Some inequalities for the ratio of gamma functions

Abstract

In this paper, we present some sharp inequalities for the ratio of gamma functions. The main tool is the multiple-correction method formulated in (Cao et al. in J. Number Theory 149:327-350, 2015; Cao in J. Math. Anal. Appl. 424:1425-1446, 2015). Some conjectures are proposed.

1 Introduction

The problem of finding new and sharp inequalities for the ratio of gamma functions and, in particular, for the Wallis ratio

$$ W(n):=\frac{(2n-1)!!}{(2n)!!}=\frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}\Gamma (n+1)}, \quad n\in\mathbb{N}=\{1,2,\ldots\} $$

(1.1)

has attracted the attention of many researchers (see, e.g., [1–13] and the references therein). Here, Γ is the gamma function, and we employ the special double factorial function as follows:

$$\begin{aligned}& (2n)!!=2\cdot4\cdot6\cdots(2n)=2^{n}\Gamma(n+1), \\& (2n-1)!!=1\cdot3\cdot5\cdots(2n-1)=\pi^{-1/2}2^{n}\Gamma \biggl(n+\frac{1}{2}\biggr). \end{aligned}$$

See [14], p.258.

Chen and Qi [15] established the following inequalities:

$$ \frac{1}{\sqrt{\pi(n+\frac{4}{\pi}-1)}}\le W(n)< \frac{1}{\sqrt{\pi (n+\frac{1}{4})}},\quad n\in\mathbb{N}, $$

(1.2)

where $\frac{4}{\pi}-1$ and $\frac{1}{4}$ are best possible.

For $n\in\mathbb{N}$, let

$$\begin{aligned}& T_{1}(n):=\frac{1\cdot4\cdots(3n-2)}{3\cdot6\cdots(3n)} =\frac{\Gamma(n+\frac{1}{3})}{\Gamma(n+1)\Gamma(\frac{1}{3})}, \end{aligned}$$

(1.3)

$$\begin{aligned}& T_{2}(n):=\frac{2\cdot5\cdots(3n-1)}{3\cdot6\cdots(3n)}=\frac{\Gamma (n+\frac{2}{3})}{\Gamma(n+1)\Gamma(\frac{2}{3})}. \end{aligned}$$

(1.4)

Very recently, Mortici et al. [7] proved that for every integer $n\ge1$,

$$ \frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n}}\exp \biggl(-\frac{2}{81n^{2}} \biggr)< T_{1}(n) < \frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n}}\exp \biggl(-\frac{2}{81n^{2}}+ \frac {2}{243n^{3}} \biggr) $$

(1.5)

and

$$ \frac{\tau}{\sqrt[3]{n+\frac{1}{3}}}\exp \biggl(-\frac{1}{81n^{2}} \biggr)< T_{2}(n) < \frac{\tau}{\sqrt[3]{n+\frac{1}{3}}}\exp \biggl(-\frac{1}{81n^{2}}+ \frac {2}{243n^{3}} \biggr), $$

(1.6)

where

$$ \beta=\frac{1}{\Gamma (\frac{1}{3} )}=\frac{\sqrt{3}}{2\pi}\Gamma \biggl( \frac{2}{3} \biggr)=0.37328\ldots\quad \mbox{and}\quad \tau= \frac {1}{\Gamma (\frac{2}{3} )}=0.73848\ldots. $$

(1.7)

From (1.5) and (1.6), we have the following asymptotic formulas:

$$\begin{aligned}& T_{1}(n)=\frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n}}\exp \biggl(-\frac {2}{81n^{2}}+ \frac{2}{243n^{3}} \biggr) \biggl(1+O\biggl(\frac{1}{n^{4}}\biggr) \biggr), \\& T_{2}(n)=\frac{\tau}{\sqrt [3]{n+\frac{1}{3}}}\exp \biggl(-\frac{1}{81n^{2}}+ \frac{2}{243n^{3}} \biggr) \biggl(1+O\biggl(\frac{1}{n^{4}}\biggr) \biggr). \end{aligned}$$

Motivated by these works, the purpose of this paper is to present some sharp inequalities for $W(n)$, $T_{1}(n)$ and $T_{2}(n)$ by using the multiple-correction method developed in [16, 17].

