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# Some exact constants for the approximation of the quantity in the Wallis’ formula

Journal of Inequalities and Applications20132013:67

https://doi.org/10.1186/1029-242X-2013-67

• Received: 14 December 2012
• Accepted: 7 February 2013
• Published:

## Abstract

In this article, a sharp two-sided bounding inequality and some best constants for the approximation of the quantity associated with the Wallis’ formula are presented.

MSC:41A44, 26D20, 33B15.

## Keywords

• approximation
• gamma function
• Stirling’s formula
• Wallis’ formula

## 1 Introduction and main result

Throughout the paper, denotes the set of all integers, denotes the set of all positive integers,
$\begin{array}{r}{\mathbb{N}}_{0}:=\mathbb{N}\cup \left\{0\right\},\\ n!!:=\prod _{i=0}^{\left[\left(n-1\right)/2\right]}\left(n-2i\right),\end{array}$
(1)
and
${W}_{n}:=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}.$
(2)

Here in (1), the floor function $\left[t\right]$ denotes the integer which is less than or equal to the number t.

The Euler gamma function is defined and denoted for $Rez>0$ by
$\mathrm{\Gamma }\left(z\right):={\int }_{0}^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}\phantom{\rule{0.2em}{0ex}}dt.$
(3)
One of the elementary properties of the gamma function is that
$\mathrm{\Gamma }\left(x+1\right)=x\mathrm{\Gamma }\left(x\right).$
(4)
In particular,
$\mathrm{\Gamma }\left(n+1\right)=n!,\phantom{\rule{1em}{0ex}}n\in {\mathbb{N}}_{0}.$
(5)
Also, note that
$\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }.$
(6)
For the approximation of n!, a well-known result is the following Stirling’s formula:
$n!\sim \sqrt{2\pi n}{n}^{n}{e}^{-n},\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty },$
(7)

which is an important tool in analytical probability theory, statistical physics and physical chemistry.

Consider the quantity ${W}_{n}$, defined by (2). This quantity is important in the probability theory - for example, the three events, (a) a return to the origin takes place at time 2n, (b) no return occurs up to and including time 2n, and (c) the path is non-negative between 0 and 2n, have the common probability ${W}_{n}$. Also, the probability that in the time interval from 0 to 2n the particle spends 2k time units on the positive side and $2n-2k$ time units on the negative side is ${W}_{k}{W}_{n-k}$. For details of these interesting results, one may see [, Chapter III].

${W}_{n}$ is closely related to the Wallis’ formula.

The Wallis’ formula
$\frac{2}{\pi }=\prod _{n=1}^{\mathrm{\infty }}\frac{\left(2n-1\right)\left(2n+1\right)}{{\left(2n\right)}^{2}}$
(8)
can be obtained by taking
$x=\frac{\pi }{2}$
in the infinite product representation of sinx (see [, p.10], [, p.211])
$sinx=x\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{{x}^{2}}{{n}^{2}{\pi }^{2}}\right),\phantom{\rule{1em}{0ex}}x\in \mathbb{R}.$
(9)
Since
$\prod _{n=1}^{\mathrm{\infty }}\frac{\left(2n-1\right)\left(2n+1\right)}{{\left(2n\right)}^{2}}=\underset{n\to \mathrm{\infty }}{lim}\left(2n+1\right){W}_{n}^{2},$
(10)
another important form of Wallis’ formula is (see [, pp.181-184])
$\underset{n\to \mathrm{\infty }}{lim}\left(2n+1\right){W}_{n}^{2}=\frac{2}{\pi }.$
(11)
The following generalization of Wallis’ formula was given in .
$\frac{\pi }{tsin\left(\pi /t\right)}=\frac{1}{t-1}\prod _{i=1}^{\mathrm{\infty }}\frac{{\left(it\right)}^{2}}{\left(it+t-1\right)\left(it-t+1\right)},\phantom{\rule{1em}{0ex}}t>1.$
(12)
In fact, by letting
$x=\left(1-1/t\right)\pi ,\phantom{\rule{1em}{0ex}}t\ne 0$
in (9), we have
$sin\frac{\pi }{t}=\frac{\pi }{t}\left(t-1\right)\prod _{i=1}^{\mathrm{\infty }}\frac{\left(it+t-1\right)\left(it-t+1\right)}{{\left(it\right)}^{2}},\phantom{\rule{1em}{0ex}}t\ne 0.$
(13)
From (13), we get
$\frac{\pi }{tsin\left(\pi /t\right)}=\frac{1}{t-1}\prod _{i=1}^{\mathrm{\infty }}\frac{{\left(it\right)}^{2}}{\left(it+t-1\right)\left(it-t+1\right)}$
(14)
for
$t\ne 0,\phantom{\rule{2em}{0ex}}t\ne \frac{1}{k},\phantom{\rule{1em}{0ex}}k\in \mathbb{Z}.$

