Open Access

Sharp inequalities and asymptotic expansion associated with the Wallis sequence

Journal of Inequalities and Applications20152015:186

https://doi.org/10.1186/s13660-015-0699-z

Received: 2 January 2015

Accepted: 18 May 2015

Published: 10 June 2015

Abstract

We present asymptotic expansion of function involving the ratio of gamma functions and provide a recurrence relation for determining the coefficients of the asymptotic expansion. As a consequence, we obtain asymptotic expansion of the Wallis sequence. Also, we establish sharp inequalities for the Wallis sequence.

Keywords

Wallis sequencegamma functionpsi functionpolygamma functioninequalityasymptotic expansion

MSC

40A0533B1541A6026D15

1 Introduction

The Wallis sequence to which the title refers is
$$ W_{n}=\prod_{k=1}^{n} \frac{4k^{2}}{4k^{2}-1},\quad n\in\mathbb{N}:=\{1,2,3,\ldots\}. $$
(1.1)
Wallis (1616-1703) discovered that
$$ \prod_{k=1}^{\infty} \frac{4k^{2}}{4k^{2}-1}=\frac{2}{1}\frac{2}{3}\frac {4}{3} \frac{4}{5}\frac{6}{5}\frac{6}{7}\frac{8}{7}=\cdots= \frac{\pi}{2} $$
(1.2)
(see [1], p.68). Several elementary proofs of (1.2) can be found (see, for example, [24]). An interesting geometric construction produces (1.2) [5]. Many formulas exist for the representation of π, and a collection of these formulas is listed [6, 7]. For more history of π see [1, 810].
Some inequalities and asymptotic formulas associated with the Wallis sequence \(W_{n}\) can be found (see, for example, [1124]). Hirschhorn [13] proved that for \(n\in\mathbb{N}\),
$$ \frac{\pi}{2} \biggl(1-\frac{1}{4n+\frac{7}{3}} \biggr)< W_{n}< \frac{\pi }{2} \biggl(1-\frac{1}{4n+\frac{8}{3}} \biggr). $$
(1.3)
Also in [13], Hirschhorn pointed out that if the \(c_{j}\) are given by
$$ \tanh \biggl(\frac{x}{4} \biggr)=\sum _{j=0}^{\infty}c_{j}\frac{x^{2j+1}}{(2j)!}, $$
(1.4)
then, as \(n\to\infty\),
$$ W_{n} \sim\frac{\pi}{2} \biggl(1+\frac{1}{2n} \biggr)^{-1}\prod_{j\geq 0}\exp \biggl( \frac{c_{j}}{n^{2j+1}} \biggr)=\frac{\pi}{2} \biggl(1+\frac {1}{2n} \biggr)^{-1}\exp \Biggl(\sum_{j=0}^{\infty} \frac {c_{j}}{n^{2j+1}} \Biggr). $$
(1.5)
Very recently, Lin et al. [17] found that
$$ c_{j}=\frac{(2^{2j+2}-1)B_{2j+2}}{2^{2j+1}(2j+1)(j+1)},\quad j\in\mathbb{N}_{0}:= \mathbb{N}\cup\{0\}, $$
(1.6)
where \(B_{n}\) (\(n\in\mathbb{N}_{0}\)) are the Bernoulli numbers defined by the following generating function:
$$ \frac{z}{e^{z}-1}=\sum_{n=0}^{\infty} B_{n} \frac{z^{n}}{n!},\quad |z|< 2\pi. $$
Also in [17], Lin et al. derived
$$ W_{n}=\frac{\pi}{2} \biggl(1-\frac{1}{4n+\frac{5}{2}} \biggr)^{1-\frac {3}{64n^{2}}+\frac{3}{64n^{3}}-\frac{23}{1\text{,}024n^{4}}+O(n^{-5})},\quad n\to\infty. $$
(1.7)
The gamma function is defined for \(x>0\) by
$$ \Gamma(x)=\int^{\infty}_{0}t^{x-1} e^{-t}\,\mathrm{d} t. $$
The logarithmic derivative of \(\Gamma(x)\), denoted by \(\psi(x)=\Gamma'(x)/\Gamma(x)\), is called psi (or digamma) function, and \(\psi^{(k)}(x)\) (\(k\in\mathbb{N}\)) are called polygamma functions. These functions play an important role in various branches of mathematics as well as in physics and engineering. For the various properties of these functions, please refer to [25], pp.255-260.
Define the function \(W(x)\) by
$$ W(x)=\frac{\pi}{2} \biggl(1+\frac{1}{2x} \biggr)^{-1}\frac{1}{x} \biggl[\frac{\Gamma(x+1)}{\Gamma(x+\frac{1}{2})} \biggr]^{2}. $$
(1.8)
It is easy to see that
$$ W_{n}=W(n). $$
The first aim of present paper is to establish sharp inequalities for \(W_{n}\). More precisely, we determine the best possible constants α, β, λ, and μ such that the double inequalities
$$ \frac{\pi}{2} \biggl(1-\frac{1}{4n+\alpha} \biggr)< W_{n}\leq \frac{\pi}{2} \biggl(1-\frac{1}{4n+\beta} \biggr) $$
and
$$ \frac{\pi}{2} \biggl(1-\frac{1}{4n+\frac{5}{2}} \biggr)^{\lambda}< W_{n}\leq\frac{\pi}{2} \biggl(1- \frac{1}{4n+\frac{5}{2}} \biggr)^{\mu} $$
hold for all \(n\in\mathbb{N}\). The second aim of present paper is to develop the formula (1.7) to produce a complete asymptotic expansion. More precisely, we provide a recurrence relation for determining the coefficients \(r_{j}\) (\(j\in\mathbb{N}_{0}\)) such that
$$ W(x)\sim\frac{\pi}{2} \biggl(1-\frac{1}{4x+\frac{5}{2}} \biggr)^{\sum _{j=0}^{\infty}r_{j}x^{-j}},\quad x\to\infty. $$