Throughout the paper, the notation $\lfloor x\rfloor$ denotes the largest integer not exceeding x. The notation $P_{k}(x)$ (or $Q_{k}(x)$) denotes a polynomial of degree k in x, while the notation $\Psi(k;x)$ means a polynomial of degree k in x with all coefficients non-negative, which may be different at each occurrence. Let $(a_{n})_{n\ge1}$ and $(b_{n})_{n\ge0}$ be two sequences of real numbers with $a_{n}\neq0$ for all $n\in\mathbb{N}$. The generalized continued fraction

$$ \tau=b_{0}+\frac{a_{1}}{b_{1}+\frac{a_{2}}{b_{2}+\ddots}}= b_{0}+ \frac{a_{1}}{b_{1}+}\frac{a_{2}}{b_{2}+}= \cdots =b_{0}+ \mathop{\mathbf{K}}_{n=1}^{\infty} \biggl( \frac{a_{n}}{b_{n}} \biggr) $$

(1.8)

is defined as the limit of the nth approximant

$$ \frac{A_{n}}{B_{n}}=b_{0}+\mathop{\mathbf{K}}_{k=1}^{n} \biggl(\frac{a_{k}}{b_{k}} \biggr) $$

(1.9)

as n tends to infinity. The numerators $A_{n}$ and denominators $B_{n}$ of the approximants satisfy the recurrence relations

$$ A_{n+2}=b_{n+2}A_{n+1}+a_{n+2}A_{n}, \qquad B_{n+2}=b_{n+2}B_{n+1}+a_{n+2}B_{n} $$

(1.10)

with initial values $A_{0}=b_{0}$, $B_{0}=1$, $A_{1}=b_{0}b_{1}+a_{1}$ and $B_{1}=b_{1}$. See [18], p.105.

2 Inequalities for $W(n)$

In this section, in order to illustrate the idea of this paper, we consider the case of Wallis ratio. We introduce some class of correction sequence $(\operatorname{MC}_{k}(n))_{n\ge1}$ such that the relative error sequence $E_{k}(n)$ has the fastest possible rate of convergence, which is defined by the relations

$$ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{\sqrt{\pi(n+\frac{1}{4})}} \exp\bigl( \operatorname{MC}_{k}(n)\bigr) \exp\bigl(E_{k}(n)\bigr),\quad k\in\mathbb{N}_{0}=\mathbb{N}\cup \{0\}. $$

(2.1)

The tool for measuring the rate of convergence is a result given by Mortici in [19], which says that a sequence $(x_{n})_{n\ge1}$ converging to zero is fastest possible when the difference $(x_{n}-x_{n+1})_{n\ge1}$ is fastest possible. More precisely, if there exists $\lim_{n\rightarrow\infty}n^{\kappa}(x_{n}-x_{n+1})=l$, with $\kappa>1$, then $\lim_{n\rightarrow\infty}n^{\kappa-1}x_{n}=\frac{l}{\kappa-1}$. Mortici’s lemma has been effectively applied in many papers such as [7, 11, 16, 17, 20–24]. If $\lim_{n\rightarrow\infty}n^{\varrho}x_{n}=l\neq0$ with constant $\varrho> 0$, we say that the sequence $(x_{n})_{n\ge1}$ is order $n^{-\varrho}$, and write the exponent of convergence $\varrho=\varrho (x_{n})$. Clearly if $\varrho(E_{k}(n))=\varrho_{k}$, we have the following asymptotic formula:

$$ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{\sqrt{\pi(n+\frac{1}{4})}} \exp\bigl(\operatorname{MC}_{k}(n) \bigr) \bigl(1+O\bigl(n^{-\varrho_{k}}\bigr) \bigr). $$

Let us briefly review the so-called multiple-correction method presented in our previous papers [16, 17]. Actually, the multiple-correction method is a recursive algorithm, and one of its advantages is that by repeating the correction process we always can accelerate the convergence, i.e., the sequence $(\varrho(E_{k}(n)))_{k\ge0}$ is strictly increasing. The key step is to find a suitable structure of $\operatorname {MC}_{k}(n)$. We often choose the kth correction sequence $\operatorname {MC}_{k}(n)$ to be a finite generalized continued fraction (see [17] or (2.6) below) or a simple rational function (see [16] or (3.1) and (3.8) below).

It follows from (2.1) that

$$\begin{aligned}& E_{k}(n)=\ln W(n)+\frac{1}{2}\ln\pi+\frac{1}{2}\ln \biggl(n+\frac{1}{4} \biggr)-\operatorname{MC}_{k}(n), \end{aligned}$$

(2.2)

$$\begin{aligned}& E_{k}(n)-E_{k}(n+1)=-\ln\frac{2n+1}{2n+2}-\frac{1}{2} \ln\frac{(n+1)+\frac{1}{4}}{n+\frac{1}{4}}+\operatorname{MC}_{k}(n+1) -\operatorname{MC}_{k}(n). \end{aligned}$$

(2.3)

