A sharp two-sided inequality for bounding the Wallis ratio

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.

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 [8–12] to be logarithmically completely monotonic.