^2+22)}{30x(21x^4+35x^2+4)}}<{\psi }^{\prime }\left(x\right)-{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{2x^2}}<{\displaystyle \frac{35x^2+18}{30x^3(7x^2+5)}}.\end{array}$$

(71)

Using the second inequality in (71), we obtain

$$\begin{array}{cc}\hfill f^{\prime }\left(x\right)& <x\left({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{2x^2}}+{\displaystyle \frac{35x^2+18}{30x^3(7x^2+5)}}\right)\hfill \\ \hfill & -{\displaystyle \frac{57600x^7+86400x^6+67200x^5+31680x^4+7860x^3+1090x^2+61x+1}{x{(240x^3+120x^2+30x+1)}^2}}\hfill \\ \hfill & =-{\displaystyle \frac{P(x-1)}{30x^2(7x^2+5){(240x^3+120x^2+30x+1)}^2}},\hfill \end{array}$$

where

$$\begin{array}{c}\hfill P\left(x\right)=22687+428180x+1360240x^2+1758765x^3+1035000x^4+231000x^5.\end{array}$$

We then see that $f^{\prime }\left(x\right)<0$ for $x\geqq 1$, and we have

$$\begin{array}{c}\hfill f\left(x\right)>\underset{t\to \infty }{lim}f\left(t\right)=0\mathrm{and}{\left({\displaystyle \frac{\vartheta \left(x\right)}{x}}\right)}^{\prime }>0.\end{array}$$

Hence, clearly, the function $x\mapsto {\displaystyle \frac{\vartheta \left(x\right)}{x}}$ is strictly increasing for $x\geqq 1$. Consequently, we have

$$\begin{array}{c}\hfill \vartheta \left(x\right)<{\displaystyle \frac{x}{x+y}}\vartheta (x+y)\mathrm{and}\vartheta \left(y\right)<{\displaystyle \frac{y}{x+y}}\vartheta (x+y)(x,y\geqq 1).\end{array}$$

Adding these two expressions, we obtain (70). The proof of Theorem 6 is thus completed. □

Remark 6. 

If we let $x_j\geqq 1$$(j=1,2,\dots ,n),$ then we have

$$\begin{array}{c}\hfill \sum _{j=1}^n\vartheta \left(x_j\right)<\vartheta \left(\sum _{j=1}^n{x}_j\right).\end{array}$$

Theorem 7. 

The remainder $\rho \left(x\right)$ of the formula (15) satisfies the following inequality:

$$\begin{array}{c}\hfill \rho \left(x\right)+\rho \left(y\right)<\rho (x+y)(x,y\geqq 1).\end{array}$$

(72)

Proof. 

It follows from (15) that

$$\begin{array}{c}\hfill {\displaystyle \frac{\rho \left(x\right)}{x}}={\displaystyle \frac{1}{x}}ln\mathsf{\Gamma }\left(x\right)-\left(1-{\displaystyle \frac{1}{2x}}\right)lnx+1-{\displaystyle \frac{ln\left(\sqrt{2\pi }\right)}{x}}-\left(x+{\displaystyle \frac{53}{210x}}\right)ln\left(1+{\displaystyle \frac{1}{12x^3+\frac{24}{7}x-\frac{1}{2}}}\right),\end{array}$$

which, upon differentiation with respect to x, yields

$$\begin{array}{cc}\hfill x^2{\left({\displaystyle \frac{\rho \left(x\right)}{x}}\right)}^{\prime }& =x\psi \left(x\right)-ln\mathsf{\Gamma }\left(x\right)-{\displaystyle \frac{1}{2}}lnx+ln\sqrt{2\pi }\hfill \\ \hfill & -{\displaystyle \frac{282240x^7-141120x^6+90720x^5-80640x^4-1488x^3-11520x^2-2186x+245}{10(168x^3+48x+7)(168x^3+48x-7)}}\hfill \\ \hfill & -\left(x^2-{\displaystyle \frac{53}{210}}\right)ln\left(1+{\displaystyle \frac{1}{12x^3+\frac{24}{7}x-\frac{1}{2}}}\right)=:I\left(x\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I^{\prime }\left(x\right)& =x{\psi }^{\prime }\left(x\right)-2xln\left(1+{\displaystyle \frac{1}{12x^3+\frac{24}{7}x-\frac{1}{2}}}\right)-{\displaystyle \frac{J\left(x\right)}{10x{(168x^3+48x-7)}^2{(168x^3+48x+7)}^2}},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill J\left(x\right)& =7965941760x^{13}+3982970880x^{12}+9103933440x^{11}+4551966720x^{10}+4015484928x^9\hfill \\ \hfill & +1950842880x^8+926860032x^7+357759360x^6+136469952x^5+18639360x^4\hfill \\ \hfill & +8619552x^3-1128960x^2+24010x+12005.\hfill \end{array}$$

