New asymptotic expansions and Padé approximants related to the triple gamma function

Abstract

In this work, our main focus is to establish asymptotic expansions for the triple gamma function in terms of the triple Bernoulli polynomials. As application, an asymptotic expansion for hyperfactorial function is also obtained. Furthermore, using these asymptotic expansions, Padé approximants related to the triple gamma function are derived as a consequence. The results obtained are new, and their importance is demonstrated by deducing several interesting remarks and corollaries.

Introduction

E.W. Barnes introduced the multiple gamma function $\Gamma _{n}$ and expressed it in terms of the multiple Hurwitz zeta-functions [1–4]. After this invention, several researchers and mathematicians have studied this function. $\Gamma _{n}$ was studied by G.H. Hardy [11, 12] in view of the theory of elliptic functions. $\Gamma _{n}$ can be used to compute summation of series and infinite products. It is also applied in the theory of elliptic functions and theta functions [1–4]. Moreover, multiple gamma functions are also useful to study the determinant of Laplacians on the n-dimensional unit sphere $S^{n}$ [7]. Also, $\Gamma _{n}$ appears in functional equations for the Selberg zeta functions [18] associated with higher rank symmetric spaces, so obtaining their asymptotic expansion is of interest. It can be observed from the literature that the $\Gamma _{n}$ plays a vital role in analytic number theory, approximation theory, mathematical physics, and several branches of science and engineering [7]. $\Gamma _{n}$ satisfies the following recurrence relations [17]:

$$\begin{aligned} \Gamma _{n+1}(z+1)=\frac {\Gamma _{n+1}(z)}{\Gamma _{n}(z)}, \qquad \Gamma _{1}(z)=\Gamma (z),\qquad \Gamma _{n}(1)=1, \quad n \in \mathbb{N}, \end{aligned}$$

(1.1)

where $\Gamma (z)$ is Euler’s gamma function. The reciprocal of $\Gamma _{2}(z)$ is the well-known Barnes G-function and is denoted by $G(z)$. For further information on $\Gamma _{n}$, we refer to [1–4, 8, 9, 15–17] and the references therein.

One of the most recent research interests in the theory of multiple gamma functions is to find their asymptotic expansions. In [6], asymptotic expansions for $G(z)$ are obtained by C.-P. Chen. Recently, Z. Xu and W. Wang [20] generalized the result. In [10], C. Ferreira and J.L. López derived an asymptotic expansion of $G(z)$. Recently, X. Li and C.-P. Chen [13] established Padé approximant involving asymptotics for the gamma function.

The above results motivate us to find asymptotic expansion for the triple gamma function $\Gamma _{3}$. In this work, we derive the asymptotic expansion for $\Gamma _{3}$. Moreover, Padé approximants related to these asymptotic expansions are also obtained. It can be noted that $\Gamma _{3}$ has the Weierstrass canonical product [7] form:

$$\begin{aligned} \Gamma _{3}(z+1)&=\exp \bigl[Dz^{3}+Ez^{2}+Fz \bigr] \\ &\quad{}\cdot \prod_{k=1}^{\infty} \biggl( \biggl(1+\frac {z}{k} \biggr)^{-\frac {k(k+1)}{2}} \exp \biggl[ \frac {k+1}{2}z- \frac {1}{4} \biggl(1+\frac {1}{k} \biggr)z^{2} +\frac {1}{6k} \biggl(1+ \frac {1}{k} \biggr)z^{3} \biggr] \biggr), \end{aligned}$$

(1.2)

where

$$\begin{aligned}& D=-\frac {1}{6} \biggl(\gamma +\frac {\pi ^{2}}{6}+ \frac {3}{2} \biggr),\qquad E=\frac {1}{4} \biggl(\gamma +\log (2\pi )+\frac {1}{2} \biggr), \quad \text{and} \\& F=\frac {3}{8}- \frac {\log (2\pi )}{4}-\log A. \end{aligned}$$

Here A denotes the Glaisher–Kinkelin constant and is defined as [7]

$$ \log A=\frac {1}{12}-\zeta '(-1) $$

with ζ being the Riemann zeta function. $\Gamma _{3}$ can also be expressed in terms of triple Bernoulli polynomials [15] as follows:

$$\begin{aligned} \log \Gamma _{3}(z)&=\frac {B_{3,3}(z)}{6}\log z- \frac {11}{36}B_{3,0}z^{3}- \frac {3}{4}B_{3,1}z^{2}- \frac {1}{2}B_{3,2}z \\ &\quad {}+\sum_{k=1}^{n} \frac {(-1)^{k-1}B_{3,k+3}}{k(k+1)(k+2)(k+3)z^{k}}+\mathcal{R}_{3,n}(z), \end{aligned}$$

(1.3)

where $B_{3,k}(x)$ are triple Bernoulli polynomials defined as

$$ \frac {t^{3}e^{xt}}{(e^{t}-1)^{3}}=\sum_{k=0}^{\infty}B_{3,k}(x) \frac {t^{k}}{k!}, \quad \vert t \vert < 2\pi , $$

and $B_{3,k}=B_{3,k}(0)$ are triple Bernoulli numbers [14] with $\mathcal{R}_{3,n}(z)$, which is the remainder of order n and has the following integral representation:

$$ \mathcal{R}_{3,n}(z)= \int _{0}^{\infty}\frac {e^{-zt}}{t^{4}} \Biggl( \frac {t^{3}}{(1-e^{-t})^{3}}-\sum_{k=0}^{n} \frac {(-1)^{k}}{k!}B_{3,k}t^{k} \Biggr)\,dt \quad \text{for }\Re z>0 \text{ and } n\geq 3. $$

The following relation [14] between triple Bernoulli polynomials and Bernoulli polynomials will be helpful for computing triple Bernoulli polynomials to prove the main results:

$$\begin{aligned} B_{3,k}(x) =\frac {k(k-1)(k-2)}{2} \biggl( (x-1) (x-2) \frac {B_{k-2}(x)}{k-2}-(2x-3)\frac {B_{k-1}(x)}{k-1}+ \frac {B_{k}(x)}{k} \biggr), \end{aligned}$$

(1.4)

where $B_{k}(x)$ are Bernoulli polynomials defined as

$$ \frac {te^{tx}}{e^{t}-1}=\sum_{k=0}^{\infty}B_{k}(x) \frac {t^{k}}{k!}, \quad \vert t \vert < 2\pi , $$

and $B_{k}=B_{k}(0)$ are known as Bernoulli numbers.