Next, we illustrate how to determine all the related coefficients in $\operatorname{MC}_{k}(n)$. The choice of the initial-correction sequence $\operatorname{MC}_{0}(n)$ is crucial. Since

$$-\ln\frac{2n+1}{2n+2}-\frac{1}{2}\ln\frac{(n+1)+\frac{1}{4}}{n+\frac{1}{4}}= - \frac{1}{ 32 n^{3}}+O\biggl(\frac{1}{n^{4}}\biggr), $$

in order to obtain the fastest possible sequence $E_{0}(n)-E_{0}(n+1)$, we choose $\operatorname{MC}_{0}(n)=\frac{-\frac{1}{64}}{n^{2}+d_{1}n+d_{2}}$. Substituting this expression into (2.3) and expanding $E_{0}(n)-E_{0}(n+1)$ into a power series in terms of $n^{-1}$, by making use of Mathematica software we have

$$\begin{aligned} E_{0}(n)-E_{0}(n+1) =&\frac{\frac{3}{128}-\frac{3 d_{1}}{64}}{n^{4}}+ \frac {-25+48d_{1}+32d_{1}^{2}-32d_{2}}{512n^{5}} \\ &{}-\frac{5(-31+64d_{1}+64d_{1}^{2}+32d_{1}^{3}-64d_{2}-64d_{1}d_{2})}{2\text{,}048 n^{6}}+O\biggl(\frac{1}{n^{7}}\biggr). \end{aligned}$$

The fastest sequence $E_{0}(n)-E_{0}(n+1)$ is obtained by enforcing the first two coefficients of this power series to be zero. In this case, we find

$$ d_{1}=\frac{1}{2}, \qquad d_{2} = \frac{7}{32}, $$

(2.4)

and hence

$$ E_{0}(n)-E_{0}(n+1)=-\frac{169}{32\text{,}768 n^{7}}+O \biggl(\frac{1}{n^{8}}\biggr). $$

(2.5)

By Mortici’s lemma, it is not difficult to verify that $\varrho(E_{0}(n))=6$.

Just as in our previous works [16, 17], we can continue the above correction process to successively determine the correction sequence $(\operatorname{MC}_{k}(n))_{n\ge1}$ until some $k^{*}$ you want. On the one hand, to find the related coefficients avoiding huge computation, we often use appropriate symbolic computations software. On the other hand, the exact expressions at each occurrence also need a lot of space. Hence in this paper we omit many related details. Interested readers may refer to our previous papers [16, 17]. Let

$$ \operatorname{MC}_{k}(n)=\mathop{\mathbf{K}}_{j=0}^{k} \biggl(\frac{\kappa_{j}}{ (n+\frac{1}{4} )^{2}+\lambda_{j}} \biggr), \quad k\in\mathbb{N}_{0}=\mathbb{N} \cup \{0\}, $$

(2.6)

where

$$\begin{aligned}& \kappa_{0}=-\frac{1}{64},\qquad \lambda_{0}= \frac{5}{32}, \\& \kappa_{1}=-\frac{169}{3\text{,}072}, \qquad \lambda_{1}= \frac{6\text{,}245}{5\text{,}408}, \\& \kappa_{2}=-\frac{106\text{,}287\text{,}409}{109\text{,}674\text{,}240}, \qquad \lambda_{2}= \frac {1\text{,}813\text{,}097\text{,}726\text{,}405}{574\text{,}802\text{,}307\text{,}872}, \\& \kappa_{3}=-\frac{2\text{,}021\text{,}129\text{,}514\text{,}296\text{,}956\text{,}422\text{,}801}{404\text{,}884\text{,}957\text{,}099\text{,}688\text{,}791\text{,}040}, \\& \lambda_{3}= \frac{27\text{,}813\text{,}569\text{,}451\text{,}306\text{,}412\text{,}085\text{,}060\text{,}531\text{,}045}{ 4\text{,}519\text{,}565\text{,}955\text{,}619\text{,}764\text{,}991\text{,}777\text{,}552\text{,}928}. \end{aligned}$$

By (2.3) and Mortici’s lemma, we can use Mathematica software to check that $\varrho(E_{k}(n))=4k+6$ for $0\le k\le3$. Thus,

$$ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{\sqrt{\pi(n+\frac{1}{4})}} \exp\bigl( \operatorname{MC}_{k}(n)\bigr) \bigl(1+O\bigl(n^{-4k-6}\bigr) \bigr). $$

(2.7)

Lastly, we remove the factor $\exp(E_{k}(n))$ in (2.1) and further prove sharp double inequalities for as small n as possible. Now we introduce another sequence $\operatorname{MD}_{k}(n)$ defined by

$$\begin{aligned}& \operatorname{MD}_{0}(n)=\frac{\kappa_{0}}{ (n+\frac{1}{4} )^{2}}, \\& \operatorname{MD}_{k}(n)=\frac{\kappa_{0}}{ (n+\frac{1}{4} )^{2}+\lambda _{0}+}\cdots\frac{\kappa_{k-1}}{ (n+\frac{1}{4} )^{2}+\lambda _{k-1}+} \frac{\kappa_{k}}{ (n+\frac{1}{4} )^{2}}\quad \mbox{for } k\ge1 \end{aligned}$$

(i.e., $\operatorname{MD}_{k}(n)$ may be defined by taking the last coefficient $\lambda_{k}=0$ in the definition of $\operatorname{MC}_{k}(n)$), and improve inequalities (1.2) as follows.

