Padé approximant related to the Wallis formula

Abstract

Based on the Padé approximation method, in this paper we determine the coefficients $a_j$ and $b_j$ such that

$$\pi ={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\{\frac{n^k+a_1{n}^{k-1}+\cdots +a_k}{n^{k+1}+b_1{n}^k+\cdots +b_{k+1}}+O\left(\frac{1}{n^{2k+3}}\right)\},n\to \mathrm{\infty },$$

where $k\ge 0$ is any given integer. Based on the obtained result, we establish a more accurate formula for approximating π, which refines some known results.

Introduction

It is well known that the number π satisfies the following inequalities:

$$\begin{array}{r}\frac{2}{2n+1}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2<\pi <\frac{1}{n}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2,n\in \mathbb{N}:=\{1,2,3,\dots \},\end{array}$$ 1.1

where

$$(2n)!!=2\cdot 4\cdot 6\cdots (2n)=2^{n}n!,(2n-1)!!=1\cdot 3\cdot 5\cdots (2n-1).$$

This result is due to Wallis (see [1]).

Based on a basic theorem in mathematical statistics concerning unbiased estimators with minimum variance, Gurland [1] yielded a closer approximation to π than that afforded by (1.1), namely,

$$\begin{array}{r}\frac{4n+3}{{(2n+1)}^2}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2<\pi <\frac{4}{4n+1}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2,n\in \mathbb{N}.\end{array}$$ 1.2

By using (1.2), Brutman [2] and Falaleev [3] established estimates of the Landau constants.

Mortici [4], Theorem 2, improved Gurland’s result (1.2) and obtained the following double inequality:

$$\begin{array}{rl}& (\frac{n+\frac{1}{4}}{n^2+\frac{1}{2}n+\frac{3}{32}}+\frac{9}{2,048n^5}-\frac{45}{8,192n^6}){\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\\ & <\pi <(\frac{n+\frac{1}{4}}{n^2+\frac{1}{2}n+\frac{3}{32}}+\frac{9}{2,048n^5}){\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2,n\in \mathbb{N}.\end{array}$$ 1.3

We see from (1.3) that

$$\pi ={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\{\frac{n+\frac{1}{4}}{n^2+\frac{1}{2}n+\frac{3}{32}}+O\left(\frac{1}{n^5}\right)\},n\to \mathrm{\infty }.$$ 1.4

Based on the Padé approximation method, in this paper we develop the approximation formula (1.4) to produce a general result. More precisely, we determine the coefficients $a_j$ and $b_j$ such that

$$\pi ={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\{\frac{n^k+a_1{n}^{k-1}+\cdots +a_k}{n^{k+1}+b_1{n}^k+\cdots +b_{k+1}}+O\left(\frac{1}{n^{2k+3}}\right)\},n\to \mathrm{\infty },$$ 1.5

where $k\ge 0$ is any given integer. Based on the obtained result, we establish a more accurate formula for approximating π, which refines some known results.

The numerical values given in this paper have been calculated via the computer program MAPLE 13.

Lemmas

Euler’s gamma function $\mathrm{\Gamma }(x)$ is one of the most important functions in mathematical analysis and has applications in diverse areas. The logarithmic derivative of $\mathrm{\Gamma }(x)$, denoted by $\psi (x)={\mathrm{\Gamma }}^{\prime }(x)/\mathrm{\Gamma }(x)$, is called the psi (or digamma) function.

The following lemmas are required in the sequel.

Lemma 2.1

[5]

Let $r\ne 0$ be a given real number and $\ell \ge 0$ be a given integer. The following asymptotic expansion holds:

$$\frac{\mathrm{\Gamma }(x+1)}{\mathrm{\Gamma }(x+\frac{1}{2})}\sim \sqrt{x}{(1+\sum _{j=1}^{\mathrm{\infty }}\frac{p_j}{x^j})}^{x^{\ell }/r},x\to \mathrm{\infty },$$ 2.1

with the coefficients $p_j\equiv p_j(\ell ,r)(j\in \mathbb{N})$ given by

$$p_j=\sum \frac{r^{k_1+k_2+\cdots +k_j}}{k_1!k_2!\cdots k_j!}{\left(\frac{(2^2-1)B_2}{1\cdot 1\cdot 2^2}\right)}^{k_1}{\left(\frac{(2^4-1)B_4}{2\cdot 3\cdot 2^4}\right)}^{k_2}\cdots {\left(\frac{(2^{2j}-1)B_{2j}}{j(2j-1)2^{2j}}\right)}^{k_j},$$ 2.2