It can be noted that putting $x=0$ in (1.4), we get the following recurrence relation, which will be useful to compute triple Bernoulli numbers:

$$\begin{aligned} B_{3,k}= \frac {k(k-1)(k-2)}{2} \biggl( \frac {2B_{k-2}}{k-2} + \frac {3B_{k-1}}{k-1}+\frac {B_{k}}{k} \biggr)\quad \text{for }k\geq 3. \end{aligned}$$

(1.5)

The rest of the paper is organized as follows. In Sect. 2, two different classes of asymptotic expansions of triple gamma functions with the formulas for determining the coefficients of each class are found. An asymptotic expansion for hyperfactorial function is also obtained as a particular case. Moreover, Padé approximants related to these asymptotic expansions are obtained in Sect. 3.

Asymptotic expansions for the triple gamma function

Let

$$\begin{aligned} g(z)= \frac {\exp ({\frac {}{}\frac{11}{36}B_{3,0}z^{3}+\frac{3}{4}B_{3,1}z^{2}+\frac{1}{2}B_{3,2}z} )\Gamma _{3}(z)}{z^{\frac{1}{6}B_{3,3}(z)}}. \end{aligned}$$

(2.1)

Then, using the expansion for $\Gamma _{3}$ given in (1.3), we obtain

$$\begin{aligned} g(z)\sim \exp{ \Biggl(\sum_{k=1}^{\infty} \frac {(-1)^{k-1}B_{3,k+3}}{k(k+1)(k+2)(k+3)z^{k}} \Biggr)} \end{aligned}$$

(2.2)

for $z\rightarrow \infty$ and $\Re z>0$. It follows from (1.5) [14, p. 187] that for $k\geq 1$,

$$\begin{aligned}& B_{3,2k+1} =\frac {3(2k+1)(2k-1)}{2}B_{2k}, \\& B_{3,2k+2} =k(2k+1) (2k+2) \biggl(\frac {B_{2k}}{k}+ \frac {B_{2k+2}}{2(k+1)} \biggr). \end{aligned}$$

Using the above expressions together with (1.5) in (2.2) gives

$$\begin{aligned} g(z) & \sim \exp{ \biggl(\frac {19}{240z}+ \frac {1}{160z^{2}}- \frac {2}{945z^{3}}-\frac {1}{672z^{4}}+ \frac {19}{50\text{,}400z^{5}}\cdots \biggr)} \end{aligned}$$

(2.3)

for $z\rightarrow \infty$ and $\Re z>0$.

Using the expansion $e^{z}=\sum_{k=0}^{\infty}\frac {z^{k}}{k!}$ in the exponential part of (2.3) containing Bernoulli numbers, a simple computation yields the following:

$$\begin{aligned} g(z) &\sim \biggl(1+\frac {19}{240z}+\frac {1081}{115\text{,}200z^{2}}- \frac {893\text{,}507}{580\text{,}608\text{,}000z^{3}} -\frac {900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000z^{4}} \\ &\quad {}+ \frac {161\text{,}466\text{,}866\text{,}293}{668\text{,}860\text{,}416\text{,}000\text{,}000z^{5}}+\cdots \biggr). \end{aligned}$$

(2.4)

Theorem 2.1

Let $p\geq 0$ be any integer and $q\neq 0$ be any real number. Then $g(z)$ defined in (2.1) has the following asymptotic expansion:

$$\begin{aligned} g(z)\sim \Biggl(1+\sum_{i=1}^{\infty} \frac {a_{i}}{z^{i}} \Biggr)^{z^{p}/q} \end{aligned}$$

(2.5)

for $z\rightarrow \infty$ and $\Re z>0$, where the coefficients $a_{i}=a_{i}(p,q)$ $(i\in \mathbb{N})$ are given by

$$\begin{aligned} a_{i}&=\sum_{(1+p)k_{1}+(2+p)k_{2}+\cdots +(i+p)k_{i}=i} \frac {q^{k_{1}+k_{2}+\cdots +k_{i}}}{k_{1}!k_{2}!\cdots k_{i}!} \\ &\quad {}\cdot \biggl(\frac {B_{3,4}}{1\cdot 2\cdot 3\cdot 4} \biggr)^{k_{1}} \biggl( \frac {-B_{3,5}}{2\cdot 3\cdot 4\cdot 5} \biggr)^{k_{2}} \cdots \biggl(\frac {(-1)^{i-1}B_{3,3+i}}{i(i+1)(i+2)(i+3)} \biggr)^{k_{i}}, \end{aligned}$$

(2.6)

where the summation is considered over all $k_{i}$, with $k_{i}$s as nonnegative integers satisfying the following relation:

$$ (1+p)k_{1}+(2+p)k_{2}+\cdots +(i+p)k_{i}=i. $$

Proof

Our main objective is to determine $a_{i}$ ($i\in \mathbb{N}$). For this we first express (2.5) as follows:

$$\begin{aligned} \bigl(g(z) \bigr)^{q/z^{p}} =1+\sum _{i=1}^{n} \frac {a_{i}}{z^{i}}+O \bigl(z^{-n-1}\bigr). \end{aligned}$$

