# Sharp inequalities related to the volume of the unit ball in $$\mathbb{R}^{n}$$

## Abstract

Let $$\Omega _{n}=\pi ^{n/2}/\Gamma (\frac{n}{2}+1)$$ ($$n \in \mathbb{N}$$) denote the volume of the unit ball in $$\mathbb{R}^{n}$$. In this paper, the logarithmically complete monotonicity of a function involving the ratio of two gamma functions is presented, which yields a sharp double inequality for the quantity $$\Omega _{n}^{2}/(\Omega _{n-1}\Omega _{n+1})$$. Also, we establish new sharp inequalities for the quantity $$\Omega _{n}^{2}/(\Omega _{n-1}\Omega _{n+1})$$.

## 1 Introduction

In the recent past, several researchers have established interesting properties of the volume $$\Omega _{n}$$ of the unit ball in $$\mathbb{R}^{n}$$,

$$\Omega _{n}=\frac{\pi ^{n/2}}{\Gamma (\frac{n}{2}+1)}, \quad n \in \mathbb{N}:=\{1, 2, \ldots \},$$

including monotonicity properties, inequalities, and asymptotic expansions.

Böhm and Hertel [1, p. 264] pointed out that the sequence $$\{\Omega _{n} \}_{n \in \mathbb{N}}$$ is not monotonic. Indeed, we have

$$\Omega _{n} < \Omega _{n+1} \quad \text{if } 1 \leq n \leq 4 \quad \text{and} \quad \Omega _{n} > \Omega _{n+1} \quad \text{if } n\geq 5.$$

Anderson et al.  showed that $$\{\Omega _{n}^{1/n} \}_{n \in \mathbb{N}}$$ is monotonically decreasing to zero, while Anderson and Qiu  proved that the sequence $$\{\Omega _{n}^{1/(n\ln n)} \}_{n\geq 2}$$ decreases to $$e^{-1/2}$$. Guo and Qi  proved that the sequence $$\{\Omega _{n}^{1/(n\ln n)} \}_{n\geq 2}$$ is logarithmically convex. Klain and Rota  proved that the sequence $$\{n\Omega _{n}/\Omega _{n-1} \}_{n \in \mathbb{N}}$$ is increasing.

Diverse sharp inequalities for the volume of the unit ball in $$\mathbb{R}^{n}$$ have been established . For example, Alzer  proved that for $$n\in \mathbb{N}$$,

\begin{aligned}& a_{1}\Omega _{n+1}^{n/(n+1)} \leq \Omega _{n}< b_{1}\Omega _{n+1}^{n/(n+1)}, \\& \sqrt{\frac{n+a_{2}}{2\pi}} < \frac{\Omega _{n-1}}{\Omega _{n}}\leq \sqrt{ \frac{n+b_{2}}{2\pi}}, \\& \biggl(1+\frac{1}{n} \biggr)^{a_{3}} \leq \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}< \biggl(1+ \frac{1}{n} \biggr)^{b_{3}}, \end{aligned}
(1.1)

with the best possible constants

\begin{aligned}& a_{1} = \frac{2}{\sqrt{\pi}}=1.1283\ldots , \qquad b_{1}= \sqrt{e}=1.6487\ldots , \\& a_{2} =\frac{1}{2}, \qquad b_{2}= \frac{\pi}{2}-1=0.5707\ldots , \\& a_{3} =2-\frac{\ln \pi}{\ln 2}=0.3485\ldots ,\qquad b_{3}= \frac{1}{2}. \end{aligned}

Merkle  improved the left-hand side of (1.1) and obtained the following result:

\begin{aligned} \biggl(1+\frac{1}{n+1} \biggr)^{1/2}\leq \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}},\quad n\in \mathbb{N}. \end{aligned}
(1.2)

Chen and Lin [10, Theorem 3.1] developed (1.2) to produce the following symmetric double inequality:

\begin{aligned} \biggl(1+\frac{1}{n+1} \biggr)^{\alpha}< \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \leq \biggl(1+ \frac{1}{n+1} \biggr)^{\beta},\quad n\in \mathbb{N}, \end{aligned}

with the best possible constants

\begin{aligned} \alpha =\frac{1}{2},\quad \beta =\frac{2\ln 2-\ln \pi}{\ln 3-\ln 2}=0.5957713 \ldots. \end{aligned}

Ban and Chen [8, Theorem 3.2] proved, for $$n\in \mathbb{N}$$,

\begin{aligned} \biggl(1+\frac{1}{n+\theta _{1}} \biggr)^{1/2}\leq \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}< \biggl(1+ \frac{1}{n+\theta _{2}} \biggr)^{1/2}, \end{aligned}
(1.3)

with the best possible constants

\begin{aligned} \theta _{1}=\frac{2\pi ^{2}-16}{16-\pi ^{2}}=0.60994576\ldots \quad \text{and}\quad \theta _{2}=\frac{1}{2}. \end{aligned}