where $B_j$ are the Bernoulli numbers summed over all nonnegative integers $k_j$ satisfying the equation

$$(1+\ell )k_1+(3+\ell )k_2+\cdots +(2j+\ell -1)k_j=j.$$

In particular, setting $(\ell ,r)=(0,-2)$ in (2.1) yields

$$x{\left(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right)}^2\sim 1+\sum _{j=1}^{\mathrm{\infty }}\frac{c_j}{x^j},x\to \mathrm{\infty },$$ 2.3

where the coefficients $c_j\equiv p_j(0,-2)(j\in \mathbb{N})$ are given by

$$c_j=\sum \frac{{(-2)}^{k_1+k_2+\cdots +k_j}}{k_1!k_2!\cdots k_j!}{\left(\frac{(2^2-1)B_2}{1\cdot 1\cdot 2^2}\right)}^{k_1}{\left(\frac{(2^4-1)B_4}{2\cdot 3\cdot 2^4}\right)}^{k_2}\cdots {\left(\frac{(2^{2j}-1)B_{2j}}{j(2j-1)2^{2j}}\right)}^{k_j},$$ 2.4

summed over all nonnegative integers $k_j$ satisfying the equation

$$k_1+3k_2+\cdots +(2j-1)k_j=j.$$

Lemma 2.2

[5]

Let $m,n\in \mathbb{N}$. Then, for $x>0$,

$$\begin{array}{rl}\sum _{j=1}^{2m}(1-\frac{1}{2^{2j}})\frac{2B_{2j}}{(2j)!}\frac{(2j+n-2)!}{x^{2j+n-1}}& <{(-1)}^n({\psi }^{(n-1)}(x+1)-{\psi }^{(n-1)}(x+\frac{1}{2}))+\frac{(n-1)!}{2x^n}\\ & <\sum _{j=1}^{2m-1}(1-\frac{1}{2^{2j}})\frac{2B_{2j}}{(2j)!}\frac{(2j+n-2)!}{x^{2j+n-1}}.\end{array}$$ 2.5

In particular, we have

$$U(x)<\psi (x+1)-\psi (x+\frac{1}{2})<V(x),$$ 2.6

where

$$\begin{array}{rl}V(x)=& \frac{1}{2x}-\frac{1}{8x^2}+\frac{1}{64x^4}-\frac{1}{128x^6}+\frac{17}{2,048x^8}-\frac{31}{2,048x^{10}}+\frac{691}{16,384x^{12}}\\ & -\frac{5,461}{32,768x^{14}}+\frac{929,569}{1,048,576x^{16}}\end{array}$$

and

$$U(x)=V(x)-\frac{3,202,291}{524,288x^{18}}.$$

For our later use, we introduce Padé approximant (see [6–11]). Let f be a formal power series

$$f(t)=c_0+c_{1}t+c_2{t}^2+\cdots .$$ 2.7

The Padé approximation of order $(p,q)$ of the function f is the rational function, denoted by

$${[p/q]}_f(t)=\frac{{\sum }_{j=0}^p{a}_j{t}^j}{1+{\sum }_{j=1}^q{b}_j{t}^j},$$ 2.8

where $p\ge 0$ and $q\ge 1$ are two given integers, the coefficients $a_j$ and $b_j$ are given by (see [6–8, 10, 11])

$$\{\begin{array}{c}{a}_0=c_0,\hfill \\ a_1=c_0{b}_1+c_1,\hfill \\ a_2=c_0{b}_2+c_1{b}_1+c_2,\hfill \\ ⋮\hfill \\ a_p=c_0{b}_p+\cdots +c_{p-1}{b}_1+c_p,\hfill \\ 0=c_{p+1}+c_p{b}_1+\cdots +c_{p-q+1}{b}_q,\hfill \\ ⋮\hfill \\ 0=c_{p+q}+c_{p+q-1}{b}_1+\cdots +c_p{b}_q,\hfill \end{array}$$ 2.9

and the following holds:

$${[p/q]}_f(t)-f(t)=O\left(t^{p+q+1}\right).$$ 2.10

Thus, the first $p+q+1$ coefficients of the series expansion of ${[p/q]}_f$ are identical to those of f. Moreover, we have (see [9])

$$\begin{array}{rl}& {[p/q]}_f(t)=\frac{\left|\begin{array}{cccc}{t}^q{f}_{p-q}(t)& t^{q-1}{f}_{p-q+1}(t)& \cdots & f_p(t)\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|}{\left|\begin{array}{cccc}{t}^q& t^{q-1}& \cdots & 1\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|},\end{array}$$ 2.11

with $f_n(x)=c_0+c_{1}x+\cdots +c_n{x}^n$, the nth partial sum of the series f in (2.7).

Main results

Let

$$f(x)=x{\left(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right)}^2.$$ 3.1

It follows from (2.3) that, as $x\to \mathrm{\infty }$,

$$\begin{array}{rl}f(x)\sim \sum _{j=0}^{\mathrm{\infty }}\frac{c_j}{x^j}=& 1-\frac{1}{4x}+\frac{1}{32x^2}+\frac{1}{128x^3}-\frac{5}{2,048x^4}-\frac{23}{8,192x^5}+\frac{53}{65,536x^6}\\ & +\frac{593}{262,144x^7}-\cdots ,\end{array}$$ 3.2

with the coefficients $c_j$ given by (2.4). In what follows, the function f is given in (3.1).

Based on the Padé approximation method, we now give a derivation of formula (1.4). To this end, we consider

$${[1/2]}_f(x)=\frac{{\sum }_{j=0}^1{a}_j{x}^{-j}}{1+{\sum }_{j=1}^2{b}_j{x}^{-j}}.$$

Noting that

$$c_0=1,c_1=-\frac{1}{4},c_2=\frac{1}{32},c_3=\frac{1}{128}$$

holds, we have, by (2.9),

$$\{\begin{array}{c}{a}_0=1,\hfill \\ a_1=b_1-\frac{1}{4},\hfill \\ 0=\frac{1}{32}-\frac{1}{4}{b}_1+b_2,\hfill \\ 0=\frac{1}{128}+\frac{1}{32}{b}_1-\frac{1}{4}{b}_2,\hfill \end{array}$$

that is,

$$a_0=1,a_1=\frac{1}{4},b_1=\frac{1}{2},b_2=\frac{3}{32}.$$

We thus obtain that

$${[1/2]}_f(x)=\frac{1+\frac{1}{4x}}{1+\frac{1}{2x}+\frac{3}{32x^2}},$$ 3.3

and we have, by (2.10),

$$x{\left(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right)}^2-\frac{1+\frac{1}{4x}}{1+\frac{1}{2x}+\frac{3}{32x^2}}=O\left(\frac{1}{x^4}\right),x\to \mathrm{\infty }.$$ 3.4

Noting that

$$\frac{\mathrm{\Gamma }(n+\frac{1}{2})}{\mathrm{\Gamma }(n+1)}=\sqrt{\pi }\cdot \frac{(2n-1)!!}{(2n)!!},n\in \mathbb{N}\text{ (the Wallis ratio)}$$ 3.5

holds, replacing x by n in (3.4) yields (1.4).

From the Padé approximation method introduced in Section 2 and the asymptotic expansion (3.2), we obtain a general result given by Theorem 3.1. As a consequence, we obtain (1.5).

Theorem 3.1

The Padé approximation of order $(p,q)$ of the asymptotic formula of the function $f(x)=x{(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)})}^2$ (at the point $x=\mathrm{\infty }$) is the following rational function:

$${[p/q]}_f(x)=\frac{1+{\sum }_{j=1}^p{a}_j{x}^{-j}}{1+{\sum }_{j=1}^q{b}_j{x}^{-j}}=x\left(\frac{x^p+a_1{x}^{p-1}+\cdots +a_p}{x^q+b_1{x}^{q-1}+\cdots +b_q}\right),$$ 3.6

where $p\ge 0$ and $q\ge 1$ are two given integers and $q=p+1$ (an empty sum is understood to be zero), the coefficients $a_j$ and $b_j$ are given by