Theorem 1

For $n\in\mathbb{N}$ and $k=0,1,2,3$, we have

$$ \frac{1}{\sqrt{\pi(n+\frac{1}{4})}}\exp\bigl(\operatorname{MD}_{k}(n) \bigr)\le\frac {(2n-1)!!}{(2n)!!} < \frac{1}{\sqrt{\pi(n+\frac{1}{4})}}\exp\bigl(\operatorname{MC}_{k}(n) \bigr). $$

(2.8)

Proof

We only give the proof in the case of $k=3$, other may be proved similarly. It suffices to show that $a(n)>0$ and $b(n)<0$, where

$$ a(n)=\ln W(n)+\frac{1}{2}\ln{\pi}+\frac{1}{2} \ln \biggl(n+\frac{1}{4} \biggr)- \operatorname{MD}_{k}(n) $$

(2.9)

and

$$ b(n)=\ln W(n)+\frac{1}{2}\ln{\pi}+\frac{1}{2} \ln \biggl(n+\frac{1}{4} \biggr)- \operatorname{MC}_{k}(n). $$

(2.10)

As both $a(n)$ and $b(n)$ converge to zero, it suffices to prove that $(a(n))_{n\ge1}$ is decreasing and $(b(n))_{n\ge1}$ is increasing. We set $a(n+1)-a(n)=f(n)$ and $b(n+1)-b(n)=g(n)$, where

$$ f(x)=\ln\frac{2x+1}{2x+2}+\ln\frac{(x+1)+\frac{1}{4}}{x+\frac{1}{4}}-\operatorname {MD}_{k}(x+1)+\operatorname{MD}_{k}(x) $$

and

$$ g(x)=\ln\frac{2x+1}{2x+2}+\ln\frac{(x+1)+\frac{1}{4}}{x+\frac{1}{4}}-\operatorname {MC}_{k}(x+1)+\operatorname{MC}_{k}(x). $$

By using Mathematica software, one may check

$$ f'(x)=\frac{\Psi_{1}(17;x)(x-1)+14164\ldots23853}{\Psi_{2}(4;x) (\Psi _{3}(7;x)(x-1)+14752\ldots04469 )^{2}\Psi_{4}(8;x)^{2}}>0\quad (x\ge1) $$

(2.11)

and

$$ g'(x)=-\frac{\Psi_{5}(16;x)^{2}}{ (1+x)(1+2x)(1+4x)(5+4x)\Psi_{6}(8;x)^{2}\Psi_{7}(8;x)^{2}}< 0, $$

(2.12)

where $14164\ldots23853$ and $14752\ldots04469$ denote two large positive constants. Thus, $f(x)$ is strictly increasing on $[1,\infty)$ with $f(\infty)=0$, while $g(x)$ is strictly decreasing on $[1,\infty)$ with $g(\infty)=0$. Hence $f<0$ and $g(x)>0$ for $x\ge1$. This finishes the proof of Theorem 1. □

3 Inequalities for $T_{1}(n)$ and $T_{2}(n)$

In this section, we will use the method of Section 2 to improve (1.5) and (1.6), respectively.

Theorem 2

Let β be defined by (1.7). For $n\in\mathbb{N}$, we have

$$\begin{aligned}& \frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n+\frac{2}{27}}}\exp \biggl(\frac {\frac{4}{2\text{,}187}}{n^{4}+b_{1}n^{3}+b_{2}n^{2}+b_{3}n} \biggr) \\& \quad < \frac{1\cdot4\cdots(3n-2)}{3\cdot6\cdots(3n)} < \frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n+\frac{2}{27}}}\exp \biggl(\frac {\frac{4}{2\text{,}187}}{n^{4}+b_{1}n^{3}+b_{2}n^{2}+b_{3}n+b_{4}} \biggr), \end{aligned}$$

(3.1)

where

$$ b_{1} = \frac{2}{3},\qquad b_{2} = \frac{61}{81}, \qquad b_{3} = \frac {52}{243}, \qquad b_{4} = -\frac{1\text{,}742}{6\text{,}561}.$$