Recently, Mortici  constructed asymptotic series associated with some expressions involving the volume of the n-dimensional unit ball. New refinements and improvements of some old and recent inequalities for $$\Omega _{n}$$ were also presented. For example, Mortici [16, Theorem 15] presented the following asymptotic expansion for the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$:

\begin{aligned} \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}&\sim 1+ \frac{1}{2n}- \frac{3}{8n^{2}}+\frac{3}{16n^{3}}+\frac{3}{128n^{4}}- \frac{33}{256n^{5}}-\frac{39}{1024n^{6}}+\cdots , \end{aligned}
(1.4)

as $$n\to \infty$$. Moreover, the author provided a recurrence relation for successively determining the coefficient of $$1/n^{j}$$ ($$j\in \mathbb{N}$$) in expansion (1.4).

Lu and Zhang  established a general continued fraction approximation for the nth root of the volume of the unit n-dimensional ball, and then obtained related inequalities. Chen and Paris  presented asymptotic expansions and inequalities related to $$\Omega _{n}$$ and the quantities:

\begin{aligned} \frac{\Omega _{n-1}}{\Omega _{n}},\qquad \frac{\Omega _{n}}{\Omega _{n-1}+\Omega _{n+1}}, \quad \text{and} \quad \frac{\Omega _{n}^{1/n}}{\Omega _{n+1}^{1/(n+1)}}. \end{aligned}

It is easy to see that

\begin{aligned} \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}= \biggl(\frac{n}{2}+ \frac{1}{2} \biggr) \biggl( \frac{\Gamma (\frac{n}{2}+\frac{1}{2})}{\Gamma (\frac{n}{2}+1)} \biggr)^{2}. \end{aligned}
(1.5)

Replacement of $$n/2$$ by x in (1.5) yields

\begin{aligned} I(x):=\frac{\Omega _{2x}^{2}}{\Omega _{2x-1}\Omega _{2x+1}}= \biggl(x+ \frac{1}{2} \biggr) \biggl(\frac{\Gamma (x+\frac{1}{2})}{\Gamma (x+1)} \biggr)^{2}, \end{aligned}
(1.6)

where $$\Omega _{x}=\pi ^{x/2}/\Gamma (\frac{x}{2}+1)$$.

From (1.5) and (1.6), we see that the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$ is closely related to the ratio of two gamma functions $$\frac{\Gamma (x+\frac{1}{2})}{\Gamma (x+1)}$$. The problem of finding new and sharp inequalities for the gamma function Γ and, in particular, for the Wallis ratio

$$\frac{(2n-1)!!}{(2n)!!}= \frac{\Gamma (n+\frac{1}{2})}{\sqrt{\pi}\Gamma (n+1)}, \quad n\in \mathbb{N}\mathbbm{,}$$

has attracted the attention of many researchers (see  and the references therein). Here, we employ the special double factorial notation as follows:

\begin{aligned} &(2n)!!=2\cdot 4\cdot 6\cdots (2n)=2^{n} n!, \\ &(2n-1)!!=1\cdot 3\cdot 5\cdots (2n-1)=\pi ^{-1/2}2^{n} \Gamma \biggl(n+ \frac {1}{2}\biggr), \\ &0!!=1,\qquad (-1)!!=1. \end{aligned}

Chen and Paris [30, Corollary 1(i)] obtained the following double inequality:

\begin{aligned}[b] \sqrt{x}\exp \Biggl(\sum _{j=1}^{2m} \biggl(1- \frac{1}{2^{2j}} \biggr)\frac{B_{2j}}{j(2j-1)x^{2j-1}} \Biggr)&< \frac{\Gamma (x+1)}{\Gamma (x+\frac{1}{2})} \\ &< \sqrt{x}\exp \Biggl(\sum_{j=1}^{2m+1} \biggl(1-\frac{1}{2^{2j}} \biggr)\frac{B_{2j}}{j(2j-1)x^{2j-1}} \Biggr) \end{aligned}
(1.7)

for $$x>0$$ and $$m\in \mathbb{N}_{0}$$, where $$B_{n}$$ ($$n \in \mathbb{N}_{0}$$) are the Bernoulli numbers defined by the following generating function:

$$\frac{t}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n} \frac{t^{n}}{n!},\quad \vert t \vert < 2 \pi .$$
(1.8)