$$\{\begin{array}{c}{a}_1=b_1+c_1,\hfill \\ a_2=b_2+c_1{b}_1+c_2,\hfill \\ ⋮\hfill \\ a_p=b_p+\cdots +c_{p-1}{b}_1+c_p,\hfill \\ 0=c_{p+1}+c_p{b}_1+\cdots +c_{p-q+1}{b}_q,\hfill \\ ⋮\hfill \\ 0=c_{p+q}+c_{p+q-1}{b}_1+\cdots +c_p{b}_q,\hfill \end{array}$$ 3.7

and $c_j$ is given in (2.4), and the following holds:

$$f(x)-{[p/q]}_f(x)=O\left(\frac{1}{x^{p+q+1}}\right),x\to \mathrm{\infty }.$$ 3.8

Moreover, we have

$$\begin{array}{rl}& {[p/q]}_f(x)=\frac{\left|\begin{array}{cccc}\frac{1}{x^q}{f}_{p-q}(x)& \frac{1}{x^{q-1}}{f}_{p-q+1}(t)& \cdots & f_p(t)\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|}{\left|\begin{array}{cccc}\frac{1}{x^q}& \frac{1}{x^{q-1}}& \cdots & 1\\ c_{p-q+1}& c_{p-q+2}& \cdots & c_{p+1}\\ ⋮& ⋮& \ddots & ⋮\\ c_p& c_{p+1}& \cdots & c_{p+q}\end{array}\right|},\end{array}$$ 3.9

with $f_n(x)={\sum }_{j=0}^n\frac{c_j}{x^j}$, the nth partial sum of the asymptotic series (3.2).

Remark 3.1

Using (3.9), we can also derive (3.3). Indeed, we have

$$\begin{array}{rl}{[1/2]}_f(x)& =\frac{\left|\begin{array}{ccc}\frac{1}{x^2}{f}_{-1}(x)& \frac{1}{x}{f}_0(x)& f_1(x)\\ c_0& c_1& c_2\\ c_1& c_2& c_3\end{array}\right|}{\left|\begin{array}{ccc}\frac{1}{x^2}& \frac{1}{x}& 1\\ c_0& c_1& c_2\\ c_1& c_2& c_3\end{array}\right|}=\frac{\left|\begin{array}{ccc}0& \frac{1}{x}& 1-\frac{1}{4x}\\ 1& -\frac{1}{4}& \frac{1}{32}\\ -\frac{1}{4}& \frac{1}{32}& \frac{1}{128}\end{array}\right|}{\left|\begin{array}{ccc}\frac{1}{x^2}& \frac{1}{x}& 1\\ 1& -\frac{1}{4}& \frac{1}{32}\\ -\frac{1}{4}& \frac{1}{32}& \frac{1}{128}\end{array}\right|}=\frac{1+\frac{1}{4x}}{1+\frac{1}{2x}+\frac{3}{32x^2}}.\end{array}$$

Replacing x by n in (3.8) applying (3.5), we obtain the following corollary.

Corollary 3.1

As $n\to \mathrm{\infty }$,

$$\pi ={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\{\frac{n^p+{\sum }_{j=1}^p{a}_j{n}^{p-j}}{n^q+{\sum }_{j=1}^q{b}_j{n}^{q-j}}+O\left(\frac{1}{n^{p+q+2}}\right)\},n\to \mathrm{\infty },$$ 3.10

where $p\ge 0$ and $q\ge 1$ are two given integers and $q=p+1$, and the coefficients $a_j$ and $b_j$ are given by (3.7).

Remark 3.2

Setting $(p,q)=(k,k+1)$ in (3.10) yields (1.5).

Setting

$$(p,q)=(4,5)\text{and}(p,q)=(5,6)$$

in (3.10), respectively, we find

$$\pi ={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\{\frac{n^4+n^3+\frac{107}{64}{n}^2+\frac{91}{128}n+\frac{789}{4,096}}{n^5+\frac{5}{4}{n}^4+\frac{125}{64}{n}^3+\frac{295}{256}{n}^2+\frac{1,689}{4,096}n+\frac{945}{16,384}}+O\left(\frac{1}{n^{11}}\right)\}$$ 3.11

and

$$\begin{array}{rl}\pi =& {\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\\ & \times \{\frac{n^5+\frac{5}{4}{n}^4+\frac{51}{16}{n}^3+\frac{133}{64}{n}^2+\frac{5,243}{4,096}n+\frac{3,867}{16,384}}{n^6+\frac{3}{2}{n}^5+\frac{113}{32}{n}^4+\frac{93}{32}{n}^3+\frac{7,729}{4,096}{n}^2+\frac{4,881}{8,192}n+\frac{10,395}{131,072}}+O\left(\frac{1}{n^{13}}\right)\}\end{array}$$ 3.12

as $n\to \mathrm{\infty }$.

