Open Access

A sharp two-sided inequality for bounding the Wallis ratio

Journal of Inequalities and Applications20152015:43

https://doi.org/10.1186/s13660-015-0560-4

Received: 12 October 2014

Accepted: 14 January 2015

Published: 4 February 2015

Abstract

In this article we establish a sharp two-sided inequality for bounding the Wallis ratio. Some best constants for the estimation of the Wallis ratio are obtained. An asymptotic formula for the Wallis ratio is also presented.

Keywords

estimationinequalityasymptotic formulaWallis ratiogamma function

MSC

11B6541A4405A1026D2033B1541A60

1 Introduction and main results

For \(n\in\mathbb{N}\) (the set of all positive integers), the double factorial \(n!!\) is defined by
$$ n!!=\prod_{i=0}^{\lfloor(n-1)/2\rfloor}(n-2i), $$
(1)
where in (1) the floor function \(\lfloor t\rfloor\) denotes the largest integer less than or equal to t. For our own convenience, in what follows, we denote the ratio of two neighboring double factorials by
$$ W_{n}=\frac{(2n-1)!!}{(2n)!!}, $$
(2)
which is called the Wallis ratio in the literature.
The Wallis ratio \(W_{n}\) can be represented as follows (see [1, p.258]):
$$ W_{n}=\frac{(2n)!}{2^{2n}(n!)^{2}}=\frac{1}{2^{2n}} \binom{2n}{n}=\frac{\Gamma (n+\frac{1}{2} )}{\sqrt{\pi}\Gamma(n+1)}, $$
(3)
where in (3) \(\Gamma(x)\) is the classical Euler’s gamma function defined for \(x>0\) by
$$ \Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t} \,d t. $$
(4)
In [2] the author proved, for all \(n\in\mathbb{N}\),
$$ \biggl[n\pi \biggl(1+\frac{1}{4n-\frac{1}{2} +\frac{3}{16n+\frac{15}{4n}}} \biggr) \biggr]^{-1/2}< W_{n}< \biggl[n\pi \biggl(1+\frac{1}{4n-\frac{1}{2} +\frac{3}{16n}} \biggr) \biggr]^{-1/2}. $$
(5)

In [3], the following result was established:

For \(n\in\mathbb{N}\) and \(\varepsilon\in(0,\frac{1}{2})\),
$$ \biggl[n\pi \biggl(1+\frac{1}{4n-\frac{1}{2}} \biggr) \biggr]^{-1/2}< W_{n}< \biggl[n\pi \biggl(1+\frac{1}{4n-\frac{1}{2} +\varepsilon} \biggr) \biggr]^{-1/2}. $$
(6)
The right-hand inequality in (6) holds for \(n > n^{*}\), where \(n^{*}\) is the maximal root on
$$32\varepsilon n^{2} + 4\varepsilon^{2}n + 32\varepsilon n-17n + 4\varepsilon ^{2}-1 = 0. $$

We also note that in [4] the authors proved the result below.

For all \(n\in\mathbb{N}\)
$$ \frac{\sqrt{\pi}}{2\sqrt{n+\frac{9\pi}{16}-1}}\le\frac{(2n)!!}{(2n+1)!!} < \frac{\sqrt{\pi}}{2\sqrt{n+\frac{3}{4}}}, $$
(7)
which is equivalent to the following:
$$ \frac{2\sqrt{n+\frac{3}{4}}}{(2n+1)\sqrt{\pi}} < W_{n}\le \frac{2\sqrt{n+\frac{9\pi}{16}-1}}{(2n+1)\sqrt{\pi}}. $$
(8)
In this article we shall establish a sharp two-sided inequality for bounding the Wallis ratio in the form
$$ C_{1}P_{n}< W_{n}<C_{2}P_{n}, $$
(9)
where in (9) the constants \(C_{1}>0\) and \(C_{2}>0\) are best possible. This means that the constant \(C_{1}\) in (9) cannot be replaced by a number which is greater than \(C_{1}\) and the constant \(C_{2}\) in (9) cannot be replaced by a number which is less than \(C_{2}\). An asymptotic formula for the Wallis ratio is also given.

Our main result may be stated as the following theorem.

Theorem 1

For all \(n\ge2\),
$$ \biggl(\frac{2}{3} \biggr)^{3/2} \biggl(1- \frac{1}{2n} \biggr)^{n+\frac{1}{2}} \biggl(n-\frac{3}{2} \biggr)^{-\frac{1}{2}}\le W_{n}< \sqrt{\frac{e}{\pi }} \biggl(1- \frac{1}{2n} \biggr)^{n+\frac{1}{2}} \biggl(n-\frac{3}{2} \biggr)^{-\frac{1}{2}}. $$
(10)
The constants \((\frac{2}{3} )^{3/2}\) and \(\sqrt{\frac{e}{\pi}} \) in (10) are best possible. Furthermore, the asymptotic formula
$$ W_{n} \sim\sqrt{\frac{e}{\pi}} \biggl(1- \frac{1}{2n} \biggr)^{n+\frac{1}{2}} \biggl(n-\frac{3}{2} \biggr)^{-\frac{1}{2}},\quad n\to\infty, $$
(11)
is valid.

2 Proof of main result

We are now in a position to prove our main result stated in Theorem 1.