From (1.7), we derive

\begin{aligned} & \biggl(1+\frac{1}{2x} \biggr)\exp \Biggl(-\sum _{j=1}^{2m} \biggl(1- \frac{1}{2^{2j}} \biggr)\frac{2B_{2j}}{j(2j-1)x^{2j-1}} \Biggr) \\ &\quad > \frac{\Omega _{2x}^{2}}{\Omega _{2x-1}\Omega _{2x+1}}= \biggl(x+ \frac{1}{2} \biggr) \biggl( \frac{\Gamma (x+\frac{1}{2})}{\Gamma (x+1)} \biggr)^{2} \\ &\quad > \biggl(1+\frac{1}{2x} \biggr) \exp \Biggl(-\sum _{j=1}^{2m+1} \biggl(1-\frac{1}{2^{2j}} \biggr) \frac{2B_{2j}}{j(2j-1)x^{2j-1}} \Biggr) \end{aligned}
(1.9)

for $$x>0$$ and $$m\in \mathbb{N}_{0}$$. Replacing x by $$n/2$$ in (1.9) yields

\begin{aligned} & \biggl(1+\frac{1}{n} \biggr)\exp \Biggl(-\sum _{j=1}^{2m} \frac{ (2^{2j}-1 )B_{2j}}{j(2j-1)n^{2j-1}} \Biggr) \\ &\quad > \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}> \biggl(1+ \frac{1}{n} \biggr)\exp \Biggl(-\sum_{j=1}^{2m+1} \frac{ (2^{2j}-1 )B_{2j}}{j(2j-1)n^{2j-1}} \Biggr) \end{aligned}

for $$n\in \mathbb{N}$$ and $$m\in \mathbb{N}_{0}$$.

In this paper, we prove that the function $$G(x)= (1+\frac{1}{2x+\frac{1}{2}} )^{1/2}/I(x)$$ is logarithmically completely monotonic on $$(0,\infty )$$ (Theorem 3.1), which yields a sharp double inequality for the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$ (see (3.5)). Also, we establish new sharp inequalities for the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$ (Theorems 4.1 and 4.2).

The numerical values given in this paper have been calculated via the computer program MAPLE 17.

## 2 Lemmas

### Lemma 2.1

()

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 $$(a, b)$$.

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.

### Lemma 2.2

()

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 $$B_{n}$$ ($$n \in \mathbb{N}_{0}$$) are the Bernoulli numbers defined by (1.8).

In particular, we obtain from (2.1) that

\begin{aligned}& \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}},\quad x>0, \end{aligned}
(2.2)
\begin{aligned}& \frac{1}{2x}-\frac{1}{8x^{2}}+ \frac{1}{64x^{4}}-\frac{1}{128x^{6}}+ \frac{17}{2048x^{8}}- \frac{31}{2048x^{10}} \\& \quad < \psi (x+1)-\psi \biggl(x+\frac{1}{2} \biggr) < \frac{1}{2x}- \frac{1}{8x^{2}}+\frac{1}{64x^{4}}- \frac{1}{128x^{6}}+ \frac{17}{2048x^{8}},\quad x>0, \end{aligned}
(2.3)

and

\begin{aligned} -\frac{1}{2x^{2}}+\frac{1}{4x^{3}}- \frac{1}{16x^{5}}< \psi '(x+1)- \psi ' \biggl(x+ \frac{1}{2} \biggr),\quad x>0. \end{aligned}
(2.4)

## 3 Logarithmically complete monotonicity of the function $$(1+\frac{1}{2x+\frac{1}{2}})^{1/2}/I(x)$$

A function f is said to be completely monotonic on an interval I if it has derivatives of all orders on I and satisfies the following inequality:

$$(-1)^{n}f^{(n)}(x)\geq 0\quad \text{for } x\in I \text{ and } n\in \mathbb{N}_{0}:= \mathbb{N}\cup \{0\}.$$
(3.1)

Dubourdieu [32, p. 98] pointed out that, if a nonconstant function f is completely monotonic on $$I=(a, \infty )$$, then strict inequality holds true in (3.1). See also  for a simpler proof of this result. It is known (Bernstein’s theorem) that f is completely monotonic on $$(0, \infty )$$ if and only if

$$f(x)= \int ^{\infty}_{0}e^{-xt}\, \mathrm{d}\mu (t),$$

where μ is a nonnegative measure on $$[0, \infty )$$ such that the integral converges for all $$x>0$$. See [34, p. 161].

Recall  that a positive function f is said to be logarithmically completely monotonic on an interval I if its logarithm lnf satisfies

$$(-1)^{k}\bigl[\ln f(x)\bigr]^{(k)}\ge 0 \quad \text{for }x \in I \text{ and } k \in \mathbb{N}.$$

A logarithmically completely monotonic function f on I must be completely monotonic on I (see, e.g., ).