(2.7)

Note that even in the absence of any information on convergence, (2.7) is still valid as we are interested only in the asymptotic expansion. From (2.2), we have

$$\begin{aligned} g(z) =\exp \Biggl(\sum_{k=1}^{n} \frac {(-1)^{k-1}B_{3,k+3}}{k(k+1)(k+2)(k+3)z^{k}}+ \mathcal{R}(z) \Biggr), \end{aligned}$$

where $\mathcal{R}(z)=O(z^{-n-1})$, which implies

$$\begin{aligned} & \bigl(g(z) \bigr)^{q/z^{p}} \\ & =e^{q\mathcal{R}(z)/z^{p}} \exp{ \Biggl(\sum_{k=1}^{n} \frac {(-1)^{k-1}qB_{3,k+3}}{k(k+1)(k+2)(k+3)z^{k+p}} \Biggr)} \\ &=e^{q\mathcal{R}(z)/z^{p}}\prod_{k=1}^{n} \biggl(1+ \biggl( \frac {(-1)^{k-1}qB_{3,k+3}}{k(k+1)(k+2)(k+3)z^{k+p}} \biggr) \\ &\quad {}+\frac {1}{2!} \biggl( \frac {(-1)^{k-1}qB_{3,k+3}}{k(k+1)(k+2)(k+3)z^{k+p}} \biggr)^{2}+\cdots \biggr) \\ &=e^{q\mathcal{R}(z)/z^{p}}\sum_{k_{1}=0}^{\infty} \sum_{k_{2}=0}^{ \infty}\cdots \sum _{k_{n}=0}^{\infty} \frac {1}{k_{1}!k_{2}!\cdots k_{n}!} \biggl( \frac {qB_{3,4}}{1\cdot 2\cdot 3\cdot 4} \biggr)^{k_{1}} \biggl( \frac {-qB_{3,5}}{2\cdot 3\cdot 4\cdot 5} \biggr)^{k_{2}} \\ & \quad {}\cdots \biggl( \frac {(-1)^{n-1}qB_{3,n+3}}{n(n+1)(n+2)(n+3)} \biggr)^{k_{n}} \cdot \frac {1}{z^{(1+p)k_{1}+(2+p)k_{2}+\cdots +(n+p)k_{n}}}. \end{aligned}$$

(2.8)

Comparing the powers of z in (2.7) and (2.8), we obtain

$$\begin{aligned} a_{i}&=\sum_{(1+p)k_{1}+(2+p)k_{2}+\cdots +(i+p)k_{i}=i} \frac {q^{k_{1}+k_{2}+\cdots +k_{i}}}{k_{1}!k_{2}!\cdots k_{i}!} \\ &\quad {}\cdot \biggl(\frac {B_{3,4}}{1\cdot 2\cdot 3\cdot 4} \biggr)^{k_{1}} \biggl( \frac {-B_{3,5}}{2\cdot 3\cdot 4\cdot 5} \biggr)^{k_{2}} \cdots \biggl(\frac {(-1)^{i-1}B_{3,3+i}}{i(i+1)(i+2)(i+3)} \biggr)^{k_{i}}, \end{aligned}$$

which completes the proof. □

Remark 2.1

$p=0$ and $q=1$ in (2.5) yield (2.4). Setting $p=0$, $q=2$ and $p=1$, $q=1$ in (2.5), respectively, we have the following two asymptotic expansions:

$$\begin{aligned} \Gamma _{3}(z)\sim {}& z^{\frac{B_{3,3}(z)}{6}}e^{-\frac{11}{36}B_{3,0}z^{3}- \frac{3}{4}B_{3,1}z^{2}-\frac{1}{2}B_{3,2}z}\biggl(1+\frac {19}{120z}+\frac {721}{28\text{,}800z^{2}}- \frac {115\text{,}547}{72\text{,}576\text{,}000z^{3}} \\ & {}-\frac {117\text{,}935\text{,}033}{34\text{,}836\text{,}480\text{,}000z^{4}} + \frac {3\text{,}885\text{,}027\text{,}493}{20\text{,}901\text{,}888\text{,}000\text{,}000z^{5}}+\cdots \biggr)^{\frac{1}{2}} \end{aligned}$$

(2.9)

and

$$\begin{aligned} \Gamma _{3}(z)\sim{}& z^{\frac{B_{3,3}(z)}{6}} e^{-\frac{11}{36}B_{3,0}z^{3}- \frac{3}{4}B_{3,1}z^{2}-\frac{1}{2}B_{3,2}z} \biggl(1+ \frac {19}{240z^{2}}+\frac {1}{160z^{3}} \\ &{}+ \frac {2461}{2\text{,}419\text{,}200z^{4}}- \frac {89}{89\text{,}600z^{5}}+\cdots \biggr)^{z}. \end{aligned}$$

(2.10)

Remark 2.2

Let $H(n)$ be the hyperfactorial function defined as $H(n):=1^{1}\cdot 2^{2}\cdot 3^{3}\cdots n^{n}$.