2 Lemmas

The following lemmas are required in our present investigation.

Lemma 1

([26], Corollary 1)

Let \(m, n\in\mathbb{N}\). Then for \(x>0\),
$$\begin{aligned}& \sum_{j=1}^{2m} \biggl(1- \frac{1}{2^{2j}} \biggr)\frac {2B_{2j}}{(2j)!}\frac{(2j+n-2)!}{x^{2j+n-1}} \\& \quad < (-1)^{n} \biggl(\psi^{(n-1)}(x+1)-\psi^{(n-1)} \biggl(x+\frac {1}{2} \biggr) \biggr)+\frac{(n-1)!}{2x^{n}} \\& \quad < \sum_{j=1}^{2m-1} \biggl(1- \frac{1}{2^{2j}} \biggr)\frac {2B_{2j}}{(2j)!}\frac{(2j+n-2)!}{x^{2j+n-1}}, \end{aligned}$$
(2.1)
where \(B_{n}\) are the Bernoulli numbers.
It follows from (2.1) that, for \(x>0\),
$$ \frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}}- \frac{1}{128x^{6}}< \psi (x+1)-\psi \biggl(x+\frac{1}{2} \biggr)< \frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}} $$
(2.2)
and
$$ -\frac{1}{2x^{2}}+\frac{1}{4x^{3}}-\frac{1}{16x^{5}}< \psi'(x+1)-\psi' \biggl(x+\frac{1}{2} \biggr)< - \frac{1}{2x^{2}}+\frac{1}{4x^{3}}-\frac {1}{16x^{5}}+\frac{3}{64x^{7}}. $$
(2.3)

Lemma 2

For all \(x\geq1\),
$$ \biggl[\frac{\Gamma(x+1)}{\Gamma(x+\frac{3}{2})} \biggr]^{2}< \frac {1}{x}-\frac{3}{4x^{2}}+\frac{17}{32x^{3}}-\frac{45}{128x^{4}}+ \frac{443}{2\text{,}048x^{5}}. $$
(2.4)