Using the asymptotic formulas for $ln\mathsf{\Gamma }\left(x\right)$ and $\psi \left(x\right)$ (see [1]), we have

$$\begin{array}{c}\hfill I\left(x\right)={\displaystyle \frac{2117}{635040x^7}}+O\left({\displaystyle \frac{1}{x^9}}\right)(x\to \infty ),\end{array}$$

which implies that

$$\begin{array}{c}\hfill \underset{x\to \infty }{lim}I\left(x\right)=0.\end{array}$$

Now, by using the second inequality in (71), we find for $x\geqq 1$ that

$$\begin{array}{cc}\hfill {\displaystyle \frac{I^{\prime }\left(x\right)}{x}}& <{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{2x^2}}+{\displaystyle \frac{35x^2+18}{30x^3(7x^2+5)}}-2ln\left(1+{\displaystyle \frac{1}{12x^3+\frac{24}{7}x-\frac{1}{2}}}\right)\hfill \\ \hfill & -{\displaystyle \frac{J\left(x\right)}{10x^2{(168x^3+48x-7)}^2{(168x^3+48x+7)}^2}}=:K\left(x\right).\hfill \end{array}$$

Differentiating this last equation, we get

$$\begin{array}{c}\hfill K^{\prime }\left(x\right)={\displaystyle \frac{49L\left(x\right)}{30x^4{(7x^2+5)}^2{(168x^3+48x-7)}^3{(168x^3+48x+7)}^3}},\end{array}$$

where

$$\begin{array}{cc}\hfill L\left(x\right)& =42547688128512x^{16}+67379520079872x^{14}+39425152757760x^{12}+10861290506304x^{10}\hfill \\ \hfill & +1597403771424x^8+138040119360x^6+4254019935x^4-89512955x^2+648270.\hfill \end{array}$$

Hence, clearly, we obtain $K^{\prime }\left(x\right)>0$ for $x\geqq 1$, and we have

$$\begin{array}{c}\hfill K\left(x\right)<\underset{t\to \infty }{lim}K\left(t\right)=0\mathrm{and}{I}^{\prime }\left(x\right)<0(x\geqq 1).\end{array}$$

Therefore, the function $I\left(x\right)$ is strictly decreasing for $x\geqq 1$, and we have

$$\begin{array}{c}\hfill I\left(x\right)>\underset{t\to \infty }{lim}I\left(t\right)=0\mathrm{and}{\left({\displaystyle \frac{\rho \left(x\right)}{x}}\right)}^{\prime }>0(x\geqq 1).\end{array}$$

Therefore, the function $x\mapsto {\displaystyle \frac{\rho \left(x\right)}{x}}$ is strictly increasing for $x\geqq 1$. So, we have

$$\begin{array}{c}\hfill \rho \left(x\right)<{\displaystyle \frac{x}{x+y}}\rho (x+y)\mathrm{and}\rho \left(y\right)<{\displaystyle \frac{y}{x+y}}\rho (x+y)(x,y\geqq 1).\end{array}$$

Adding these two expressions, we obtain (72). This completes the proof of Theorem 7. □

Remark 7. 

If we let $x_j\geqq 1$$(j=1,2,\dots ,n),$ then we have

$$\begin{array}{c}\hfill \sum _{j=1}^n\rho \left(x_j\right)<\rho \left(\sum _{j=1}^n{x}_j\right).\end{array}$$