(12) is a special case of (14). The proof of (12) in  involves integrating powers of a generalized sine function.

There is a close relationship between Stirling’s formula and Wallis’ formula. The determination of the constant $\sqrt{2\pi }$ in the usual proof of Stirling’s formula (7) or Stirling’s asymptotic formula
$\mathrm{\Gamma }\left(x\right)\sim \sqrt{2\pi }{x}^{x-1/2}{e}^{-x},\phantom{\rule{1em}{0ex}}x\to \mathrm{\infty },$
(15)

relies on Wallis’ formula (see [, pp.18-20], [, pp.213-215], [, pp.181-184]).

Also, note that
${W}_{n}={\left[\left(2n+1\right){\int }_{0}^{\pi /2}{sin}^{2n+1}x\phantom{\rule{0.2em}{0ex}}dx\right]}^{-1}$
(16)
$={\left[\left(2n+1\right){\int }_{0}^{\pi /2}{cos}^{2n+1}x\phantom{\rule{0.2em}{0ex}}dx\right]}^{-1}$
(17)
and Wallis’ sine (cosine) formula (see [, p.258])
${W}_{n}=\frac{2}{\pi }{\int }_{0}^{\pi /2}{sin}^{2n}x\phantom{\rule{0.2em}{0ex}}dx$
(18)
$=\frac{2}{\pi }{\int }_{0}^{\pi /2}{cos}^{2n}x\phantom{\rule{0.2em}{0ex}}dx.$
(19)

Some inequalities involving ${W}_{n}$ were given in .

In this article, we give a sharp two-sided bounding inequality and some exact constants for the approximation of ${W}_{n}$, defined by (2). The main result of the paper is as follows.

Theorem 1 For all $n\in \mathbb{N}$, $n\ge 2$,
$\sqrt{\frac{e}{\pi }}{\left(1-\frac{1}{2n}\right)}^{n}\frac{\sqrt{n-1}}{n}<{W}_{n}\le \frac{4}{3}{\left(1-\frac{1}{2n}\right)}^{n}\frac{\sqrt{n-1}}{n}.$
(20)

The constants $\sqrt{e/\pi }$ and $4/3$ in (20) are best possible.

Moreover,
${W}_{n}\sim \sqrt{\frac{e}{\pi }}{\left(1-\frac{1}{2n}\right)}^{n}\frac{\sqrt{n-1}}{n},\phantom{\rule{1em}{0ex}}n\to \mathrm{\infty }.$
(21)

Remark 1 By saying that the constants $\sqrt{e/\pi }$ and $4/3$ in (20) are best possible, we mean that the constant $\sqrt{e/\pi }$ in (20) cannot be replaced by a number which is greater than $\sqrt{e/\pi }$ and the constant $4/3$ in (20) cannot be replaced by a number which is less than $4/3$.

## 2 Lemmas

We need the following lemmas to prove our result.

Lemma 1 ([, Theorem 1.1])

The function
$f\left(x\right):=\frac{{x}^{x+\frac{1}{2}}}{{e}^{x}\mathrm{\Gamma }\left(x+1\right)}$
(22)

is strictly logarithmically concave and strictly increasing from $\left(0,\mathrm{\infty }\right)$ onto $\left(0,\frac{1}{\sqrt{2\pi }}\right)$.

Lemma 2 ([, Theorem 1.3])

The function
$h\left(x\right):=\frac{{e}^{x}\sqrt{x-1}\mathrm{\Gamma }\left(x+1\right)}{{x}^{x+1}}$
(23)

is strictly logarithmically concave and strictly increasing from $\left(1,\mathrm{\infty }\right)$ onto $\left(0,\sqrt{2\pi }\right)$.