Proof

We consider the function \(G(x)\) defined by
$$\begin{aligned} G(x) =&2\ln\Gamma(x+1)-2 \biggl[\ln\Gamma \biggl(x+\frac{1}{2} \biggr)+ \ln \biggl(x+\frac{1}{2} \biggr) \biggr] \\ &{} -\ln \biggl(\frac{1}{x}-\frac{3}{4x^{2}}+\frac{17}{32x^{3}}- \frac {45}{128x^{4}}+\frac{443}{2\text{,}048x^{5}} \biggr). \end{aligned}$$
From the asymptotic expansion ([25], p.257):
$$\begin{aligned} x^{b-a}\frac{\Gamma(x+a)}{\Gamma(x+b)} =&1+\frac{(a-b)(a+b-1)}{2x} \\ &{}+\frac{1}{12}\binom{a-b}{2} \bigl(3(a+b-1)^{2}-a+b-1 \bigr)\frac {1}{x^{2}}+\cdots \quad \text{as } x \to\infty, \end{aligned}$$
(2.5)
we conclude that
$$ \lim_{x\to\infty}G(x)=0. $$
Differentiating and applying the first inequality in (2.2) yield, for \(x\geq1\),
$$\begin{aligned} G'(x) =&2 \biggl[\psi(x+1)-\psi \biggl(x+\frac{1}{2} \biggr) \biggr] \\ &{} -\frac {4\text{,}096x^{5}-2\text{,}048x^{4}+896x^{3}-384x^{2}+222x-2\text{,}215}{x(2x+1)(2\text{,}048x^{4}-1\text{,}536x^{3}+1\text{,}088x^{2}-720x+443)} \\ >&2 \biggl(\frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}}- \frac {1}{128x^{6}} \biggr) \\ &{} -\frac {4\text{,}096x^{5}-2\text{,}048x^{4}+896x^{3}-384x^{2}+222x-2\text{,}215}{x(2x+1)(2\text{,}048x^{4}-1\text{,}536x^{3}+1\text{,}088x^{2}-720x+443)} \\ =& \bigl(158\text{,}193+797\text{,}514(x-1)+1\text{,}606\text{,}106(x-1)^{2}+1\text{,}619\text{,}020(x-1)^{3} \\ &{}+816\text{,}432(x-1)^{4}+164\text{,}640(x-1)^{5}\bigr) /\bigl(64x^{6}(2x+1) \bigl(1\text{,}323+5\text{,}040(x-1) \\ &{}+8\text{,}768(x-1)^{2} +6\text{,}656(x-1)^{3}+2\text{,}048(x-1)^{4} \bigr)\bigr) \\ >&0. \end{aligned}$$
This leads to
$$\begin{aligned} G(x) =&\ln \biggl[\frac{\Gamma(x+1)}{\Gamma(x+\frac{3}{2})} \biggr]^{2}-\ln \biggl( \frac{1}{x}-\frac{3}{4x^{2}}+\frac{17}{32x^{3}}-\frac {45}{128x^{4}}+ \frac{443}{2\text{,}048x^{5}} \biggr) \\ < &\lim_{x\to\infty}G(x)=0,\quad x\geq1. \end{aligned}$$
The proof of Lemma 2 is complete. □
By (2.2), we obtain
$$\begin{aligned} (2x+1) \biggl[\psi(x+1)-\psi \biggl(x+\frac{1}{2} \biggr) \biggr]-1 < &(2x+1) \biggl(\frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}} \biggr)-1 \\ =&\frac{1}{4x}-\frac{1}{8x^{2}}+\frac{1}{32x^{3}}+\frac{1}{64x^{4}}. \end{aligned}$$
(2.6)
By (2.4), we get
$$\begin{aligned} 1- \biggl(x+\frac{1}{2} \biggr) \biggl[\frac{\Gamma(x+1)}{\Gamma(x+\frac {3}{2})} \biggr]^{2} >&1- \biggl(x+\frac{1}{2} \biggr) \biggl( \frac{1}{x}-\frac {3}{4x^{2}}+\frac{17}{32x^{3}}-\frac{45}{128x^{4}}+ \frac{443}{2\text{,}048x^{5}} \biggr) \\ =&\frac{1}{4x}-\frac{5}{32x^{2}}+\frac{11}{128x^{3}}-\frac{83}{2\text{,}048x^{4}}- \frac {443}{4\text{,}096x^{5}}. \end{aligned}$$
(2.7)
The proof of Theorem 1 makes use of (2.6) and (2.7).

Lemma 3

([27])

Let \(-\infty\leq a< b\leq\infty\). Let f and g be differentiable functions on an interval \((a, b)\). Assume that either \(g'>0\) everywhere on \((a, b)\) or \(g'<0\) on \((a, b)\). Suppose that \(f(a+)=g(a+)=0\) or \(f(b-)=g(b-)=0\). Then
  1. (1)

    if \(\frac{f'}{g'}\) is increasing on \((a, b)\), then \((\frac{f}{g} )'>0\) on \((a, b)\);

     
  2. (2)

    if \(\frac{f'}{g'}\) is decreasing on \((a, b)\), then \((\frac{f}{g} )'<0\) on \((a, b)\).

     

3 Sharp inequalities

Theorem 1

For all \(n \in\mathbb{N}\),
$$ \frac{\pi}{2} \biggl(1-\frac{1}{4n+\alpha} \biggr)< W_{n}\leq \frac{\pi}{2} \biggl(1-\frac{1}{4n+\beta} \biggr) $$
(3.1)
with the best possible constants
$$ \alpha=\frac{5}{2} \quad \textit{and}\quad \beta=\frac{32-9\pi}{3\pi-8}=2.614909986 \ldots. $$
Equality in (3.1) occurs for \(n=1\).