### Theorem 3.1

The function

$$G(x)=\frac{ (1+\frac{1}{2x+\frac{1}{2}} )^{1/2}}{I(x)}= \frac{ (1+\frac{1}{2x+\frac{1}{2}} )^{1/2}}{(x+\frac{1}{2})} \biggl[ \frac{\Gamma (x+1)}{\Gamma (x+\frac{1}{2})} \biggr]^{2}$$
(3.2)

is logarithmically completely monotonic on $$(0,\infty )$$.

### Proof

The logarithm of the gamma function has the following integral representation (see [39, p. 258]):

\begin{aligned} \ln \Gamma (z) = \int _{0}^{\infty} \biggl[(z-1)e^{-t}+ \frac{e^{-zt}-e^{-t}}{1-e^{-t}} \biggr]\frac{\mathrm{d} t}{t}. \end{aligned}
(3.3)

Using (3.3) and

\begin{aligned} \ln x= \int _{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}\, \mathrm{d} t, \end{aligned}

we obtain

\begin{aligned} \ln G(x)&=\frac{1}{2}\ln \frac{x+\frac{3}{4}}{x+\frac{1}{4}} - \ln \biggl(x+\frac{1}{2} \biggr)+2 \biggl[\ln \Gamma (x+1)-\ln \Gamma \biggl(x+\frac{1}{2} \biggr) \biggr] \\ &= \int _{0}^{\infty} \biggl(\frac{1}{2}e^{-(x+\frac{1}{4})t}- \frac{1}{2}e^{-(x+\frac{3}{4})t}+e^{-(x+\frac{1}{2})t}+ \frac{2[e^{-(x+1)t}-e^{-(x+\frac{1}{2})t}]}{1-e^{-t}} \biggr) \frac{\mathrm{d} t}{t} \\ &= \int _{0}^{\infty} \biggl(\frac{1}{2e^{t/4}}- \frac{1}{2e^{3t/4}}+ \frac{1}{e^{t/2}}-\frac{2}{e^{t/2}+1} \biggr) \frac{e^{-xt}}{t} \,\mathrm{d} t \\ &= \int _{0}^{\infty}q(t)e^{-xt}\,\mathrm{d} t, \end{aligned}
(3.4)

where

\begin{aligned} q(t)=\frac{(e^{t/4}+1)(e^{t/4}-1)^{3}}{2t e^{3t/4}(e^{t/2}+1)}>0, \quad t>0. \end{aligned}

We conclude from (3.4) that

$$(-1)^{n} \bigl(\ln G(x) \bigr)^{(n)}= \int _{0}^{\infty}t^{n} q(t) e^{-xt} \,\mathrm{d} t>0 \quad \text{for } x>0 \text{ and } n \in \mathbb{N}.$$

The proof of Theorem 3.1 is complete. □

### Remark 3.1

The function $$G(x)$$, defined by (3.2), is completely monotonic on $$(0,\infty )$$. In particular, the sequence $$\{G(n/2)\}$$ is strictly decreasing for $$n\in \mathbb{N}$$, and we have

$$1=G(\infty )< G \biggl(\frac{n}{2} \biggr)= \frac{ (1+\frac{1}{n+\frac{1}{2}} )^{1/2}}{I(\frac{n}{2})}\leq G \biggl(\frac{1}{2} \biggr)=\frac{\sqrt{15} \pi}{12},\quad n\in \mathbb{N},$$

which yields the following double inequality for the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$:

$$p \biggl(1+\frac{1}{n+\frac{1}{2}} \biggr)^{1/2}\leq \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}< q \biggl(1+ \frac{1}{n+\frac{1}{2}} \biggr)^{1/2}, \quad n\in \mathbb{N},$$
(3.5)

with the best possible constants

$$p=\frac{12}{\sqrt{15} \pi}=0.986247\ldots \quad \text{and}\quad q=1.$$

## 4 Sharp inequalities for $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$

### Theorem 4.1

For $$n\in \mathbb{N}$$, the following double inequality holds:

\begin{aligned} \biggl(1+\frac{1}{n+\frac{1}{2}} \biggr)^{\lambda}\leq \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}< \biggl(1+ \frac{1}{n+\frac{1}{2}} \biggr)^{\mu}, \end{aligned}
(4.1)

where the constants

\begin{aligned} \lambda =\frac{2\ln 2-\ln \pi}{\ln 5-\ln 3}=0.47289\ldots \quad \textit{and}\quad \mu = \frac{1}{2} \end{aligned}

are the best possible.