Remark 1

It should be stressed that two kinds of correction processes are applied in (3.1). For comparison, we have

$$\begin{aligned}& T_{1}(n)=\frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n+\frac{2}{27}}} \biggl(1+O\biggl(\frac{1}{n^{4}} \biggr) \biggr), \\& T_{1}(n)=\frac{\beta}{\sqrt[3]{n^{2}+\frac{1}{3}n+\frac{2}{27}}}\exp \biggl(\frac{\frac{4}{2\text{,}187}}{n^{4}+b_{1}n^{3}+b_{2}n^{2}+b_{3}n+b_{4}} \biggr) \biggl(1+O\biggl(\frac{1}{n^{10}}\biggr) \biggr). \end{aligned}$$

Remark 2

We note that

$$\begin{aligned}& n^{2}+\frac{1}{3}n+\frac{2}{27}=\biggl(n+ \frac{1}{6}\biggr)^{2}+\frac{5}{108}, \\& n^{4}+b_{1}n^{3}+b_{2}n^{2}+b_{3}n+b_{4}= \biggl(n+\frac{1}{6}\biggr)^{4}+\frac{95}{162}\biggl(n+ \frac{1}{6}\biggr)^{2}-\frac{29\text{,}663}{104\text{,}976}. \end{aligned}$$

Proof

Since the proof of Theorem 2 is similar to that of Theorem 1, we only outline the idea of the proof here. Inequalities (3.1) are equivalent to

$$\begin{aligned}& u(n)= \ln T_{1}(n)-\ln{\beta}+\frac{1}{3}\ln \biggl(n^{2}+\frac{1}{3}n+\frac {2}{27} \biggr)- \frac{4/2\text{,}187}{n^{4}+b_{1}n^{3}+b_{2}n^{2}+b_{3}n}>0, \end{aligned}$$

(3.2)

$$\begin{aligned}& v(n)= \ln T_{1}(n)-\ln{\beta}+\frac{1}{3}\ln \biggl(n^{2}+\frac{1}{3}n+\frac {2}{27} \biggr)- \frac{4/2\text{,}187}{n^{4}+b_{1}n^{3}+b_{2}n^{2}+b_{3}n+b_{4}}< 0. \end{aligned}$$

(3.3)

As both $u(n)$ and $v(n)$ converge to zero, it suffices to prove that $(u(n))_{n\ge1}$ is decreasing and $(v(n))_{n\ge1}$ is increasing. We write $u(n+1)-u(n)=\phi(n)$ and $v(n+1)-v(n)=\varphi (n)$, where

$$\begin{aligned}& \phi(x)=\ln\frac{3x+1}{3x+3}+\frac{1}{3}\ln\frac{(x+1)^{2} +\frac{1}{3}(x+1)+\frac{2}{27}}{x^{2}+\frac{1}{3}x+\frac{2}{27}} \\& \hphantom{\phi(x)=}{}-\frac{4/2\text{,}187}{(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2} +b_{3}(x+1)}+ \frac{4/2\text{,}187}{x^{4}+b_{1}x^{3}+b_{2}x^{2}+b_{3}x}, \end{aligned}$$

(3.4)

$$\begin{aligned}& \varphi(x)=\ln\frac{3x+1}{3x+3}+\frac{1}{3}\ln \frac{(x+1)^{2}+\frac{1}{3}(x+1)+\frac{2}{27}}{x^{2} +\frac{1}{3}x+\frac{2}{27}} \\& \hphantom{\varphi(x)=}{}-\frac{4/2\text{,}187}{(x+1)^{4}+b_{1}(x+1)^{3} +b_{2}(x+1)^{2}+b_{3}(x+1)+b_{4}} \\& \hphantom{\varphi(x)=}{}+ \frac{4/2\text{,}187}{x^{4}+b_{1}x^{3}+b_{2}x^{2}+b_{3}x+b_{4}}. \end{aligned}$$

(3.5)

By using Mathematica software again, we have for $x\ge1$,

$$\begin{aligned}& \phi'(x)= -\frac{64 (\Psi_{1}(9;x)(x-1) +1\text{,}252\text{,}698\text{,}974\text{,}961\text{,}648\text{,}556 )}{ \Psi_{2}(6;x)(-1\text{,}742+ 1\text{,}404x + 4\text{,}941x^{2} + 4\text{,}374 x^{3} + 6\text{,}561 x^{4})^{2}\Psi _{3}(4;x)^{2}}>0, \end{aligned}$$

(3.6)

$$\begin{aligned}& \varphi'(x)= \frac{64 (\Psi_{4}(9;x)(x-1)+1\text{,}514\text{,}366\text{,}592\text{,}000 )}{9x^{2}(1 + x)^{2}(1+3 x)^{2} (2 + 9 x + 27 x^{2}) \Psi_{5}(12;x)}< 0. \end{aligned}$$