Proof

The inequality (3.1) can be written as
$$ \alpha\leq F(n)< \beta, $$
where
$$ F(x)=\frac{1}{1-\frac{1}{x+1/2} [\frac{\Gamma(x+1)}{\Gamma (x+1/2)} ]^{2}}-4x. $$
Using (2.5), we conclude that
$$ \lim_{x\to\infty}F(x)=\frac{5}{2}. $$
Differentiating \(F(x)\) and applying (2.4), (2.6), and (2.7) yield, for \(x\geq6\),
$$\begin{aligned}& \biggl(1- \biggl(x+\frac{1}{2} \biggr) \biggl[\frac{\Gamma(x+1)}{\Gamma (x+\frac{3}{2})} \biggr]^{2} \biggr)^{2}F'(x) \\& \quad = \biggl\{ (2x+1) \biggl[\psi(x+1)-\psi \biggl(x+\frac{1}{2} \biggr) \biggr]-1 \biggr\} \biggl[\frac{\Gamma(x+1)}{\Gamma(x+\frac{3}{2})} \biggr]^{2} \\& \qquad {} -4 \biggl(1- \biggl(x+\frac{1}{2} \biggr) \biggl[ \frac{\Gamma (x+1)}{\Gamma(x+\frac{3}{2})} \biggr]^{2} \biggr)^{2} \\& \quad < \biggl(\frac{1}{4x}-\frac{1}{8x^{2}}+\frac{1}{32x^{3}}+ \frac {1}{64x^{4}} \biggr) \biggl(\frac{1}{x}-\frac{3}{4x^{2}}+ \frac{17}{32x^{3}}-\frac {45}{128x^{4}}+\frac{443}{2\text{,}048x^{5}} \biggr) \\& \qquad {} -4 \biggl(\frac{1}{4x}-\frac{5}{32x^{2}}+\frac{11}{128x^{3}}- \frac {83}{2\text{,}048x^{4}}-\frac{443}{4\text{,}096x^{5}} \biggr)^{2} \\& \quad =-\frac{1}{4\text{,}194\text{,}304x^{10}} \bigl(248\text{,}771\text{,}713+769\text{,}183 \text{,}956(x-6)+510\text{,}154\text{,}660(x-6)^{2} \\& \qquad {} +149\text{,}038\text{,}464(x-6)^{3}+22\text{,}221 \text{,}824(x-6)^{4} \\& \qquad {}+1\text{,}658\text{,}880(x-6)^{5}+49 \text{,}152(x-6)^{6} \bigr) \\& \quad < 0. \end{aligned}$$
Straightforward calculation produces
$$\begin{aligned}& F(1)=\frac{32-9\pi}{3\pi-8}=2.6149\ldots, \\& F(2)=\frac{-315\pi+1\text{,}024}{45\pi-128}=2.5724\ldots, \\& F(3)=\frac{-1\text{,}925\pi+6\text{,}144}{175\pi-512}=2.5526\ldots, \\& F(4)=\frac{-165\text{,}375\pi+524\text{,}288}{11\text{,}025\pi-32\text{,}768}=2.5412\ldots, \\& F(5)=\frac{-829\text{,}521\pi+2\text{,}621\text{,}440}{43\text{,}659\pi-131\text{,}072}=2.5338\ldots, \\& F(6)=\frac{-15\text{,}954\text{,}939\pi+50\text{,}331\text{,}648}{693\text{,}693\pi-2\text{,}097\text{,}152}=2.5286\ldots. \end{aligned}$$
Thus, the sequence \((F(n) )_{n\in\mathbb{N}}\) is strictly decreasing. This leads to
$$ \frac{5}{2}< \lim_{x \to\infty}F(x) < F(n)\leq F(1)=\frac{32-9\pi}{3\pi-8}, \quad n\in\mathbb{N}. $$
The proof of Theorem 1 is complete. □

Remark 1

In fact, Elezović et al. [12] have previously shown that \(\frac{5}{2}\) is the best possible constant for a lower bound of \(W_{n}\) of the type \(\frac{\pi}{2} (1-\frac{1}{4n+\alpha} )\). Moreover, the authors pointed out that
$$ W_{n}=\frac{\pi}{2} \biggl(1-\frac{1}{4n+\frac{5}{2}} \biggr)+O \biggl( \frac{1}{n^{3}} \biggr), \quad n\to\infty. $$

Theorem 2

For all \(n\in\mathbb{N}\),
$$ \frac{\pi}{2} \biggl(1-\frac{1}{4n+\frac{5}{2}} \biggr)^{\lambda}< W_{n}\leq\frac{\pi}{2} \biggl(1- \frac{1}{4n+\frac{5}{2}} \biggr)^{\mu} $$
(3.2)
with the best possible constants
$$ \lambda=1\quad \textit{and}\quad \mu=\frac{\ln(3\pi/8)}{\ln(13/11)}=0.98112316\ldots. $$
Equality in (3.2) occurs for \(n=1\).