### Proof

Inequality (4.1) can be written as

\begin{aligned} \lambda \leq x_{n}< \mu , \end{aligned}

where the sequence $$\{x_{n} \}_{n\in \mathbb{N}}$$ is defined by

\begin{aligned} x_{n}= \frac{\ln ( (\frac{n}{2}+\frac{1}{2} ) (\frac{\Gamma (\frac{n}{2}+\frac{1}{2})}{\Gamma (\frac{n}{2}+1)} )^{2} )}{\ln (1+\frac{1}{n+\frac{1}{2}} )}. \end{aligned}

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

\begin{aligned} f(x)= \frac{2\ln \Gamma (x+\frac{1}{2} )-2\ln \Gamma (x+1)+\ln (x+\frac{1}{2} )}{\ln (1+\frac{1}{2x+\frac{1}{2}} )}= \frac{f_{1}(x)}{f_{2}(x)}, \end{aligned}

where

\begin{aligned} f_{1}(x)=2\ln \Gamma \biggl(x+\frac{1}{2} \biggr)-2\ln \Gamma (x+1)+ \ln \biggl(x+\frac{1}{2} \biggr) \end{aligned}

and

\begin{aligned} f_{2}(x)=\ln \biggl(1+\frac{1}{2x+\frac{1}{2}} \biggr). \end{aligned}

We conclude from the asymptotic formula of $$\ln \Gamma (z)$$ (see [39, p. 257, Eq. (6.1.41)]) that

$$f_{1}(\infty )=\lim_{x\to \infty}f_{1}(x)=0.$$

Elementary calculations show that

$$\frac{4f'_{1}(x)}{f'_{2}(x)}=(4x+3) (4x+1) \biggl[\psi (x+1)-\psi \biggl(x+ \frac{1}{2} \biggr)-\frac{1}{2x+1} \biggr]=:f_{3}(x).$$

By using inequalities (2.2) and (2.4), we obtain, for $$x\geq 2$$,

\begin{aligned} f'_{3}(x)&=(32x+16) \biggl[\psi (x+1)-\psi \biggl(x+ \frac{1}{2} \biggr)- \frac{1}{2x+1} \biggr] \\ &\quad{}+(4x+3) (4x+1) \biggl[\psi '(x+1)-\psi ' \biggl(x+\frac{1}{2} \biggr)+\frac{2}{(2x+1)^{2}} \biggr] \\ &>(32x+16) \biggl[\frac{1}{2x}-\frac{1}{8x^{2}}+ \frac{1}{64x^{4}}- \frac{1}{128x^{6}}-\frac{1}{2x+1} \biggr] \\ &\quad{}+(4x+3) (4x+1) \biggl[-\frac{1}{2x^{2}}+\frac{1}{4x^{3}}- \frac{1}{16x^{5}}+\frac{2}{(2x+1)^{2}} \biggr] \\ &= \frac{352+2001(x-2)+2784(x-2)^{2}+1656(x-2)^{3}+456(x-2)^{4}+48(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 $$x\geq 2$$. By Lemma 2.1, 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 $$x\geq 2$$. Therefore, the sequence $$\{x_{n} \}$$ is strictly increasing for $$n\geq 4$$. Direct computation yields

\begin{aligned} &x_{1}=\frac{2\ln 2-\ln \pi}{\ln 5-\ln 3}=0.47289\ldots ,\qquad x_{2}=\frac{\ln 3-3\ln 2+\ln \pi}{\ln 7-\ln 5}=0.48711 \ldots , \\ &x_{3}=\frac{5\ln 2-2\ln 3-\ln \pi}{2\ln 3-\ln 7}=0.49253\ldots , \\ & x_{4}=\frac{2\ln 3+\ln 5-7\ln 2+\ln \pi}{\ln 11-2\ln 3}=0.49515 \ldots. \end{aligned}

Consequently, the sequence $$\{x_{n} \}_{n\in \mathbb{N}}$$ is strictly increasing. This leads to

$$\frac{2\ln 2-\ln \pi}{\ln 5-\ln 3}=x_{1}\leq x_{n}< \lim_{n \to \infty}x_{n}\quad \text{for } n\in \mathbb{N}.$$

It remains to prove that

$$\lim_{n \to \infty}x_{n}= \frac{1}{2}.$$
(4.2)

We conclude from the asymptotic formula of $$\ln \Gamma (z)$$ (see [39, p. 257, Eq. (6.1.41)]) that

$$x_{n}= \frac{\frac{1}{2 n}-\frac{1}{2 n^{2}}+O(n^{-3})}{\frac{1}{n}-\frac{1}{n^{2}}+O(n^{-3})}= \frac{\frac{1}{2}+O(n^{-1})}{1+O(n^{-1})}\to \frac{1}{2}\quad \text{as } n\to \infty .$$

Hence, (4.2) holds. This completes the proof of Theorem 4.1. □

### Theorem 4.2

For $$n\in \mathbb{N}$$, the following double inequality holds:

\begin{aligned} & \biggl(1+\frac{1}{n+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16n^{3}+48n^{2}+60n+a} \biggr) \leq \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \\ &\quad < \biggl(1+\frac{1}{n+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16n^{3}+48n^{2}+60n+b} \biggr), \end{aligned}
(4.3)

where the constants

\begin{aligned} a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}=21.42398\ldots \quad \textit{and}\quad b=29 \end{aligned}

are the best possible.