Formulas (3.11) and (3.12) motivate us to establish the following theorem.

Theorem 3.2

The following inequality holds:

$$\begin{array}{rl}& \frac{x^5+\frac{5}{4}{x}^4+\frac{51}{16}{x}^3+\frac{133}{64}{x}^2+\frac{5,243}{4,096}x+\frac{3,867}{16,384}}{x^6+\frac{3}{2}{x}^5+\frac{113}{32}{x}^4+\frac{93}{32}{x}^3+\frac{7,729}{4,096}{x}^2+\frac{4,881}{8,192}x+\frac{10,395}{131,072}}\\ & <{\left(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right)}^2\\ & <\frac{x^4+x^3+\frac{107}{64}{x}^2+\frac{91}{128}x+\frac{789}{4,096}}{x^5+\frac{5}{4}{x}^4+\frac{125}{64}{x}^3+\frac{295}{256}{x}^2+\frac{1,689}{4,096}x+\frac{945}{16,384}}.\end{array}$$ 3.13

The left-hand side inequality holds for $x\ge 4$, while the right-hand side inequality is valid for $x\ge 3$.

Proof

It suffices to show that

$$F(x)>0\text{for }x\ge 4\text{and}G(x)<0\text{for }x\ge 3,$$

where

$$F(x)=2ln\left(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right)-ln\frac{x^5+\frac{5}{4}{x}^4+\frac{51}{16}{x}^3+\frac{133}{64}{x}^2+\frac{5,243}{4,096}x+\frac{3,867}{16,384}}{x^6+\frac{3}{2}{x}^5+\frac{113}{32}{x}^4+\frac{93}{32}{x}^3+\frac{7,729}{4,096}{x}^2+\frac{4,881}{8,192}x+\frac{10,395}{131,072}}$$

and

$$G(x)=2ln\left(\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right)-ln\frac{x^4+x^3+\frac{107}{64}{x}^2+\frac{91}{128}x+\frac{789}{4,096}}{x^5+\frac{5}{4}{x}^4+\frac{125}{64}{x}^3+\frac{295}{256}{x}^2+\frac{1,689}{4,096}x+\frac{945}{16,384}}.$$

Using the following asymptotic expansion (see [12]):

$$\begin{array}{rl}{\left[\frac{\mathrm{\Gamma }(x+\frac{1}{2})}{\mathrm{\Gamma }(x+1)}\right]}^2\sim & \frac{1}{x}exp(-\frac{1}{4x}+\frac{1}{96x^3}-\frac{1}{320x^5}+\frac{17}{7,168x^7}-\frac{31}{9,216x^9}\\ & +\frac{691}{90,112x^{11}}-\frac{5,461}{212,992x^{13}}+\frac{929,569}{7,864,320x^{15}}-\cdots ),x\to \mathrm{\infty },\end{array}$$ 3.14

we obtain that

$$\underset{x\to \mathrm{\infty }}{lim}F(x)=0\text{and}\underset{x\to \mathrm{\infty }}{lim}G(x)=0.$$

Differentiating $F(x)$ and applying the first inequality in (2.6), we find

$$\begin{array}{rl}{F}^{\prime }(x)& =-2[\psi (x+1)-\psi (x+\frac{1}{2})]+\frac{P_{10}(x)}{P_{11}(x)}\\ & <-2U(x)+\frac{P_{10}(x)}{P_{11}(x)}=-\frac{P_{16}(x-4)}{524,288x^{18}{P}_{11}(x)},\end{array}$$