Proof

Inequality (3.2) can be written as
$$ \lambda>x_{n}\geq\mu, $$
where the sequence \((x_{n} )_{n\in\mathbb{N}}\) is defined by
$$ x_{n}=\frac{\ln (\frac{1}{n+\frac{1}{2}} (\frac{\Gamma (n+1)}{\Gamma(n+\frac{1}{2})} )^{2} )}{\ln (1-\frac {1}{4n+\frac{5}{2}} )}. $$
We are now in a position to show that the sequence \((x_{n} )_{n\in\mathbb{N}}\) is strictly increasing. To this end, we consider the function \(f(x)\) defined by
$$ f(x)=\frac{2\ln\Gamma(x+1)-2\ln\Gamma (x+\frac{1}{2} )-\ln (x+\frac{1}{2} )}{\ln (1-\frac{1}{4x+\frac{5}{2}} )}=\frac {f_{1}(x)}{f_{2}(x)}, $$
where
$$ f_{1}(x)=2\ln\Gamma(x+1)-2\ln\Gamma \biggl(x+\frac{1}{2} \biggr)-\ln \biggl(x+\frac{1}{2} \biggr) $$
and
$$ f_{2}(x)=\ln \biggl(1-\frac{1}{4x+\frac{5}{2}} \biggr). $$
We conclude from the asymptotic formula of \(\ln\Gamma(z)\) ([25], p.257) that
$$ f_{1}(\infty)=\lim_{x\to\infty}f_{1}(x)=0. $$
Elementary calculations show that
$$ \frac{8f'_{1}(x)}{f'_{2}(x)}=\bigl(64x^{2}+64x+15\bigr) \biggl[\psi(x+1)-\psi \biggl(x+\frac{1}{2} \biggr)-\frac{1}{2x+1} \biggr]=: f_{3}(x). $$
By using inequalities (2.2) and (2.3), we obtain, for \(x\geq2\),
$$\begin{aligned} f'_{3}(x) =&(128x+64) \biggl[\psi(x+1)-\psi \biggl(x+ \frac{1}{2} \biggr)-\frac {1}{2x+1} \biggr] \\ &{} +\bigl(64x^{2}+64x+15\bigr) \biggl[\psi'(x+1)- \psi' \biggl(x+\frac{1}{2} \biggr)+\frac{2}{(2x+1)^{2}} \biggr] \\ >&(128x+64) \biggl[\frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}}- \frac {1}{128x^{6}}-\frac{1}{2x+1} \biggr] \\ &{} +\bigl(64x^{2}+64x+15\bigr) \biggl[-\frac{1}{2x^{2}}+ \frac{1}{4x^{3}}-\frac {1}{16x^{5}}+\frac{2}{(2x+1)^{2}} \biggr] \\ =&\frac{202+4\text{,}881(x-2)+7\text{,}860(x-2)^{2}+4\text{,}896(x-2)^{3} +1\text{,}368(x-2)^{4}+144(x-2)^{5}}{16x^{6}(2x+1)^{2}} \\ >&0. \end{aligned}$$
Hence, \(f_{3}(x)\) and \(\frac{f'_{1}(x)}{f'_{2}(x)}\) are both strictly increasing for \(x\geq2\). By Lemma 3, the function
$$ f(x)=\frac{f_{1}(x)}{f_{2}(x)}=\frac{f_{1}(x)-f_{1}(\infty )}{f_{2}(x)-f_{2}(\infty)} $$
is strictly increasing for \(x\geq2\). Therefore, the sequence \((x_{n} )\) is strictly increasing for \(n\geq2\). Direct computation would yield
$$ x_{1}=\frac{\ln(3\pi/8)}{\ln(13/11)}=0.9811\ldots,\qquad x_{2}= \frac{-7\ln2+2\ln3+\ln\pi+\ln5}{-\ln19+\ln3+\ln7}=0.9927\ldots. $$
Consequently, the sequence \((x_{n} )_{n\in\mathbb{N}}\) is strictly increasing. This leads to
$$ \lim_{n \to\infty}x_{n}>x_{n}\geq x_{1}=\frac{\ln(3\pi/8)}{\ln(13/11)} \quad \text{for } n\in\mathbb{N}. $$
It remains to prove that
$$ \lim_{n \to\infty}x_{n}=1. $$
(3.3)
We conclude from the asymptotic formula of \(\ln\Gamma(z)\) ([25], p.257) that
$$ f(x)= \frac{1+O(x^{-1})}{1+O(x^{-1})}\to1\quad \text{as } x\to \infty, $$
which implies (3.3). This completes the proof of Theorem 2. □