Lemma 3 ([, p.258])

For all $n\in \mathbb{N}$,
$\mathrm{\Gamma }\left(n+\frac{1}{2}\right)=\sqrt{\pi }n!{W}_{n},$
(24)

where ${W}_{n}$ is defined by (2).

Remark 2 Some functions associated with the functions $f\left(x\right)$ and $h\left(x\right)$, defined by (22) and (23) respectively, were proved to be logarithmically completely monotonic in . For more recent work on (logarithmically) completely monotonic functions, please see, for example, .

## 3 Proof of the main result

Proof of Theorem 1

By Lemma 1, we have
$\frac{3}{e\sqrt{e\pi }}=f\left(\frac{3}{2}\right)\le f\left(n-\frac{1}{2}\right)=\frac{{\left(n-\frac{1}{2}\right)}^{n}}{{e}^{n-1/2}\mathrm{\Gamma }\left(n+1/2\right)}<\frac{1}{\sqrt{2\pi }},\phantom{\rule{1em}{0ex}}n\ge 2,$
(25)
and
$\underset{n\to \mathrm{\infty }}{lim}\frac{{\left(n-\frac{1}{2}\right)}^{n}}{{e}^{n-1/2}\mathrm{\Gamma }\left(n+1/2\right)}=\frac{1}{\sqrt{2\pi }}.$
(26)

The lower and upper bounds in (25) are best possible.

By Lemma 3, (25) and (26) can be rewritten respectively as
$\frac{3}{{e}^{2}}\le \frac{{\left(n-\frac{1}{2}\right)}^{n}}{{W}_{n}{e}^{n}n!}<\frac{1}{\sqrt{2e}},\phantom{\rule{1em}{0ex}}n\ge 2,$
(27)
and
$\underset{n\to \mathrm{\infty }}{lim}\frac{{\left(n-\frac{1}{2}\right)}^{n}}{{W}_{n}{e}^{n}n!}=\frac{1}{\sqrt{2e}}.$
(28)

The constants $3/{e}^{2}$ and $1/\sqrt{2e}$ in (27) are best possible.

By Lemma 2, we get
${\left(\frac{e}{2}\right)}^{2}=h\left(2\right)\le h\left(n\right)=\frac{{e}^{n}n!\sqrt{n-1}}{{n}^{n+1}}<\sqrt{2\pi },\phantom{\rule{1em}{0ex}}n\ge 2,$
(29)
and
$\underset{n\to \mathrm{\infty }}{lim}\frac{{e}^{n}n!\sqrt{n-1}}{{n}^{n+1}}=\sqrt{2\pi }.$
(30)

The lower bound ${\left(e/2\right)}^{2}$ and the upper bound $\sqrt{2\pi }$ in (29) are best possible.

From (27) and (29), we obtain that for all $n\ge 2$,
$\frac{3}{4}\le \frac{\sqrt{n-1}{\left(n-\frac{1}{2}\right)}^{n}}{{W}_{n}{n}^{n+1}}<\sqrt{\frac{\pi }{e}}.$
(31)
The constants $3/4$ and $\sqrt{\pi /e}$ in (31) are best possible. From (31) we get that for all $n\ge 2$,
$\sqrt{\frac{e}{\pi }}{\left(1-\frac{1}{2n}\right)}^{n}\frac{\sqrt{n-1}}{n}<{W}_{n}\le \frac{4}{3}{\left(1-\frac{1}{2n}\right)}^{n}\frac{\sqrt{n-1}}{n}.$
(32)

The constants $\sqrt{e/\pi }$ and $4/3$ in (32) are best possible.

From (28) and (30), we see that
$\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt{n-1}{\left(n-\frac{1}{2}\right)}^{n}}{{W}_{n}{n}^{n+1}}=\sqrt{\frac{\pi }{e}},$
(33)

which is equivalent to (21).

The proof is thus completed. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

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 Henan Province of China under Grant 112300410022.

## Authors’ Affiliations

(1)
Department of Mathematics, Zhongyuan University of Technology, Zhengzhou, Henan, 450007, China
(2)
School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, Henan, 454010, China

## References 