where

$$\begin{array}{rl}{P}_{10}(x)=& 4(20,998,323+301,244,208x+1,329,622,624x^2+3,532,111,872x^3\\ & +6,831,390,720x^4+8,950,906,880x^5+9,510,060,032x^6\\ & +6,476,005,376x^7+4,244,635,648x^8+1,342,177,280x^9+536,870,912x^{10}),\\ P_{11}(x)=& (16,384x^5+20,480x^4+52,224x^3+34,048x^2+20,972x+3,867)\\ & \times (131,072x^6+196,608x^5+462,848x^4+380,928x^3+247,328x^2\\ & +78,096x+10,395)\end{array}$$

and

$$\begin{array}{rl}{P}_{16}(x)=& 73,399,302,245,132,658,732,474+401,687,666,421,636,714,876,048x\\ & +882,663,824,965,187,436,960,169x^2\\ & +1,129,813,735,156,766,429,414,420x^3\\ & +975,385,167,000,268,446,720,384x^4\\ & +611,802,531,654,753,268,270,848x^5\\ & +290,696,674,545,996,984,221,376x^6\\ & +107,149,026,028,490,487,475,968x^7\\ & +31,018,031,026,615,120,693,760x^8\\ & +7,080,024,048,117,231,228,928x^9\\ & +1,270,066,473,244,063,756,800x^{10}+177,136,978,237,041,715,200x^{11}\\ & +18,824,726,793,935,462,400x^{12}+1,473,208,721,923,276,800x^{13}\\ & +80,051,720,723,251,200x^{14}+2,698,074,228,326,400x^{15}\\ & +42,489,357,926,400x^{16}.\end{array}$$

Hence, $F^{\prime }(x)<0$ for $x\ge 4$, and we have

$$F(x)>\underset{t\to \mathrm{\infty }}{lim}F(t)=0,x\ge 4.$$

Differentiating $G(x)$ and applying the second inequality in (2.6), we find

$$\begin{array}{rl}{G}^{\prime }(x)& =-2[\psi (x+1)-\psi (x+\frac{1}{2})]+\frac{4P_8(x)}{P_9(x)}>-2V(x)+\frac{4P_8(x)}{P_9(x)}\\ & =\frac{P_{14}(x-3)}{524,288x^{16}{P}_9(x)},\end{array}$$

where

$$\begin{array}{rl}{P}_8(x)=& 16,777,216x^8+33,554,432x^7+72,351,744x^6+79,167,488x^5+75,583,488x^4\\ & +45,043,712x^3+18,211,328x^2+4,212,480x+644,661,\\ P_9(x)=& (4,096x^4+4,096x^3+6,848x^2+2,912x+789)\\ & \times (16,384x^5+20,480x^4+32,000x^3+18,880x^2+6,756x+945)\end{array}$$

and

$$\begin{array}{rl}{P}_{14}(x)=& 427,884,340,806,856,575+5,508,337,280,234,438,700x\\ & +16,278,641,070,340,979,232x^2\\ & +25,110,186,749,213,013,376x^3+25,009,399,125,661,680,960x^4\\ & +17,642,792,222,808,253,696x^5\\ & +9,230,356,959,310,493,184x^6+3,661,094,552,739,530,752x^7\\ & +1,108,535,832,992,448,000x^8\\ & +255,024,028,762,675,200x^9+43,854,087,132,979,200x^{10}\\ & +5,462,018,666,496,000x^{11}\\ & +465,495,496,704,000x^{12}+24,287,993,856,000x^{13}\\ & +585,252,864,000x^{14}.\end{array}$$

Hence, $G^{\prime }(x)>0$ for $x\ge 3$, and we have

$$G(x)<\underset{t\to \mathrm{\infty }}{lim}G(t)=0,x\ge 3.$$

The proof is complete. □

Corollary 3.2

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

$$a_n<\pi <b_n,$$ 3.15

where

$$a_n=\frac{n^5+\frac{5}{4}{n}^4+\frac{51}{16}{n}^3+\frac{133}{64}{n}^2+\frac{5,243}{4,096}n+\frac{3,867}{16,384}}{n^6+\frac{3}{2}{n}^5+\frac{113}{32}{n}^4+\frac{93}{32}{n}^3+\frac{7,729}{4,096}{n}^2+\frac{4,881}{8,192}n+\frac{10,395}{131,072}}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2$$ 3.16

and

$$b_n=\frac{n^4+n^3+\frac{107}{64}{n}^2+\frac{91}{128}n+\frac{789}{4,096}}{n^5+\frac{5}{4}{n}^4+\frac{125}{64}{n}^3+\frac{295}{256}{n}^2+\frac{1,689}{4,096}n+\frac{945}{16,384}}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2.$$ 3.17