Then, using the relations $\Gamma _{3}(n)=G(n-1)\Gamma _{3}(n-1)$ and $G(n+1)=\frac {(n!)^{n}}{H(n)}$, we have

$$\begin{aligned} \Gamma _{3}(n)&=G(3)\cdot G(4)\cdots G(n-1) = \frac { \prod_{k=1}^{n-2}(k!)^{k}}{ \prod_{k=1}^{n-2}H(k)} \\ &= \frac { \prod_{k=1}^{n}(k!)^{k}}{((n-1)!)^{n-1}(n!)^{n}H_{n-2}},\quad \text{where } \prod_{k=1}^{n}H(k)=H_{n}. \end{aligned}$$

With the help of Remark 2.1, the following asymptotic expansions can be obtained:

$$\begin{aligned} (1!)^{1}(2!)^{2}\cdots (n!)^{n} & \sim (n!)^{n} \bigl((n-1)!\bigr)^{n-1} H_{n-2} n^{ \frac{B_{3,3}(n)}{6}}e^{-\frac{11}{36}B_{3,0}n^{3}-\frac{3}{4}B_{3,1}n^{2}- \frac{1}{2}B_{3,2}n} \\ &\quad {}\times \biggl( 1+\frac {19}{240n}+\frac {1081}{115\text{,}200n^{2}}- \frac {893\text{,}507}{580\text{,}608\text{,}000n^{3}}-\frac {900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000n^{4}} \\ &\quad{}+\frac {161\text{,}466\text{,}866\text{,}293}{668\text{,}860\text{,}416\text{,}000\text{,}000n^{5}}+\cdots \biggr) \end{aligned}$$

(2.11)

and

$$\begin{aligned} (1!)^{1}(2!)^{2}\cdots (n!)^{n} & \sim (n!)^{n} \bigl((n-1)!\bigr)^{n-1} H_{n-2} n^{ \frac{B_{3,3}(n)}{6}}e^{-\frac{11}{36}B_{3,0}n^{3}-\frac{3}{4}B_{3,1}n^{2}- \frac{1}{2}B_{3,2}n} \\ &\quad {} \times \biggl( 1+\frac {19}{240n^{2}}+\frac {1}{160n^{3}}+ \frac {2461}{2\text{,}419\text{,}200n^{4}}-\frac {89}{89\text{,}600n^{5}}+\cdots \biggr)^{n}. \end{aligned}$$

(2.12)

Theorem 2.2

Let $p\geq 0$ be any integer and $q\neq 0$ be any real number. Then $g(z)$ defined in (2.1) satisfies the following asymptotic expansion:

$$\begin{aligned} g(z)\sim \Biggl(1+\ln \Biggl(1+\sum _{i=1}^{\infty} \frac {b_{i}}{z^{i}} \Biggr) \Biggr)^{z^{p}/q} \end{aligned}$$

(2.13)

for $z\rightarrow \infty$ and $\Re z>0$, where the coefficients $b_{i}=b_{i}(p,q)$ $(i\in \mathbb{N})$ are given by

$$ b_{i}=\sum_{k_{1}+2k_{2}+\cdots +ik_{i}=i} \frac {1}{k_{1}!k_{2}!\cdots k_{i}!}a_{1}^{k_{1}}a_{2}^{k_{2}} \cdots a_{i}^{k_{i}}, $$

$a_{i}$ ($i\in \mathbb{N}$) are determined in (2.6).

Proof

To prove this theorem, we will follow similar techniques as Theorem 2.1. To do so, first we express (2.13) as follows:

$$\begin{aligned} \exp{ \bigl( \bigl(g(z) \bigr)^{q/z^{p}}-1 \bigr)} =1+ \sum_{i=1}^{n}\frac {b_{i}}{z^{i}}+O \bigl(z^{-n-1}\bigr). \end{aligned}$$

(2.14)

Note that, similar to (2.7), the discussion of asymptotic expansion does not require any information on the convergence of the series in (2.14). Writing (2.5) as

$$ \bigl(g(z) \bigr)^{q/z^{p}}-1 =\mathcal{R}(z)=O\bigl(z^{-m-1} \bigr), $$

we have

$$\begin{aligned} &\exp{ \bigl( \bigl(g(z) \bigr)^{q/z^{p}}-1 \bigr)} \\ &\quad =e^{ \mathcal{R}(z)}\exp{ \Biggl(\sum_{k=1}^{n} \frac {a_{k}}{z^{k}} \Biggr)} \\ &\quad =e^{\mathcal{R}(z)}\prod_{k=1}^{n} \biggl(1+ \biggl( \frac {a_{k}}{z^{k}} \biggr)+\frac {1}{2!} \biggl( \frac {a_{k}}{z^{k}} \biggr)^{2} +\cdots \biggr) \\ &\quad =e^{\mathcal{R}(z)}\sum_{k_{1}=0}^{\infty} \sum_{k_{2}=0}^{\infty} \cdots \sum _{k_{n}=0}^{\infty}\frac {1}{k_{1}!k_{2}!\cdots k_{n}!} a_{1}^{k_{1}} a_{2}^{k_{2}}\cdots a_{n}^{k_{n}} \frac {1}{z^{k_{1}+2k_{2}+\cdots +nk_{n}}}. \end{aligned}$$

(2.15)

Equating the coefficients of powers of z in (2.14) and (2.15), we have

$$ b_{i}=\sum_{k_{1}+2k_{2}+\cdots +ik_{i}=i} \frac {1}{k_{1}!k_{2}!\cdots k_{i}!}a_{1}^{k_{1}}a_{2}^{k_{2}} \cdots a_{i}^{k_{i}}, $$

where $a_{i}$ ($i\in \mathbb{N}$) are determined in (2.6), which completes the proof. □

Remark 2.3

Putting $p=0$, $q=1$; $p=0$, $q=2$; and $p=1$, $q=1$ respectively in (2.13), the following asymptotic expansions can be obtained:

$$\begin{aligned} \Gamma _{3}(z)\sim {}& z^{\frac{B_{3,3}(z)}{6}}e^{-\frac{11}{36}B_{3,0}z^{3}- \frac{3}{4}B_{3,1}z^{2}-\frac{1}{2}B_{3,2}z} \\ &{}\times \biggl(1+\ln \biggl(1+\frac {19}{240z}+ \frac {721}{57\text{,}600z^{2}}-\frac {16\text{,}567}{23\text{,}224\text{,}320z^{3}}- \frac {245\text{,}178\text{,}299}{185\text{,}794\text{,}560\text{,}000z^{4}} \\ &\quad{}+\frac {5\text{,}299\text{,}227\text{,}679}{55\text{,}738\text{,}368\text{,}000\text{,}000z^{5}}+ \cdots \biggr) \biggr), \end{aligned}$$