### Proof

First of all, we show that the double inequality (4.3) with $$a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}$$ and $$b=29$$ is valid for $$n=1, 2, 3, 4$$, and 5. For $$n\in \mathbb{N}$$, let

\begin{aligned} &L_{n}= \biggl(1+\frac{1}{n+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16n^{3}+48n^{2}+60n+\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}} \biggr), \\ &U_{n}= \biggl(1+\frac{1}{n+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16n^{3}+48n^{2}+60n+29} \biggr). \end{aligned}

Direct computation yields

\begin{aligned} &L_{1}=\frac{4}{\pi},\qquad \biggl[ \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \biggr]_{n=1}= \frac{4}{\pi}=1.2732\ldots ,\qquad U_{1}=1.2755\ldots , \\ &L_{2}=1.178064357\ldots ,\qquad \biggl[ \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \biggr]_{n=2}=1.17809724510 \ldots , \\ & U_{2}=1.178246681 \ldots , \\ &L_{3}=1.131758795\ldots ,\qquad \biggl[ \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \biggr]_{n=3}=1.13176848421 \ldots , \\ & U_{3}=1.131789661 \ldots , \\ &L_{4}=1.104462901\ldots ,\qquad \biggl[ \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \biggr]_{n=4}=1.10446616728 \ldots , \\ & U_{4}=1.104470767 \ldots , \\ &L_{5}=1.086496467\ldots ,\qquad \biggl[ \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}} \biggr]_{n=5}=1.08649774484 \ldots , \\ & U_{5}=1.086499056 \ldots. \end{aligned}

Clearly, the double inequality (4.3) with $$a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}$$ and $$b=29$$ is valid for $$n=1, 2, 3, 4$$, and 5. For $$n=1$$, the equality on the left-hand side of (4.3) holds.

We now prove that the double inequality (4.3) with $$a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}$$ and $$b=29$$ is valid for $$n\geq 6$$. It suffices to show that for $$x\geq 3$$,

\begin{aligned} & \biggl(1+\frac{1}{2x+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16(2x)^{3}+48(2x)^{2}+60(2x)+a} \biggr) \\ &\quad \leq \frac{\Omega _{2x}^{2}}{\Omega _{2x-1}\Omega _{2x+1}} < \biggl(1+\frac{1}{2x+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16(2x)^{3}+48(2x)^{2}+60(2x)+29} \biggr), \end{aligned}

which can be written as

\begin{aligned} & \biggl(1+\frac{1}{2x+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16(2x)^{3}+48(2x)^{2}+60(2x)+a} \biggr) \\ &\quad \leq \biggl(x+ \frac{1}{2} \biggr) \biggl[\frac{\Gamma (x+\frac{1}{2})}{\Gamma (x+1)} \biggr]^{2} \\ &\quad < \biggl(1+\frac{1}{2x+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16(2x)^{3}+48(2x)^{2}+60(2x)+29} \biggr). \end{aligned}
(4.4)

In order to prove the double inequality (4.4) for $$x\geq 3$$, it suffices to show that

\begin{aligned} f(x)>0\quad \text{and}\quad g(x)< 0 \quad \text{for } x\geq 3, \end{aligned}

where

\begin{aligned}& \begin{aligned} f(x)&=2 \biggl[\ln \Gamma \biggl(x+\frac{1}{2} \biggr)-\ln \Gamma (x+1) \biggr]+\ln \biggl(x+\frac{1}{2} \biggr)-\frac{1}{2}\ln \biggl(1+ \frac{1}{2x+\frac{1}{2}} \biggr) \\ &\quad{}-\ln \biggl(1-\frac{2}{16(2x)^{3}+48(2x)^{2}+60(2x)+a} \biggr), \end{aligned} \\& \begin{aligned} g(x)&=2 \biggl[\ln \Gamma \biggl(x+\frac{1}{2} \biggr)-\ln \Gamma (x+1) \biggr]+\ln \biggl(x+\frac{1}{2} \biggr)-\frac{1}{2}\ln \biggl(1+ \frac{1}{2x+\frac{1}{2}} \biggr) \\ &\quad{}-\ln \biggl(1-\frac{2}{16(2x)^{3}+48(2x)^{2}+60(2x)+29} \biggr). \end{aligned} \end{aligned}