4 Asymptotic expansion

Theorem 3

The function \(W(x)\), defined by (1.8), has the following asymptotic expansion:
$$ W(x)\sim\frac{\pi}{2} \biggl(1-\frac{1}{4x+\frac{5}{2}} \biggr)^{\sum _{j=0}^{\infty}r_{j}x^{-j}},\quad x\to\infty, $$
(4.1)
with the coefficients \(r_{j}\) given by the recurrence relation
$$ r_{0}=1, \qquad r_{j}=4\sum _{k=0}^{j-1}r_{k}q_{j-k+1}-4p_{j+1}, \quad j\in\mathbb{N}, $$
(4.2)
where
$$ p_{j}=(-1)^{j-1} \biggl(-\frac{1}{j2^{j}}+ \frac{2 ((-1)^{j+1}-(2^{-j}-1) )B_{j+1}}{j(j+1)} \biggr),\quad j\in\mathbb{N} $$
(4.3)
and
$$ q_{j}=-\sum_{k=0}^{j-1} \frac{1}{(k+1)\cdot4^{k+1}}\binom{j-1}{j-k-1} \biggl(-\frac{5}{8} \biggr)^{j-k-1},\quad j\in\mathbb{N}. $$
(4.4)
Here, \(B_{n}\) are the Bernoulli numbers.