(2.16)

$$\begin{aligned} \Gamma _{3}(z)\sim {}& z^{\frac{B_{3,3}(z)}{6}}e^{-\frac{11}{36}B_{3,0}z^{3}- \frac{3}{4}B_{3,1}z^{2}-\frac{1}{2}B_{3,2}z} \\ &{}\times \biggl(1+\ln \biggl(1+\frac {19}{240z^{2}}+ \frac {1}{160z^{3}}-\frac {15\text{,}281\text{,}579}{11\text{,}612\text{,}160\text{,}000z^{4}} \\ &{}+ \frac {1\text{,}613\text{,}002\text{,}879}{1\text{,}741\text{,}824\text{,}000\text{,}000z^{5}}+\cdots \biggr) \biggr)^{1/2}, \end{aligned}$$

(2.17)

and

$$\begin{aligned} \Gamma _{3}(z)\sim {}& z^{\frac{B_{3,3}(z)}{6}}e^{-\frac{11}{36}B_{3,0}z^{3}- \frac{3}{4}B_{3,1}z^{2}-\frac{1}{2}B_{3,2}z} \\ &{}\times \biggl(1+\ln \biggl(1+\frac {19}{240z^{2}}+ \frac {1}{160z^{3}}+\frac {5021}{1\text{,}209\text{,}600z^{4}}- \frac {67}{134\text{,}400z^{5}}+\cdots \biggr) \biggr)^{z}. \end{aligned}$$

(2.18)

Padé approximants related to the triple gamma function

Let $h(t)=\sum_{k=0}^{\infty}c_{k}t^{k}$ be a formal power series and $h_{n}(t)=\sum_{k=0}^{n}c_{k}t^{k}$ be its nth partial sum ($g_{n}$ is identically zero for $n<0$). Then the Padé approximation [19] of order $(m,n)$ of the function g is defined as the rational function

$$\begin{aligned}{} [m/n]_{h}(t):= \frac {\sum_{k=0}^{m}a_{k}t^{k}}{1+\sum_{k=1}^{n}b_{k}t^{k}}, \end{aligned}$$

(3.1)

where $m\geq 0$ and $n\geq 1$ are two given integers, and the coefficients $a_{k}$ and $b_{k}$ are given by [13]

$$ \begin{aligned} &a_{0} =c_{0}, \\ &a_{1} =c_{0}b_{1}+c_{1}, \\ &a_{2} =c_{0}b_{2}+c_{1}b_{1}+c_{2}, \\ &\quad \vdots \\ &a_{m} =c_{0}b_{m}+\cdots +c_{m-1}b_{1}+c_{m}, \\ &0 =c_{m+1}+c_{m}b_{1}+\cdots +c_{m-n+1}b_{n}, \\ &\quad \vdots \\ &0 =c_{m+n}+c_{m+n-1}b_{1}+\cdots +c_{m}b_{n}, \end{aligned} $$

(3.2)

with the following property:

$$\begin{aligned}{} [m/n]_{h}(t)-h(t)=O\bigl(t^{m+n+1}\bigr). \end{aligned}$$

(3.3)

Clearly, the first $m+n+1$ coefficients of the series expansion of $[m/n]_{g}$ are identical to those of g. Further, we have

$$\begin{aligned}{} [m/n]_{h}(t)= \frac { \begin{vmatrix} t^{n}h_{m-n}(t)& t^{n-1}h_{m-n+1}(t)& \cdots & h_{m}(t) \\ c_{m-n+1} & c_{m-n+2} & \cdots & c_{m+1} \\ \vdots & \vdots & \ddots & \vdots \\ c_{m} & c_{m+1} & \cdots & c_{m+n} \end{vmatrix} }{ \begin{vmatrix} t^{n}& t^{n-1}& \cdots & 1 \\ c_{m-n+1} & c_{m-n+2} & \cdots & c_{m+1} \\ \vdots & \vdots & \ddots & \vdots \\ c_{m} & c_{m+1} & \cdots & c_{m+n} \end{vmatrix}}. \end{aligned}$$

(3.4)

Padé approximants related to the gamma function are discussed in [13]. In this section, we are interested in finding the Padé approximants for the function $g(x)$, which is defined in (2.1).

From (2.4) and (2.1), it can be noted that, as $x\rightarrow \infty$,

$$\begin{aligned} g(x)&\sim \sum_{k=0}^{\infty} \frac {c_{k}}{x^{k}} \\ &=1+\frac {19}{240x}+ \frac {1081}{115\text{,}200x^{2}}- \frac {893\text{,}507}{580\text{,}608\text{,}000x^{3}}- \frac {900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000x^{4}} +\cdots , \end{aligned}$$

(3.5)

where the coefficients $c_{k}$ satisfy (2.6) for $p=0$ and $q=1$.