(3.7)

Thus, $\phi(x)$ is strictly increasing on $[1,\infty)$ with $\phi(\infty )=0$, while $\varphi(x)$ is strictly decreasing on $[1,\infty)$ with $\varphi(\infty)=0$. Hence $\phi(x)<0$ and $\varphi(x)>0$ for $x\ge1$. This completes the proof of Theorem 2. □

Theorem 3

Let τ be defined by (1.7). For $n\in\mathbb{N}$, we have

$$ \frac{\tau}{\sqrt[3]{n+\frac{1}{3}}}\exp \biggl(-\frac{\frac {1}{81}}{n^{2}+\frac{2}{3}n} \biggr)< \frac{2\cdot5\cdots(3n-1)}{3\cdot 6\cdots(3n)}< \frac{\tau}{\sqrt[3]{n+\frac{1}{3}}}\exp \biggl(-\frac{\frac {1}{81}}{(n+\frac{1}{3})^{2}+\frac{1}{6}} \biggr). $$

(3.8)

Proof

Similarly to the proof of Theorem 2, it suffices to show that for every integer $n\ge1$,

$$\begin{aligned}& r(n)= \ln T_{2}(n)-\ln{\tau}+\frac{1}{3}\ln \biggl(n+ \frac{1}{3} \biggr)+\frac {\frac{1}{81}}{n^{2}+\frac{2}{3}n}>0, \end{aligned}$$

(3.9)

$$\begin{aligned}& s(n)= \ln T_{2}(n)-\ln{\tau}+\frac{1}{3}\ln \biggl(n+ \frac{1}{3} \biggr)+\frac {\frac{1}{81}}{(n+\frac{1}{3})^{2}+\frac{1}{6}}< 0. \end{aligned}$$

(3.10)

We let $r(n+1)-r(n)=\eta(n)$ and $s(n+1)-s(n)=\theta(n)$, where

$$\begin{aligned}& \eta(x)= \ln\frac{3x+2}{3x+3}+\frac{1}{3}\ln\frac{(x+1)+\frac{1}{3}}{x+\frac{1}{3}}+ \frac{\frac{1}{81}}{(x+1)^{2}+\frac{2}{3}(x+1)}-\frac{\frac {1}{81}}{x^{2}+\frac{2}{3}x}, \end{aligned}$$

(3.11)

$$\begin{aligned}& \theta(x)= \ln\frac{3x+2}{3x+3}+\frac{1}{3}\ln\frac{(x+1)+\frac{1}{3}}{x+\frac{1}{3}}+ \frac{\frac{1}{81}}{(x+1+\frac{1}{6})^{2}+\frac{1}{6}}-\frac {\frac{1}{81}}{(x+\frac{1}{6})^{2}+\frac{1}{6}}. \end{aligned}$$

(3.12)

By using Mathematica software, we deduce that for every $x\ge1$,

$$\begin{aligned}& \eta'(x)= \frac{2(100+995x+2\text{,}472x^{2}+2\text{,}250x^{3}+675x^{4})}{27x^{2}(1+ x)^{2} (1+3x)(2+3x)^{2} (4+3x)(5+3x)^{2}}>0, \end{aligned}$$

(3.13)

$$\begin{aligned}& \theta'(x)= -\frac{2(18\text{,}625+86\text{,}880x+171\text{,}828x^{2}+143\text{,}640 x^{3} +43\text{,}092 x^{4})}{(1+x)(1+3x)(2+3x)(4+3x)(5+12x+18 x^{2})^{2} (35+ 48x+18 x^{2})^{2}}< 0. \end{aligned}$$

(3.14)

We observe that both $r(n)$ and $s(n)$ converge to zero. Thus, $\eta (x)$ is strictly increasing on $[1,\infty)$ with $\eta(\infty)=0$, while $\theta(x)$ is strictly decreasing on $[1,\infty)$ with $\theta (\infty)=0$. Hence $\eta(x)<0$ and $\theta(x)>0$ for $x\ge1$. This completes the proof of Theorem 3. □

Remark 3

In the proof of Theorems 2 and 3, we only use the initial-correction to improve (1.5) and (1.6). Certainly, we may further improve these inequalities by using multiple-correction.

