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

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.

1 Introduction and main result

Throughout the paper, ℤ denotes the set of all integers, ℕ denotes the set of all positive integers,

$$\begin{array}{r}{\mathbb{N}}_0:=\mathbb{N}\cup \{0\},\\ n!!:=\prod _{i=0}^{[(n-1)/2]}(n-2i),\end{array}$$

(1)

and

$$W_n:=\frac{(2n-1)!!}{(2n)!!}.$$

(2)

Here in (1), the floor function $[t]$ 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

$$\mathrm{\Gamma }(z):={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt.$$

(3)

One of the elementary properties of the gamma function is that

$$\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }(x).$$

(4)

In particular,

$$\mathrm{\Gamma }(n+1)=n!,n\in {\mathbb{N}}_0.$$

(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:

$$n!\sim \sqrt{2\pi n}{n}^n{e}^{-n},n\to \mathrm{\infty },$$

(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 [[1], Chapter III].

$W_n$ is closely related to the Wallis’ formula.

The Wallis’ formula

$$\frac{2}{\pi }=\prod _{n=1}^{\mathrm{\infty }}\frac{(2n-1)(2n+1)}{{(2n)}^2}$$

(8)

can be obtained by taking

$$x=\frac{\pi }{2}$$

in the infinite product representation of sinx (see [[2], p.10], [[3], p.211])

$$sinx=x\prod _{n=1}^{\mathrm{\infty }}(1-\frac{x^2}{n^2{\pi }^2}),x\in \mathbb{R}.$$

(9)

Since

$$\prod _{n=1}^{\mathrm{\infty }}\frac{(2n-1)(2n+1)}{{(2n)}^2}=\underset{n\to \mathrm{\infty }}{lim}(2n+1)W_n^2,$$

(10)

another important form of Wallis’ formula is (see [[4], pp.181-184])

$$\underset{n\to \mathrm{\infty }}{lim}(2n+1)W_n^2=\frac{2}{\pi }.$$

(11)

The following generalization of Wallis’ formula was given in [5].

$$\frac{\pi }{tsin(\pi /t)}=\frac{1}{t-1}\prod _{i=1}^{\mathrm{\infty }}\frac{{(it)}^2}{(it+t-1)(it-t+1)},t>1.$$

(12)

In fact, by letting

$$x=(1-1/t)\pi ,t\ne 0$$

in (9), we have

$$sin\frac{\pi }{t}=\frac{\pi }{t}(t-1)\prod _{i=1}^{\mathrm{\infty }}\frac{(it+t-1)(it-t+1)}{{(it)}^2},t\ne 0.$$

(13)

From (13), we get

$$\frac{\pi }{tsin(\pi /t)}=\frac{1}{t-1}\prod _{i=1}^{\mathrm{\infty }}\frac{{(it)}^2}{(it+t-1)(it-t+1)}$$

(14)

for

$$t\ne 0,t\ne \frac{1}{k},k\in \mathbb{Z}.$$

(12) is a special case of (14). The proof of (12) in [5] 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

$$\mathrm{\Gamma }(x)\sim \sqrt{2\pi }{x}^{x-1/2}{e}^{-x},x\to \mathrm{\infty },$$

(15)

relies on Wallis’ formula (see [[2], pp.18-20], [[3], pp.213-215], [[4], pp.181-184]).

Also, note that

$$W_n={[(2n+1){\int }_0^{\pi /2}{sin}^{2n+1}xdx]}^{-1}$$

(16)

$$={[(2n+1){\int }_0^{\pi /2}{cos}^{2n+1}xdx]}^{-1}$$

(17)

and Wallis’ sine (cosine) formula (see [[6], p.258])

$$W_n=\frac{2}{\pi }{\int }_0^{\pi /2}{sin}^{2n}xdx$$

(18)

$$=\frac{2}{\pi }{\int }_0^{\pi /2}{cos}^{2n}xdx.$$

(19)

Some inequalities involving $W_n$ were given in [7–12].