Now we will proceed for the Padé approximation of $g(x)$. Let us consider

$$ [1/1]_{g}(x)= \frac {\sum_{k=0}^{1}\frac{a_{k}}{x^{k}}}{1+\sum_{k=1}^{1}\frac{b_{k}}{x^{k}}}. $$

It follows that

$$\begin{aligned} \begin{aligned} &c_{0}=1, \qquad c_{1}=\frac {19}{240}, \qquad c_{2}= \frac {1081}{115\text{,}200}, \\ &c_{3}=- \frac {893\text{,}507}{580\text{,}608\text{,}000},\qquad c_{4}=- \frac {900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000}. \end{aligned} \end{aligned}$$

(3.6)

From (3.2), we obtain

$$\begin{aligned}& a_{0} =1, \\& a_{1} =b_{1}+\frac {19}{240}, \\& 0 =\frac {19}{240}b_{1}+\frac {1081}{115\text{,}200}, \end{aligned}$$

which gives $a_{0}=1$, $a_{1}=-\frac{359}{9120}$, and $b_{1}=-\frac{1081}{9120}$. Hence, we have

$$\begin{aligned}{} [1/1]_{g}(x)=\frac {1-\frac{359}{9120x}}{1-\frac{1081}{9120x}}= \frac {x-\frac{359}{9120}}{x-\frac{1081}{9120}}, \end{aligned}$$

(3.7)

and using (3.3), it follows that

$$ g(x) = \frac {x-\frac{359}{9120}}{x-\frac{1081}{9120}}+O \biggl( \frac {1}{x^{3}} \biggr). $$

Now we will derive a Padé approximant of order $(2,2)$. For this, let us consider

$$ [2/2]_{g}(x)= \frac {\sum_{k=0}^{2}\frac{a_{k}}{x^{k}}}{1+\sum_{k=1}^{2}\frac{b_{k}}{x^{k}}}. $$

Using (3.2), we obtain

$$\begin{aligned}& a_{0} =1 \\& a_{1} =b_{1}+\frac {19}{240}, \\& a_{2} =b_{2}+\frac {19}{240}b_{1}+ \frac {1081}{115\text{,}200}, \\& 0 =\frac {1081}{115\text{,}200}b_{1}+\frac {19}{240}b_{2}- \frac {893\text{,}507}{580\text{,}608\text{,}000}, \\& 0 =\frac {893\text{,}507}{580\text{,}608\text{,}000}b_{1}-\frac {1081}{115\text{,}200}b_{2}+ \frac {900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000}, \end{aligned}$$

which implies that

$$\begin{aligned}& a_{1} =-\frac {12\text{,}947\text{,}658\text{,}827}{28\text{,}076\text{,}662\text{,}560},\qquad a_{2}= \frac {14\text{,}176\text{,}622\text{,}313\text{,}529}{283\text{,}012\text{,}758\text{,}604\text{,}800}, \\& b_{1} =-\frac {15\text{,}170\text{,}394\text{,}613}{28\text{,}076\text{,}662\text{,}560},\qquad b_{2}= \frac {23\text{,}626\text{,}895\text{,}894\text{,}809}{283\text{,}012\text{,}758\text{,}604\text{,}800}, \end{aligned}$$

and

$$\begin{aligned}{} [2/2]_{g}(x)&= \frac {1-\frac {12\text{,}947\text{,}658\text{,}827}{28\text{,}076\text{,}662\text{,}560x}+\frac {14\text{,}176\text{,}622\text{,}313\text{,}529}{283\text{,}012\text{,}758\text{,}604\text{,}800x^{2}}}{ 1-\frac {15\text{,}170\text{,}394\text{,}613}{28\text{,}076\text{,}662\text{,}560x}+\frac {23\text{,}626\text{,}895\text{,}894\text{,}809}{283\text{,}012\text{,}758\text{,}604\text{,}800x^{2}}} \\ &= \frac {x^{2}-\frac {12\text{,}947\text{,}658\text{,}827}{28\text{,}076\text{,}662\text{,}560}x+\frac {14\text{,}176\text{,}622\text{,}313\text{,}529}{283\text{,}012\text{,}758\text{,}604\text{,}800}}{ x^{2}-\frac {15\text{,}170\text{,}394\text{,}613}{28\text{,}076\text{,}662\text{,}560}x+\frac {23\text{,}626\text{,}895\text{,}894\text{,}809}{283\text{,}012\text{,}758\text{,}604\text{,}800}}. \end{aligned}$$

(3.8)

Therefore, using (3.3), it follows that

$$ g(x) = \frac {x^{2}-\frac {12\text{,}947\text{,}658\text{,}827}{28\text{,}076\text{,}662\text{,}560}x+\frac {14\text{,}176\text{,}622\text{,}313\text{,}529}{283\text{,}012\text{,}758\text{,}604\text{,}800}}{ x^{2}-\frac {15\text{,}170\text{,}394\text{,}613}{28\text{,}076\text{,}662\text{,}560}x+\frac {23\text{,}626\text{,}895\text{,}894\text{,}809}{283\text{,}012\text{,}758\text{,}604\text{,}800}}+O \biggl(\frac {1}{x^{5}} \biggr). $$

Using the Padé approximation method and expansion (3.5), the following theorem can be derived.

Theorem 3.1

The Padé approximation of order $(m,n)$ of the asymptotic formula of the function $g(x)$ (at the point $x=\infty$) is given by the rational function:

$$ [m/n]_{g}(x)= \frac {1+\sum_{k=1}^{m}\frac{a_{k}}{x^{k}}}{1+\sum_{k=1}^{n}\frac{b_{k}}{x^{k}}} =x^{n-m} \biggl( \frac {x^{m}+a_{1}x^{m-1}+\cdots +a_{m}}{x^{n}+b_{1}x^{n-1}+\cdots +b_{n}} \biggr), $$

where $m\geq 1$ and $n\geq 1$ are given integers and the coefficients $a_{k}$ and $b_{k}$ satisfy the conditions (3.2) with $a_{0}=c_{0}=1$ and $c_{k}$ satisfies (2.6) for $p=0$ and $q=1$, and the following relation holds:

$$ g(x)-[m/n]_{g}(x)=O \biggl(\frac {1}{x^{m+n+1}} \biggr),\quad x \rightarrow \infty . $$