Proof

Write (4.1) as
$$ \frac{\ln (\frac{2}{\pi}W(x) )}{\ln (1-\frac{1}{4x+\frac {5}{2}} )}\sim \sum_{j=0}^{\infty} \frac{r_{j}}{x^{j}},\quad x \to\infty. $$
(4.5)
The logarithm of gamma function has asymptotic expansion (see [28], p.32):
$$ \ln\Gamma(x+t)\sim \biggl(x+t-\frac{1}{2} \biggr)\ln x-x+ \frac{1}{2}\ln(2\pi)+\sum_{n=1}^{\infty} \frac {(-1)^{n+1}B_{n+1}(t)}{n(n+1)}\frac{1}{x^{n}} $$
(4.6)
as \(x\to\infty\), where \(B_{n}(t)\) denotes the Bernoulli polynomials defined by the following generating function:
$$ \frac{xe^{tx}}{e^{x}-1}=\sum_{n=0}^{\infty}B_{n}(t) \frac{x^{n}}{n!}. $$
From (4.6), we obtain, as \(x\to\infty\),
$$ \biggl[\frac{\Gamma(x+t)}{\Gamma(x+s)} \biggr]^{1/(t-s)}\sim x\exp \Biggl(\frac{1}{t-s}\sum_{j=1}^{\infty} \frac{(-1)^{j+1} (B_{j+1}(t)-B_{j+1}(s) )}{j(j+1)}\frac{1}{x^{j}} \Biggr). $$
(4.7)
Setting \((s, t)=(\frac{1}{2}, 1)\) in (4.7) and noting that
$$ B_{n}(0)=(-1)^{n}B_{n}(1)=B_{n} \quad \text{and}\quad B_{n} \biggl(\frac{1}{2} \biggr)= \bigl(2^{1-n}-1\bigr)B_{n}\quad \text{for } n \in \mathbb{N}_{0} $$
(see [25], p.805), we obtain, as \(x\to\infty\),
$$ \biggl[\frac{\Gamma(x+1)}{\Gamma(x+\frac{1}{2})} \biggr]^{2}\sim x\exp \Biggl( \sum_{j=1}^{\infty}\frac{2 (1-(-1)^{j+1}(2^{-j}-1) )B_{j+1}}{j(j+1)} \frac{1}{x^{j}} \Biggr). $$
(4.8)
By using the Maclaurin expansion of \(\ln(1+t)\),
$$ \ln(1+t)=\sum_{j=1}^{\infty} \frac{(-1)^{j-1}}{j}t^{j} \quad \text{for } {-}1< t\leq1, $$
we obtain
$$ \biggl(1+\frac{1}{2x} \biggr)^{-1} \sim\exp \Biggl(\sum_{j=1}^{\infty}\frac {(-1)^{j}}{j2^{j}} \frac{1}{x^{j}} \Biggr)\quad \text{as } x\to\infty. $$
(4.9)
Applying (4.8) and (4.9) yields
$$ \ln \biggl(\frac{2}{\pi}W(x) \biggr)\sim\sum _{j=1}^{\infty}\frac{p_{j}}{x^{j}},\quad x \to\infty $$
(4.10)
with
$$ p_{j}=(-1)^{j-1} \biggl(-\frac{1}{j2^{j}}+ \frac{2 ((-1)^{j+1}-(2^{-j}-1) )B_{j+1}}{j(j+1)} \biggr),\quad j\in\mathbb{N}. $$
(4.11)
The Maclaurin expansion of \(\ln(1+t)\) with \(t=-\frac{1}{4x+\frac{5}{2}}\), yields
$$\begin{aligned} \ln \biggl(1-\frac{1}{4x+\frac{5}{2}} \biggr)&\sim-\sum_{j=1}^{\infty} \frac {1}{j\cdot4^{j}x^{j}} \biggl(1+\frac{5}{8x} \biggr)^{-j} \\ &\sim-\sum_{j=1}^{\infty}\frac{1}{j\cdot4^{j}x^{j}} \sum_{k=0}^{\infty }\binom{-j}{k} \frac{5^{k}}{8^{k}x^{k}} \\ &\sim-\sum_{j=1}^{\infty}\frac{1}{j\cdot4^{j}x^{j}} \sum_{k=0}^{\infty }(-1)^{k} \binom{k+j-1}{k}\frac{5^{k}}{8^{k}x^{k}} \\ &\sim-\sum_{j=1}^{\infty}\sum _{k=0}^{j-1}\frac{1}{(k+1)\cdot 4^{k+1}}\binom{j-1}{j-k-1} \biggl(-\frac{5}{8} \biggr)^{j-k-1}\frac{1}{x^{j}}. \end{aligned}$$
That is,
$$ \ln \biggl(1-\frac{1}{4x+\frac{5}{2}} \biggr) \sim\sum_{j=1}^{\infty} \frac{q_{j}}{x^{j}} $$
with
$$ q_{j}=-\sum_{k=0}^{j-1} \frac{1}{(k+1)\cdot4^{k+1}}\binom{j-1}{j-k-1} \biggl(-\frac{5}{8} \biggr)^{j-k-1},\quad j\in\mathbb{N}. $$
It follows from (4.5) that
$$\begin{aligned}& \frac{\sum_{j=1}^{\infty}p_{j}x^{-j}}{\sum_{j=1}^{\infty}q_{j}x^{-j}}\sim \sum_{j=0}^{\infty} \frac{r_{j}}{x^{j}}, \\& \sum_{j=1}^{\infty} \frac{p_{j}}{x^{j}} \sim \sum_{j=0}^{\infty} \frac{r_{j}}{x^{j}}\sum_{k=1}^{\infty} \frac{q_{k}}{x^{k}}, \\& \sum_{j=1}^{\infty} \frac{p_{j}}{x^{j}} \sim\sum_{j=1}^{\infty} \Biggl(\sum_{k=0}^{j-1}r_{k}q_{j-k} \Biggr)\frac{1}{x^{j}}. \end{aligned}$$
We then obtain
$$\begin{aligned}& p_{j}=\sum_{k=0}^{j-1}r_{k}q_{j-k}, \quad j\in\mathbb{N}, \\& p_{j}=\sum_{k=0}^{j-2}r_{k}q_{j-k}+r_{j-1}q_{1}, \quad j\geq2. \end{aligned}$$
Noting that \(q_{1}=-\frac{1}{4}\), we obtain
$$ r_{j-1}=4\sum_{k=0}^{j-2}r_{k}q_{j-k}-4p_{j}, \quad j\geq2, $$
and an empty sum (as usual) is understood to be nil. Noting that \(p_{1}=-\frac{1}{4}\), we then obtain the recurrence relation
$$ r_{0}=1, \qquad r_{j}=4\sum _{k=0}^{j-1}r_{k}q_{j-k+1}-4p_{j+1}, \quad j\in\mathbb{N}. $$
The proof of Theorem 3 is complete. □
Here, from (4.1), we give the following explicit asymptotic expansion:
$$ W_{n}=\frac{\pi}{2} \biggl(1-\frac{1}{4n+\frac{5}{2}} \biggr)^{1-\frac {3}{64n^{2}}+\frac{3}{64n^{3}}-\frac{23}{1\text{,}024n^{4}}-\frac{1}{512n^{5}}+\cdots },\quad n\to\infty, $$
(4.12)
which develops the formula (1.7) to produce a complete asymptotic expansion.

Declarations

Acknowledgements

The authors thank the referees for their careful reading of the manuscript and insightful comments.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Informatics, Henan Polytechnic University