Proof

Noting that (3.5) holds, we see by (3.13) that the left-hand side of (3.15) holds for $n\ge 4$, while the right-hand side of (3.15) is valid for $n\ge 3$. Elementary calculations show that the left-hand side of (3.15) is also valid for $n=1,2$ and 3, and the right-hand side of (3.15) is valid for $n=1$ and 2. The proof is complete. □

Comparison

Recently, Lin [12] improved Mortici’s result (1.3) and obtained the following inequalities:

$${\lambda }_n<\pi <{\mu }_n$$ 4.1

and

$${\delta }_n<\pi <{\omega }_n,$$ 4.2

where

$$\begin{array}{rl}& {\lambda }_n=(1+\frac{1}{4n}-\frac{3}{32n^2}+\frac{3}{128n^3}+\frac{3}{2,048n^4}-\frac{33}{8,192n^5}-\frac{39}{65,536n^6})\\ & \phantom{{\lambda }_n=}\times \frac{2}{2n+1}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2,\end{array}$$ 4.3

$${\mu }_n=(1+\frac{1}{4n}-\frac{3}{32n^2}+\frac{3}{128n^3}+\frac{3}{2,048n^4})\frac{2}{2n+1}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2,$$ 4.4

$${\delta }_n={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\frac{1}{n}exp(-\frac{1}{4n}+\frac{1}{96n^3}-\frac{1}{320n^5}+\frac{17}{7,168n^7}-\frac{31}{9,216n^9}),$$ 4.5

$${\omega }_n={\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\frac{1}{n}exp(-\frac{1}{4n}+\frac{1}{96n^3}-\frac{1}{320n^5}+\frac{17}{7,168n^7}).$$ 4.6

Direct computation yields

$$\begin{array}{rl}& a_n-{\lambda }_n\\ & =\frac{3(7,634,944n^5+12,928,000n^4+18,895,616n^3+9,755,072n^2+1,930,008n+135,135)}{32,768n^6(2n+1)(131,072n^6+196,608n^5+462,848n^4+380,928n^3+247,328n^2+78,096n+10,395)}\\ & \times {\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2>0\end{array}$$

and

$$\begin{array}{rl}& b_n-{\mu }_n\\ & =-\frac{3(45,056n^4+62,976n^3+66,496n^2+21,876n+945)}{1,024n^4(2n+1)(16,384n^5+20,480n^4+32,000n^3+18,880n^2+6,756n+945)}{\left(\frac{(2n)!!}{(2n-1)!!}\right)}^2\\ & <0.\end{array}$$

Hence, (3.15) improves (4.1).

The following numerical computations (see Table 1) would show that ${\delta }_n<a_n$ and $b_n<{\omega }_n$ for $n\in \mathbb{N}$. That is to say, inequalities (3.15) are sharper than inequalities (4.2).

Table 1.

Comparison between inequalities (3.15) and (4.2)

n	${\mathit{a}}_{\mathit{n}}\mathbf{-}{\mathit{\delta }}_{\mathit{n}}$	${\mathit{\omega }}_{\mathit{n}}\mathbf{-}{\mathit{b}}_{\mathit{n}}$
1	6.673798 × 10−3	3.789512 × 10−3
10	2.264856 × 10−13	9.947434 × 10−12
100	2.398663 × 10−24	1.051407 × 10−20
1,000	2.408054 × 10−35	1.056218 × 10−29
10,000	2.408948 × 10−46	1.056690 × 10−38

In fact, we have

$$\begin{array}{rl}& {\lambda }_n=\pi +O\left(\frac{1}{n^7}\right),{\mu }_n=\pi +O\left(\frac{1}{n^5}\right),\\ & {\delta }_n=\pi +O\left(\frac{1}{n^{11}}\right),{\omega }_n=\pi +O\left(\frac{1}{n^9}\right),\\ & a_n=\pi +O\left(\frac{1}{n^{12}}\right),b_n=\pi +O\left(\frac{1}{n^{10}}\right).\end{array}$$