Further, we have (see [5])

$$\begin{aligned}{} [m/n]_{g}(x)= \frac { \begin{vmatrix} x^{n}g_{m-n}(x)& x^{n-1}g_{m-n+1}(x)& \cdots & g_{m}(x) \\ c_{m-n+1} & c_{m-n+2} & \cdots & c_{m+1} \\ \vdots & \vdots & \ddots & \vdots \\ c_{m} & c_{m+1} & \cdots & c_{m+n} \end{vmatrix} }{ \begin{vmatrix} x^{n}& x^{n-1}& \cdots & 1 \\ c_{m-n+1} & c_{m-n+2} & \cdots & c_{m+1} \\ \vdots & \vdots & \ddots & \vdots \\ c_{m} & c_{m+1} & \cdots & c_{m+n} \end{vmatrix}}, \end{aligned}$$

(3.9)

where $g_{n}(x)=\sum_{k=0}^{n}\frac{c_{k}}{x^{k}}$ is the nth partial sum of the asymptotic series (3.5).

Remark 3.1

Using (3.9), it is also possible to derive (3.7) and (3.8). Let us verify these results.

$$\begin{aligned}{} [1/1]_{g}(x)= \frac { \begin{vmatrix} \frac{1}{x}g_{0}(x) & g_{1}(x) \\ c_{1} & c_{2} \end{vmatrix} }{ \begin{vmatrix} \frac{1}{x}& 1 \\ c_{1} & c_{2} \end{vmatrix}} = \frac { \begin{vmatrix} \frac{1}{x} & 1+\frac{19}{240x} \\ \frac{19}{240} & \frac{1081}{115\text{,}200} \end{vmatrix} }{ \begin{vmatrix} \frac{1}{x}& 1\\ \frac{19}{240} & \frac{1081}{115\text{,}200} \end{vmatrix}} = \frac {x-\frac{359}{9120}}{x-\frac{1081}{9120}} \end{aligned}$$

and

$$\begin{aligned}{} [2/2]_{g}(x)&= \frac { \begin{vmatrix} \frac{1}{x^{2}}g_{0}(x) & \frac{1}{x}g_{1}(x)&g_{2}(x) \\ c_{1} & c_{2} & c_{3}\\ c_{2} & c_{3} & c_{4} \end{vmatrix} }{ \begin{vmatrix} \frac{1}{x^{2}}&\frac{1}{x} &1 \\ c_{1} & c_{2} &c_{3}\\ c_{2} & c_{3} & c_{4} \end{vmatrix}} = \frac { \begin{vmatrix} \frac{1}{x^{2}} & \frac{1}{x} (1+\frac{19}{240x} )& 1+\frac{19}{240x}+\frac{1081}{115\text{,}200x^{2}}\\ \frac{19}{240} & \frac{1081}{115\text{,}200}&-\frac{893\text{,}507}{580\text{,}608\text{,}000} \\ \frac{1081}{115\text{,}200}&-\frac{893\text{,}507}{580\text{,}608\text{,}000}& -\frac{900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000} \end{vmatrix} }{ \begin{vmatrix} \frac{1}{x^{2}} & \frac{1}{x}& 1\\ \frac{19}{240} & \frac{1081}{115\text{,}200}&-\frac{893\text{,}507}{580\text{,}608\text{,}000} \\ \frac{1081}{115\text{,}200}&-\frac{893\text{,}507}{580\text{,}608\text{,}000}& -\frac{900\text{,}113\text{,}513}{557\text{,}383\text{,}680\text{,}000} \end{vmatrix} } \\ &= \frac {x^{2}-\frac {12\text{,}947\text{,}658\text{,}827}{28\text{,}076\text{,}662\text{,}560}x+\frac {14\text{,}176\text{,}622\text{,}313\text{,}529}{283\text{,}012\text{,}758\text{,}604\text{,}800}}{ x^{2}-\frac {15\text{,}170\text{,}394\text{,}613}{28\text{,}076\text{,}662\text{,}560}x+\frac {23\text{,}626\text{,}895\text{,}894\text{,}809}{283\text{,}012\text{,}758\text{,}604\text{,}800}}. \end{aligned}$$

Setting $m=n=r$ in (3.3), the following result is immediate.

Corollary 3.1

$$\begin{aligned} g(x) = \frac {x^{r}+a_{1}x^{r-1}+\cdots +a_{r}}{x^{r}+b_{1}x^{r-1}+\cdots +b_{r}}+O \biggl(\frac {1}{x^{2r+1}} \biggr) \end{aligned}$$

(3.10)

for $x\rightarrow \infty$ and $r\geq 1$ is any given integer, the coefficients $a_{i}$ and $b_{i}$ ($1\leq i\leq r$) satisfy (3.2) with $a_{0}=c_{0}=1$, and $c_{r}$ is given in (2.6) (for $p=0$ and $q=1$).

Remark 3.2

Setting $r=3$ in (3.10), we obtain

$$\begin{aligned} &g(x) = \frac {x^{3}-a_{1}x^{2}+a_{2}x-a_{3}}{ x^{3}-b_{1}x^{2}+b_{2}x-b_{3}} +O \biggl(\frac {1}{x^{7}} \biggr), \end{aligned}$$