References

  1. Berggren, L, Borwein, J, Borwein, P (eds.): Pi: A Source Book, 2nd edn. Springer, New York (2000) Google Scholar
  2. Amdeberhan, T, Espinosa, O, Moll, VH, Straub, A: Wallis-Ramanujan-Schur-Feynman. Am. Math. Mon. 117, 618-632 (2010) MATHMathSciNetView ArticleGoogle Scholar
  3. Miller, SJ: A probabilistic proof of Wallis’ formula for π. Am. Math. Mon. 115, 740-745 (2008) MATHGoogle Scholar
  4. Wästlund, J: An elementary proof of the Wallis product formula for pi. Am. Math. Mon. 114, 914-917 (2007) MATHGoogle Scholar
  5. Myerson, G: The limiting shape of a sequence of rectangles. Am. Math. Mon. 99, 279-280 (1992) MathSciNetView ArticleGoogle Scholar
  6. Sofo, A: Some representations of π. Aust. Math. Soc. Gaz. 31, 184-189 (2004) MATHMathSciNetGoogle Scholar
  7. Sofo, A: π and some other constants. J. Inequal. Pure Appl. Math. 6(5), Article 138 (2005) (electronic) MathSciNetGoogle Scholar
  8. Beckmann, P: A History of Pi. St. Martin’s, New York (1971) MATHGoogle Scholar
  9. Dunham, W: Journey Through Genius: The Great Theorems of Mathematics. Penguin, Baltimore (1990) MATHGoogle Scholar
  10. Agarwal, RP, Agarwal, H, Sen, SK: Birth, growth and computation of pi to ten trillion digits. Adv. Differ. Equ. 2013, 100 (2013). doi:10.1186/1687-1847-2013-100 MathSciNetView ArticleGoogle Scholar
  11. Burić, T, Elezović, N, Šimić, R: Asymptotic expansions of the multiple quotients of gamma functions with applications. Math. Inequal. Appl. 16, 1159-1170 (2013) MATHMathSciNetGoogle Scholar
  12. Elezović, N, Lin, L, Vukšić, L: Inequalities and asymptotic expansions for the Wallis sequence and the sum of the Wallis ratio. J. Math. Inequal. 7, 679-695 (2013) MATHMathSciNetView ArticleGoogle Scholar
  13. Hirschhorn, MD: Comments on the paper: ‘Wallis sequence estimated through the Euler-Maclaurin formula: even from the Wallis product π could be computed fairly accurately’ by V. Lampret. Aust. Math. Soc. Gaz. 32, 194 (2005) MATHMathSciNetGoogle Scholar
  14. Lampret, V: Wallis sequence estimated through the Euler-Maclaurin formula: even from the Wallis product π could be computed fairly accurately. Aust. Math. Soc. Gaz. 31, 328-339 (2004) MathSciNetGoogle Scholar
  15. Lampret, V: An asymptotic approximation of Wallis’ sequence. Cent. Eur. J. Math. 10, 775-787 (2012) MATHMathSciNetView ArticleGoogle Scholar
  16. Lin, L: Further refinements of Gurland’s formula for π. J. Inequal. Appl. 2013, 48 (2013) View ArticleGoogle Scholar
  17. Lin, L, Deng, J-E, Chen, C-P: Inequalities and asymptotic expansions associated with the Wallis sequence. J. Inequal. Appl. 2014, 251 (2014) MathSciNetView ArticleGoogle Scholar
  18. Mortici, C: Estimating π from the Wallis sequence. Math. Commun. 17, 489-495 (2012) MATHMathSciNetGoogle Scholar
  19. Mortici, C, Cristea, VG, Lu, D: Completely monotonic functions and inequalities associated to some ratio of gamma function. Appl. Math. Comput. 240, 168-174 (2014) MathSciNetView ArticleGoogle Scholar
  20. Mortici, C: Completely monotone functions and the Wallis ratio. Appl. Math. Lett. 25, 717-722 (2012) MATHMathSciNetView ArticleGoogle Scholar
  21. Mortici, C: Sharp inequalities and complete monotonicity for the Wallis ratio. Bull. Belg. Math. Soc. Simon Stevin 17, 929-936 (2010) MATHMathSciNetGoogle Scholar
  22. Mortici, C: A new method for establishing and proving new bounds for the Wallis ratio. Math. Inequal. Appl. 13, 803-815 (2010) MathSciNetGoogle Scholar
  23. Mortici, C: New approximation formulas for evaluating the ratio of gamma functions. Math. Comput. Model. 52, 425-433 (2010) MATHMathSciNetView ArticleGoogle Scholar
  24. Păltănea, E: On the rate of convergence of Wallis’ sequence. Aust. Math. Soc. Gaz. 34, 34-38 (2007) Google Scholar
  25. Abramowitz, M, Stegun, IA (eds.): Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th edn. National Bureau of Standards: Applied Mathematics Series, vol. 55. Dover, New York (1972) MATHGoogle Scholar
  26. Chen, CP, Paris, RB: Inequalities, asymptotic expansions and completely monotonic functions related to the gamma function. Appl. Math. Comput. 250, 514-529 (2015) MathSciNetView ArticleGoogle Scholar
  27. Pinelis, I: L’Hospital type rules for monotonicity, with applications. J. Inequal. Pure Appl. Math. 3(1), Article 5 (2002) (electronic) MathSciNetGoogle Scholar
  28. Luke, YL: The Special Functions and Their Approximations, vol. I. Academic Press, New York (1969) MATHGoogle Scholar

Copyright

© Deng et al. 2015