We conclude from the asymptotic formula of $$\ln \Gamma (z)$$ (see [39, p. 257, Eq. (6.1.41)]) that

\begin{aligned} \lim_{x\to \infty}f(x)=\lim_{x\to \infty}g(x)=0. \end{aligned}

Differentiating $$f(x)$$ and applying the left-hand side of (2.3), and noting that

\begin{aligned} a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}< \frac{43}{2}, \end{aligned}

we obtain for $$x\geq 3$$,

\begin{aligned} f'(x)&=-2 \biggl[\psi (x+1)-\psi \biggl(x+\frac{1}{2} \biggr) \biggr]+ \frac{2(16x^{2}+20x+5)}{(4x+3)(4x+1)(2x+1)} \\ &\quad{}- \frac{48(16x^{2}+16x+5)}{(128x^{3}+192x^{2}+120x+a-2)(128x^{3}+192x^{2}+120x+a)} \\ &< -2 \biggl(\frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}}- \frac{1}{128x^{6}}+\frac{17}{2048x^{8}}-\frac{31}{2048x^{10}} \biggr) \\ &\quad{}+\frac{2(16x^{2}+20x+5)}{(4x+3)(4x+1)(2x+1)} \\ &\quad{}- \frac{48(16x^{2}+16x+5)}{(128x^{3}+192x^{2}+120x+\frac{43}{2}-2)(128x^{3}+192x^{2}+120x+\frac{43}{2})} \\ &=- \frac {P_{12}(x-3)}{1024x^{10}(4x+3)(4x+1)(2x+1)(256x^{3}+384x^{2}+240x+39)(256x^{3}+384x^{2}+240x+43)}, \end{aligned}

where

\begin{aligned} P_{12}(x)&=2{,}312{,}798{,}031{,}594+12{,}277{,}183{,}388{,}658x+26{,}310{,}509{,}734{,}485x^{2} \\ &\quad{}+32{,}318{,}240{,}921{,}214x^{3}+26{,}087{,}077{,}081{,}952x^{4}+14{,}780{,}270{,}044{,}224x^{5} \\ &\quad{}+6{,}067{,}872{,}771{,}744x^{6}+1{,}824{,}299{,}158{,}976x^{7}+399{,}070{,}033{,}152x^{8} \\ &\quad{}+61{,}948{,}727{,}808x^{9}+6{,}475{,}038{,}720x^{10} +408{,}944{,}640x^{11}+11{,}796{,}480x^{12}. \end{aligned}

Hence, $$f'(x)<0$$ for $$x\geq 3$$. So, $$f(x)$$ is strictly decreasing for $$x\geq 3$$, and we have

\begin{aligned} f(x)>\lim_{t\to \infty}f(t)=0,\quad x\geq 3. \end{aligned}

Therefore, the left-hand side of (4.3) with $$a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}$$ is valid for $$n\in \mathbb{N}$$.

Differentiating $$g(x)$$ and applying the right-hand side of (2.3), we obtain for $$x\geq 3$$,

\begin{aligned} g'(x)&=-2 \biggl[\psi (x+1)-\psi \biggl(x+\frac{1}{2} \biggr) \biggr]+ \frac{2(16x^{2}+20x+5)}{(4x+3)(4x+1)(2x+1)} \\ &\quad{}- \frac{48(16x^{2}+16x+5)}{(128x^{3}+192x^{2}+120x+27)(128x^{3}+192x^{2}+120x+29)} \\ &>-2 \biggl(\frac{1}{2x}-\frac{1}{8x^{2}}+\frac{1}{64x^{4}}- \frac{1}{128x^{6}}+\frac{17}{2048x^{8}} \biggr)+ \frac{2(16x^{2}+20x+5)}{(4x+3)(4x+1)(2x+1)} \\ &\quad{}- \frac{48(16x^{2}+16x+5)}{(128x^{3}+192x^{2}+120x+27)(128x^{3}+192x^{2}+120x+29)} \\ &=- \frac {P_{9}(x-3)}{1024x^{8}(4x+3)(4x+1)(2x+1)(128x^{3}+192x^{2}+120x+27)(128x^{3}+192x^{2}+120x+29)}, \end{aligned}

where

\begin{aligned} P_{9}(x)&=23{,}529{,}054{,}501+184{,}258{,}816{,}470x+357{,}871{,}998{,}912x^{2} \\ &\quad{}+340{,}974{,}002{,}496x^{3}+191{,}948{,}408{,}224x^{4}+68{,}526{,}376{,}128x^{5} \\ &\quad{}+15{,}780{,}445{,}440x^{6}+2{,}282{,}252{,}800x^{7}+189{,}235{,}200x^{8}+6{,}881{,}280x^{9}. \end{aligned}

Hence, $$g'(x)<0$$ for $$x\geq 3$$. So, $$g(x)$$ is strictly increasing for $$x\geq 3$$, and we have

\begin{aligned} g(x)< \lim_{t\to \infty}f(t)=0,\quad x\geq 3. \end{aligned}

Therefore, the right-hand side of (4.3) with $$b=29$$ is valid for $$n\in \mathbb{N}$$.