4 Continued fraction inequalities

In this section, by changing the location of the correction sequence, we will employ the method of Section 2 to find the better continued fraction approximation. To do that, we introduce another correction sequence $(\operatorname {MC}_{k}(\frac{1}{2};n))_{n\ge1}$ and the relative error sequence $E_{k}(\frac{1}{2};n)$ with the following form:

$$ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{\sqrt{\pi}}\frac{1}{\sqrt{n+\frac {1}{4}+\operatorname{MC}_{k}(\frac{1}{2};n)}}\exp \biggl(E_{k}\biggl(\frac{1}{2};n\biggr)\biggr). $$

(4.1)

Hence

$$ E_{k}\biggl(\frac{1}{2};n\biggr)-E_{k} \biggl(\frac{1}{2};n+1\biggr)=-\ln\frac{2n+1}{2n+2}-\frac{1}{2}\ln \frac {(n+1)+\frac{1}{4}+\operatorname{MC}_{k}(\frac{1}{2};n+1)}{n+\frac{1}{4}+\operatorname {MC}_{k}(\frac{1}{2};n)}. $$

(4.2)

Let

$$ \operatorname{MC}_{k}\biggl(\frac{1}{2};n\biggr):= \mathop{\mathbf{K}}_{j=0}^{k} \biggl(\frac{\kappa_{j}}{n+\lambda _{j}} \biggr),\quad k\in \mathbb{N}_{0}=\mathbb{N}\cup \{0\}, $$

(4.3)

where $\kappa_{0}=\frac{1}{32}$, $\kappa_{j}=\frac{(2j+1)^{2}}{64}$ for $j\ge1$, and $\lambda_{j}=\frac{1}{4}$ for $j\ge0$. By applying the multiple-correction method described in Section 2, it is not difficult to check that if $0\le k\le20$, then

$$ \lim_{n\rightarrow\infty}n^{2k+4}E_{k}\biggl( \frac{1}{2};n\biggr)=C_{k}\biggl(\frac{1}{2}\biggr), $$

(4.4)

here $C_{k}(\frac{1}{2})$ is a computable constant depending only on k.

Now we will state the general inequalities and their rates of convergence as follows.

Theorem 4

For $n\in\mathbb{N}$ and $k\in\mathbb{N}\cup\{0\}$, we have

$$ \frac{1}{\sqrt{\pi}}\frac{1}{\sqrt{n+\frac{1}{4}+\operatorname{MC}_{2k}(\frac{1}{2};n)}} < \frac{(2n-1)!!}{(2n)!!}< \frac{1}{\sqrt{\pi}}\frac{1}{\sqrt{n+\frac {1}{4}+\operatorname{MC}_{2k+1}(\frac{1}{2};n)}} $$

(4.5)

and

$$ \varrho\biggl(E_{k}\biggl(\frac{1}{2};n\biggr) \biggr)\ge2k+4. $$

(4.6)

Proof

By taking $x=4n+1$ in Corollary 1 of Berndt [18], p.145, we obtain the so-called Lord Brouncker’s continued fraction formula

$$ q(n):= \biggl(\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \biggr)^{2}= \frac{4}{4n+1+\frac{1^{2}}{2(4n+1)+\frac{3^{2}}{2(4n+1) +\frac{5^{2}}{2(4n+1)+\ddots}}}}. $$

(4.7)

After some simplification, we see that if $k\ge2$, its kth approximant $q_{k}(n)=\frac{A_{k}(n)}{B_{k}(n)}$ equals

$$ \biggl(\frac{1}{\sqrt{n+\frac{1}{4}+\operatorname{MC}_{k-2}(\frac{1}{2};n)}} \biggr)^{2}. $$

As the partial coefficients of the continued fraction $q(n)$ are positive, we deduce inequalities (4.3) from (1.1) and (4.7) at once. Now, we are in a position to prove (4.4).

By the recurrence relations (1.10) we get

$$ q_{k+1}(n)-q_{k}(n)=\frac{A_{k+1}(n)}{B_{k+1}(n)}-\frac{A_{k}(n)}{B_{k}(n)} = \frac{(-1)^{k} 4\prod_{j=1}^{k}(2j-1)^{2}}{B_{k}(n)B_{k+1}(n)}. $$

(4.8)

It follows from (4.8) and (4.7) that

$$ \bigl\vert q_{k}(n)-q(n)\bigr\vert \le\bigl\vert q_{k+1}(n)-q_{k}(n)\bigr\vert = \frac{4\prod_{j=1}^{k}(2j-1)^{2}}{B_{k}(n)B_{k+1}(n)}. $$

(4.9)

Also see equation (6) in [25]. The above use of alternating series technique is originated from Alf van der Poorten’s lectures in [26]. Hence

$$ \bigl\vert \sqrt{q_{k}(n)}-\sqrt{q(n)}\bigr\vert < \frac{1}{\sqrt{q_{k}(n)}+\sqrt{q(n)}} \frac{4\prod_{j=1}^{k}(2j-1)^{2}}{B_{k}(n)B_{k+1}(n)}, $$

i.e.,

$$ \biggl\vert \frac{W(n)}{\sqrt{q_{k}(n)}}-1\biggr\vert < \frac{1}{\sqrt{q_{k}(n)}(\sqrt{q_{k}(n)}+W(n))} \frac{4\prod_{j=1}^{k}(2j-1)^{2}}{B_{k}(n)B_{k+1}(n)}. $$