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 $n\in \mathbb{N}$, $n\ge 2$,

$$\sqrt{\frac{e}{\pi }}{(1-\frac{1}{2n})}^n\frac{\sqrt{n-1}}{n}<W_n\le \frac{4}{3}{(1-\frac{1}{2n})}^n\frac{\sqrt{n-1}}{n}.$$

(20)

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

Moreover,

$$W_n\sim \sqrt{\frac{e}{\pi }}{(1-\frac{1}{2n})}^n\frac{\sqrt{n-1}}{n},n\to \mathrm{\infty }.$$

(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 ([[13], Theorem 1.1])

The function

$$f(x):=\frac{x^{x+\frac{1}{2}}}{e^x\mathrm{\Gamma }(x+1)}$$

(22)

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

Lemma 2 ([[13], Theorem 1.3])

The function

$$h(x):=\frac{e^x\sqrt{x-1}\mathrm{\Gamma }(x+1)}{x^{x+1}}$$

(23)

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

Lemma 3 ([[6], p.258])

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

$$\mathrm{\Gamma }(n+\frac{1}{2})=\sqrt{\pi }n!W_n,$$

(24)

where $W_n$ is defined by (2).

Remark 2 Some functions associated with the functions $f(x)$ and $h(x)$, defined by (22) and (23) respectively, were proved to be logarithmically completely monotonic in [14–16]. For more recent work on (logarithmically) completely monotonic functions, please see, for example, [17–43].

3 Proof of the main result

Proof of Theorem 1

By Lemma 1, we have

$$\frac{3}{e\sqrt{e\pi }}=f\left(\frac{3}{2}\right)\le f(n-\frac{1}{2})=\frac{{(n-\frac{1}{2})}^n}{e^{n-1/2}\mathrm{\Gamma }(n+1/2)}<\frac{1}{\sqrt{2\pi }},n\ge 2,$$

(25)

and

$$\underset{n\to \mathrm{\infty }}{lim}\frac{{(n-\frac{1}{2})}^n}{e^{n-1/2}\mathrm{\Gamma }(n+1/2)}=\frac{1}{\sqrt{2\pi }}.$$

(26)

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

By Lemma 3, (25) and (26) can be rewritten respectively as

$$\frac{3}{e^2}\le \frac{{(n-\frac{1}{2})}^n}{W_n{e}^{n}n!}<\frac{1}{\sqrt{2e}},n\ge 2,$$

(27)

and

$$\underset{n\to \mathrm{\infty }}{lim}\frac{{(n-\frac{1}{2})}^n}{W_n{e}^{n}n!}=\frac{1}{\sqrt{2e}}.$$

(28)

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

By Lemma 2, we get

$${\left(\frac{e}{2}\right)}^2=h(2)\le h(n)=\frac{e^{n}n!\sqrt{n-1}}{n^{n+1}}<\sqrt{2\pi },n\ge 2,$$

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

From (27) and (29), we obtain that for all $n\ge 2$,

$$\frac{3}{4}\le \frac{\sqrt{n-1}{(n-\frac{1}{2})}^n}{W_n{n}^{n+1}}<\sqrt{\frac{\pi }{e}}.$$

(31)

The constants $3/4$ and $\sqrt{\pi /e}$ in (31) are best possible. From (31) we get that for all $n\ge 2$,

$$\sqrt{\frac{e}{\pi }}{(1-\frac{1}{2n})}^n\frac{\sqrt{n-1}}{n}<W_n\le \frac{4}{3}{(1-\frac{1}{2n})}^n\frac{\sqrt{n-1}}{n}.$$

(32)

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

From (28) and (30), we see that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt{n-1}{(n-\frac{1}{2})}^n}{W_n{n}^{n+1}}=\sqrt{\frac{\pi }{e}},$$

(33)

which is equivalent to (21).

The proof is thus completed. □