Proof of Theorem 1

Define
$$ f(x):= \frac{x^{x+{1}/{2}}}{e^{x}\Gamma(x+1)}. $$
(12)
Taking the logarithm of \(f(x)\) and then differentiating yield
$$\begin{aligned} &\ln f(x)= \biggl(x-\frac{1}{2} \biggr)\ln x-x-\ln \Gamma(x), \end{aligned}$$
(13)
$$\begin{aligned} &{ \bigl[}\ln f(x) \bigr]'=\ln x-\frac{1}{2x}- \psi(x). \end{aligned}$$
(14)
In (14)
$$\psi(x):= \frac{\Gamma'(x)}{\Gamma(x)}. $$
It is well known that (see [5, p.892])
$$ \psi(x)=\ln x-\frac{1}{2x}-2\int_{0}^{\infty} \frac{t\,dt}{(t^{2} +x^{2})(e^{2\pi t}-1)},\quad x>0. $$
(15)
From (14) and (15) we get
$$ \bigl[\ln f(x) \bigr]'=2\int_{0}^{\infty} \frac{t\,dt}{(t^{2} +x^{2})(e^{2\pi t}-1)},\quad x>0. $$
(16)
Hence,
$$\bigl[\ln f(x) \bigr]'>0,\quad x\in(0,\infty), $$
which means that \(\ln f(x)\), and thus \(f(x)\), is strictly increasing on \((0,\infty)\).
It is easy to see that
$$\lim_{x\to0+}f(x)=0. $$
Since (see [6, p.20])
$$ \ln\Gamma(x)= \biggl(x-\frac{1}{2} \biggr)\ln x-x+\ln \sqrt{2 \pi}+O \biggl(\frac{1}{x} \biggr),\quad \mbox{as } x\to\infty, $$
(17)
from (13) and (17), we have
$$ \ln f(x)=-\ln\sqrt{2\pi} +O \biggl(\frac{1}{x} \biggr) , \quad\mbox{as } x\to \infty, $$
(18)
which implies
$$ \lim_{x\to\infty}f(x)=\frac{1}{\sqrt{2\pi}} . $$
(19)
Since the function \(f(x)\) is strictly increasing from \((0,\infty)\) onto \((0,\frac{1}{\sqrt{2\pi} } )\) and
$$ \Gamma(n+1)=n!, $$
(20)
we obtain
$$ \frac{2\sqrt{2} }{e^{2}}=f(2) \le f(n)=\frac{n^{n+1/2}}{e^{n}n!} < \frac{1}{\sqrt{2\pi} },\quad n\ge2, $$
(21)
and
$$ \lim_{n\to\infty}\frac{n^{n+1/2}}{e^{n}n!} = \frac{1}{\sqrt{2\pi} }. $$
(22)
The lower and upper bounds in (21) are best possible.
Also we define
$$ h(x):= \frac{e^{x}\sqrt{x-1} \Gamma(x+1)}{x^{x+1}}. $$
(23)
Since the function \(h(x)\) is strictly increasing from \((1,\infty)\) onto \((0,\sqrt{2\pi} )\) (see [7, Theorem 1.3]) and in view of (20), we obtain
$$ \sqrt{\pi} \biggl(\frac{e}{3} \biggr)^{3/2}=h \biggl(\frac{3}{2} \biggr)\le h \biggl(n-\frac{1}{2} \biggr) < \sqrt{2 \pi} ,\quad n\ge2, $$
(24)
and
$$ \lim_{n\to\infty}h \biggl(n-\frac{1}{2} \biggr)= \sqrt{2\pi} . $$
(25)
It is well known that [1, p.258] for all \(n\in\mathbb{N}\),
$$ \Gamma \biggl(n+\frac{1}{2} \biggr)= n!\sqrt{\pi} W_{n}. $$
(26)
By using (26), after some algebra, (24) and (25) can be rewritten, respectively, as
$$ \frac{e^{2}}{3\sqrt{3} }\le\frac{e^{n}n!\sqrt{n-\frac{3}{2}} W_{n}}{ (n-\frac{1}{2} )^{n+1/2}}< \sqrt{2e} , \quad n\ge2, $$
(27)
and
$$ \lim_{n\to\infty}\frac{e^{n}n!\sqrt{n-\frac{3}{2}} W_{n}}{ (n-\frac{1}{2} )^{n+1/2}}= \sqrt{2e} . $$
(28)
The constants \(\frac{e^{2}}{3\sqrt{3} }\) and \(\sqrt{2e} \) in (27) are best possible.
Combining (21) and (27) yields
$$ \biggl(\frac{2}{3} \biggr)^{3/2}\le \frac{n^{n+1/2}\sqrt{n-\frac{3}{2}} W_{n}}{ (n-\frac{1}{2} )^{n+1/2}}< \sqrt{\frac{e}{\pi}} ,\quad n\ge2. $$
(29)
The constants \((\frac{2}{3} )^{3/2}\) and \(\sqrt{\frac{e}{\pi}} \) in (29) are best possible. From (29) the inequality (10) follows.
Combining (22) and (28) gives
$$ \lim_{n\to\infty} \frac{n^{n+1/2}\sqrt{n-\frac{3}{2}} W_{n}}{ (n-\frac{1}{2} )^{n+1/2}}= \sqrt{ \frac{e}{\pi}} , $$
(30)
which is equivalent to the asymptotic formula (11). The proof of Theorem 1 is thus completed. □

Remark 1

Some related functions associated with \(f(x)\), defined by (12), were proved [812] to be logarithmically completely monotonic.

Declarations

Acknowledgements

The authors thank the editor and the referees for their valuable suggestions to improve the quality of this paper. The present investigation was supported, in part, by the Natural Science Foundation of China under Grant 11401604.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Zhongyuan University of Technology
(2)
Department of Computational Mathematics, Zhongyuan University of Technology
(3)
Department of Library, Chongqing Normal University
(4)
Department of Mathematics, Chongqing Normal University

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Copyright

© Guo et al.; licensee Springer. 2015