(4.10)

We observe that $B_{k}(n)$ is a polynomial of degree k in n, and $W(n),\sqrt{q_{k}(n)}\sim\frac{1}{2\sqrt{n}}$. Now the assertion (4.4) follows from (1.1), (4.1), (4.8) and $\lim_{t\rightarrow0}\frac{e^{t}-1}{t}=1$. □

Theorem 4 tells us that Lord Brouncker’s continued fraction formula also provides a better approximation for $W(n)$. It is a natural question whether there exists a similar continued fraction approximation formula for $\frac{\Gamma^{3}(x+\frac{1}{3})}{\Gamma^{3}(x+1)}$ or $\frac{\Gamma^{3}(x+\frac{2}{3})}{\Gamma^{3}(x+1)}$. To the best knowledge of authors, no one has even obtained such a kind of formulas, for example, see Berndt [18], Chapter 12. By the multiple-correction method and a large number of measurement data, we propose the following conjectures.

Open problems

(i) There exist two sequences $(\kappa_{j})_{j\ge0}$ and $(\lambda _{j})_{j\ge0}$ of positive numbers such that

$$ \frac{\Gamma^{3}(x+\frac{1}{3})}{\Gamma^{3}(x+1)}=\mathop{\mathbf{K}}_{j=0}^{\infty} \biggl( \frac{\kappa_{j}}{ (x+\frac{1}{6} )^{2}+\lambda_{j}} \biggr),\quad x\ge1, $$

(4.11)

where

$$\begin{aligned}& \kappa_{0}=1, \qquad (\kappa_{1},\kappa_{2}, \kappa_{3},\kappa_{4},\ldots)=-\frac {2}{3^{6}} \biggl(1, \frac{2^{3} 5^{3}}{3^{2}},\frac{4^{3} 7^{3}}{5^{2}},\frac{5^{3} 11^{3}}{7^{2}}, \frac{7^{3} 13^{3}}{9^{2}}, \ldots \biggr), \end{aligned}$$

(4.12)

$$\begin{aligned}& (\lambda_{0},\lambda_{1},\lambda_{2}, \lambda_{3},\lambda_{4},\ldots)= \frac{1}{2^{2}3^{3}}\biggl(5, \frac{5^{2}}{1\cdot3},\frac{3\text{,}307}{3\cdot5},\frac{17\text{,}167}{5\cdot7},\frac {5\cdot31\cdot353}{7\cdot9}, \ldots\biggr). \end{aligned}$$

(4.13)

(ii) We have

$$ \frac{\Gamma^{3}(x+\frac{2}{3})}{\Gamma^{3}(x+1)}=\mathop{\mathbf{K}}_{j=0}^{\infty} \biggl( \frac{\nu_{j}}{x+\frac{1}{3}} \biggr), \quad x\ge1, $$

(4.14)

where $\nu_{0}=1$, $\nu_{1}=\frac{1}{27}$, and for $j\ge2$,

$$\begin{aligned}& \nu_{j}=\frac{1}{54}\frac{ (j+\lfloor\frac{j-1}{2} \rfloor )^{3}}{2\lfloor\frac{j-1}{2}\rfloor+1}, \\& \quad \textit{i.e. } (\nu_{2},\nu_{3},\nu_{4}, \ldots)=\frac{1}{54} \biggl(\frac{2^{3}}{1}, \frac{4^{3}}{3}, \frac{5^{3}}{3},\frac{7^{3}}{5}, \frac{8^{3}}{5},\frac{10^{3}}{7}, \frac{11^{3}}{7}, \frac{13^{3}}{9},\frac{14^{3}}{9},\frac{16^{3}}{11}, \ldots \biggr). \end{aligned}$$

(4.15)

Remark 4

It seems difficult to prove the above conjectures. However, by the same method as the proof of Theorem 1, we can prove many sharp inequalities, e.g., for $n\in\mathbb{N}$ and $0\le k\le10$, we have

$$ \frac{\tau}{\sqrt[3]{n+\frac{1}{3}+\mathop{\mathbf{K}}_{j=1}^{2k+1} (\frac{\nu_{j}}{n+\frac{1}{3}} )}}< T_{2}(n)< \frac{\tau}{\sqrt [3]{n+\frac{1}{3}+\mathop{\mathbf{K}}_{j=1}^{2k} (\frac{\nu_{j}}{n+\frac{1}{3}} )}}. $$

(4.16)