where

$$\begin{aligned}& a_{1} =\frac {564\text{,}264\text{,}707\text{,}394\text{,}045\text{,}441\text{,}291\text{,}137}{777\text{,}288\text{,}995\text{,}553\text{,}534\text{,}212\text{,}638\text{,}560}, \\& a_{2} = \frac {507\text{,}337\text{,}383\text{,}020\text{,}418\text{,}800\text{,}355\text{,}971\text{,}917}{932\text{,}746\text{,}794\text{,}664\text{,}241\text{,}055\text{,}166\text{,}272\text{,}000}, \\& a_{3} = \frac {524\text{,}560\text{,}286\text{,}160\text{,}441\text{,}450\text{,}378\text{,}984\text{,}935\text{,}759}{18\text{,}804\text{,}175\text{,}380\text{,}431\text{,}099\text{,}672\text{,}152\text{,}043\text{,}520\text{,}000}, \\& b_{1} =\frac {625\text{,}800\text{,}086\text{,}208\text{,}700\text{,}233\text{,}125\text{,}023}{777\text{,}288\text{,}995\text{,}553\text{,}534\text{,}212\text{,}638\text{,}560}, \\& b_{2} =\frac {50\text{,}730\text{,}526\text{,}659\text{,}081\text{,}610\text{,}299\text{,}631\text{,}847}{84\text{,}795\text{,}163\text{,}151\text{,}294\text{,}641\text{,}378\text{,}752\text{,}000}, \\& b_{3} = \frac {1\text{,}244\text{,}184\text{,}486\text{,}344\text{,}122\text{,}340\text{,}305\text{,}441\text{,}760\text{,}881}{18\text{,}804\text{,}175\text{,}380\text{,}431\text{,}099\text{,}672\text{,}152\text{,}043\text{,}520\text{,}000}. \end{aligned}$$

Remark 3.3

Setting $r=4$ in (3.10), we obtain

$$\begin{aligned} g(x)= \frac {x^{4}-a_{1}x^{3}+a_{2}x^{2}-a_{3}x+a_{4}}{ x^{4}-b_{1}x^{3}+b_{2}x^{2}-b_{3}x+b^{4}} +O \biggl(\frac {1}{x^{9}} \biggr), \end{aligned}$$

where

$$\begin{aligned}& a_{1} = \frac {215\text{,}703\text{,}766\text{,}781\text{,}168\text{,}876\text{,}220\text{,}885\text{,}513\text{,}532\text{,}207\text{,}573\text{,}037\text{,}853\text{,}391}{156\text{,}833\text{,}412\text{,}819\text{,}388\text{,}306\text{,}354\text{,}787\text{,}238\text{,}937\text{,}752\text{,}295\text{,}790\text{,}827\text{,}680}, \\& a_{2} = \frac {439\text{,}021\text{,}821\text{,}607\text{,}824\text{,}678\text{,}358\text{,}124\text{,}666\text{,}374\text{,}361\text{,}977\text{,}917\text{,}184\text{,}670\text{,}503}{276\text{,}026\text{,}806\text{,}562\text{,}123\text{,}419\text{,}184\text{,}425\text{,}540\text{,}530\text{,}444\text{,}040\text{,}591\text{,}856\text{,}716\text{,}800}, \\& a_{3} = \frac {2\text{,}697\text{,}600\text{,}344\text{,}247\text{,}344\text{,}205\text{,}104\text{,}432\text{,}196\text{,}561\text{,}229\text{,}100\text{,}202\text{,}237\text{,}946\text{,}568\text{,}861}{4\text{,}173\text{,}525\text{,}315\text{,}219\text{,}306\text{,}098\text{,}068\text{,}514\text{,}172\text{,}820\text{,}313\text{,}893\text{,}748\text{,}873\text{,}558\text{,}016\text{,}000}, \\& a_{4} = \frac {940\text{,}070\text{,}303\text{,}236\text{,}823\text{,}651\text{,}814\text{,}961\text{,}986\text{,}110\text{,}532\text{,}058\text{,}170\text{,}155\text{,}544\text{,}925\text{,}558\text{,}921}{20\text{,}032\text{,}921\text{,}513\text{,}052\text{,}669\text{,}270\text{,}728\text{,}868\text{,}029\text{,}537\text{,}506\text{,}689\text{,}994\text{,}593\text{,}078\text{,}476\text{,}800\text{,}000}, \\& b_{1} = \frac {228\text{,}119\text{,}745\text{,}296\text{,}037\text{,}117\text{,}140\text{,}639\text{,}503\text{,}281\text{,}446\text{,}296\text{,}454\text{,}627\text{,}249}{156\text{,}833\text{,}412\text{,}819\text{,}388\text{,}306\text{,}354\text{,}787\text{,}238\text{,}937\text{,}752\text{,}295\text{,}790\text{,}827\text{,}680}, \\& b_{2} = \frac {468\text{,}216\text{,}358\text{,}741\text{,}523\text{,}424\text{,}525\text{,}741\text{,}582\text{,}917\text{,}397\text{,}922\text{,}238\text{,}128\text{,}383\text{,}943}{276\text{,}026\text{,}806\text{,}562\text{,}123\text{,}419\text{,}184\text{,}425\text{,}540\text{,}530\text{,}444\text{,}040\text{,}591\text{,}856\text{,}716\text{,}800}, \\& b_{3} = \frac {290\text{,}424\text{,}419\text{,}148\text{,}610\text{,}158\text{,}964\text{,}741\text{,}537\text{,}112\text{,}783\text{,}630\text{,}996\text{,}531\text{,}879\text{,}384\text{,}489}{379\text{,}411\text{,}392\text{,}292\text{,}664\text{,}190\text{,}733\text{,}501\text{,}288\text{,}438\text{,}210\text{,}353\text{,}977\text{,}170\text{,}323\text{,}456\text{,}000}, \\& b_{4} = \frac {1\text{,}822\text{,}684\text{,}110\text{,}010\text{,}671\text{,}047\text{,}977\text{,}349\text{,}196\text{,}992\text{,}991\text{,}024\text{,}335\text{,}965\text{,}841\text{,}858\text{,}437\text{,}321}{20\text{,}032\text{,}921\text{,}513\text{,}052\text{,}669\text{,}270\text{,}728\text{,}868\text{,}029\text{,}537\text{,}506\text{,}689\text{,}994\text{,}593\text{,}078\text{,}476\text{,}800\text{,}000}. \end{aligned}$$