If we write (4.3) as

\begin{aligned} a\leq x_{n}< b, \quad x_{n}= \frac{2}{1-\frac{\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}}{ (1+\frac{1}{n+\frac{1}{2}} )^{1/2}}}- \bigl(16n^{3}+48n^{2}+60n\bigr), \end{aligned}

we find that

\begin{aligned} x_{1}=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}} \end{aligned}

and

\begin{aligned} \lim_{n\to \infty}x_{n}&=\lim_{n\to \infty} \biggl\{ \frac{2}{1-\frac{\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}}{ (1+\frac{1}{n+\frac{1}{2}} )^{1/2}}}-\bigl(16n^{3}+48n^{2}+60n \bigr) \biggr\} \\ &=\lim_{n\to \infty} \biggl\{ \frac{2}{\frac{1}{8n^{3}}-\frac{3}{8n^{4}}+\frac{21}{32n^{5}}-\frac{101}{128n^{6}}+O (\frac{1}{n^{7}} )}- \bigl(16n^{3}+48n^{2}+60n\bigr) \biggr\} \\ &=\lim_{n\to \infty} \biggl\{ 29+O \biggl(\frac{1}{n} \biggr) \biggr\} =29. \end{aligned}

This limit is obtained by using the asymptotic expansion (1.4).

Hence, the double inequality (4.3) holds for $$n\in \mathbb{N}$$, and the constants $$a=\frac{2(248\sqrt{15}-305\pi )}{5\pi -4\sqrt{15}}$$ and $$b=29$$ are the best possible. The proof of Theorem 4.2 is complete. □

## 5 Comparison

It follows form (1.1), (1.2) and (1.3) and (4.3) that

\begin{aligned}& \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}\sim \biggl(1+ \frac{1}{n} \biggr)^{1/2}=u_{n} \quad ( \text{Alzer }), \end{aligned}
(5.1)
\begin{aligned}& \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}\sim \biggl(1+ \frac{1}{n+1} \biggr)^{1/2}=v_{n} \quad ( \text{Merkle }), \end{aligned}
(5.2)
\begin{aligned}& \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}\sim \biggl(1+ \frac{1}{n+\frac{1}{2}} \biggr)^{1/2}=w_{n} \quad (\text{Ban and Chen }), \end{aligned}
(5.3)
\begin{aligned}& \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}\sim \biggl(1+ \frac{1}{n+\frac{1}{2}} \biggr)^{1/2} \biggl(1- \frac{2}{16n^{3}+48n^{2}+60n+29} \biggr)=r_{n} \quad (\text{New}). \end{aligned}
(5.4)

We here offer some numerical computations (see Table 1) to show the superiority of our sequence $$\{r_{n}\}_{n\geq 1}$$ over the sequences $$\{u_{n}\}_{n\geq 1}$$, $$\{v_{n}\}_{n\geq 1}$$, and $$\{w_{n}\}_{n\geq 1}$$.

Here $$V_{n}:=\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$. In fact, we have, as $$n\to \infty$$,

\begin{aligned} &\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}= u_{n}+O \biggl( \frac{1}{n^{2}} \biggr),\qquad \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}= v_{n}+O \biggl( \frac{1}{n^{2}} \biggr), \\ &\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}= w_{n}+O \biggl( \frac{1}{n^{3}} \biggr),\qquad \frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}= r_{n}+O \biggl( \frac{1}{n^{7}} \biggr). \end{aligned}

These formulas are obtained by using the computer program MAPLE 17.

## 6 Conclusion

Here, in our present investigation, we have first revisited several interesting properties of the volume $$\Omega _{n}$$ of the unit ball in $$\mathbb{R}^{n}$$, including monotonicity properties, inequalities, and asymptotic expansions. We have then shown that the function $$G(x)= (1+\frac{1}{2x+\frac{1}{2}} )^{1/2}/I(x)$$ is logarithmically completely monotonic on $$(0,\infty )$$ (Theorem 3.1), which yielded a double inequality for the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$, see (3.5). Also, we have established new sharp inequalities for the quantity $$\frac{\Omega _{n}^{2}}{\Omega _{n-1}\Omega _{n+1}}$$, see (4.1) and (4.3). We have also considered a number of related developments on the subject of this paper.

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## Acknowledgements

The authors express their gratitude to the referee for very helpful and detailed comments.

## Funding

Supported by the Fundamental Research Funds for the Universities of the Henan Province (Grant No. NSFRF210446).

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