A Series Expansion of a Logarithmic Expression and a Decreasing Property of the Ratio of Two Logarithmic Expressions Containing Sine

Abstract

In the paper, by virtue of a derivative formula for the ratio of two differentiable functions and with the help of a monotonicity rule, the authors expand a logarithmic expression involving the sine function into the Maclaurin power series in terms of specific determinants and prove a decreasing property of the ratio of two logarithmic expressions containing the sine function. These results are interesting purely in pure mathematics.

1. Motivations

In [1] (pp. 42 and 55), we find the Maclaurin power series expansions

$$sinx=\sum _{k=0}^{\infty }{(-1)}^k\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362,880}-\cdots ,x\in \mathbb{R}$$

(1)

and

$$lnsinx=lnx-\sum _{k=1}^{\infty }\frac{2^{2k-1}}{\left(2k\right)!k}\left|B_{2k}\right|x^{2k}=lnx-\frac{x^2}{6}-\frac{x^4}{180}-\frac{x^6}{2835}-\cdots ,0<x<\pi ,$$

(2)

where $B_{2k}$ denotes the Bernoulli numbers, which can be generated in [2] (p. 3) by

$$\frac{z}{e^z-1}=\sum _{k=0}^{\infty }{B}_k\frac{z^k}{k!}=1-\frac{z}{2}+\sum _{k=1}^{\infty }{B}_{2k}\frac{z^{2k}}{\left(2k\right)!},\left|z\right|<2\pi .$$

For more information about $B_{2k}$, please refer to the paper [3], and closely related references therein. From the series expansion (2), we acquire

$$ln\frac{sinx}{x}=-\sum _{k=1}^{\infty }\frac{2^{2k-1}}{\left(2k\right)!k}|B_{2k}|x^{2k}=-\frac{x^2}{6}-\frac{x^4}{180}-\frac{x^6}{2835}-\cdots ,0<\left|x\right|<\pi .$$

Motivated by the questions at the sites https://mathoverflow.net/q/444321 (accessed on 15 June 2023), https://mathoverflow.net/q/444490 (accessed on 12 April 2023), https://mathoverflow.net/q/444525 (accessed on 16 April 2023), and their answers therein, we consider the following two problems in this paper:

• What is the Maclaurin power series expansion of the even function

  $$Q\left(x\right)=\{\begin{array}{ll}ln\left[\frac{6}{x^2}\left(1-\frac{sinx}{x}\right)\right],& 0<\left|x\right|<\infty \\ 0,& x=0\end{array}$$

  (3)

  around $x=0$?
• Is the even function

  $$R\left(x\right)=\{\begin{array}{ll}\frac{ln\left[\frac{6}{x^2}\left(1-\frac{sinx}{x}\right)\right]}{ln\frac{sinx}{x}},& \left|x\right|\in (0,\pi )\\ \frac{3}{10},& x=0\\ 0,& x=\pm \pi \end{array}$$

  (4)

  decreasing on the close interval $[0,\pi ]$?

It is clear that the logarithmic expression $Q\left(x\right)$ defined in (3) is more complicated than the logarithmic expression $ln\frac{sinx}{x}$. Therefore, it is interesting purely in pure mathematics to expand $Q\left(x\right)$ into a Maclaurin power series expansion around the origin $x=0$ and to compare these two logarithmic expressions by considering the decreasing property of their quotient $R\left(x\right)$.

In this paper, after preparing the general formula for higher derivatives of the quotient of two differentiable functions and reciting two monotonicity rules in Section 2, we will give an answer to the first problem by presenting a Maclaurin power series expansion of the even function $Q\left(x\right)$ around $x=0$ in Section 3, as well as provide an answer to the second problem by verifying the decreasing property of the quotient $R\left(x\right)$ in Section 4.

2. Lemmas

For smoothly solving the above two problems, we need the following lemmas.

Lemma 1.

Let $u\left(x\right)$ and $v\left(x\right)\ne 0$ be two n-time differentiable functions on an interval I for a given integer $n\ge 0$. Then, the nth derivative of the ratio $\frac{u\left(x\right)}{v\left(x\right)}$ is

$$\frac{d^n}{d{x}^n}\left[\frac{u\left(x\right)}{v\left(x\right)}\right]={(-1)}^n\frac{\left|W_{(n+1)\times (n+1)}\left(x\right)\right|}{v^{n+1}\left(x\right)},n\ge 0,$$

(5)

where the matrix

$$W_{(n+1)\times (n+1)}\left(x\right)={\left(\begin{array}{cc}{U}_{(n+1)\times 1}\left(x\right)& V_{(n+1)\times n}\left(x\right)\end{array}\right)}_{(n+1)\times (n+1)},$$

the matrix $U_{(n+1)\times 1}\left(x\right)$ is an $(n+1)\times 1$ matrix whose elements satisfy $u_{k,1}\left(x\right)=u^{(k-1)}\left(x\right)$ for $1\le k\le n+1$, the matrix $V_{(n+1)\times n}\left(x\right)$ is an $(n+1)\times n$ matrix whose elements are

$$v_{\ell ,j}\left(x\right)=\left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{\ell -1}{j-1}}\right)v^{(\ell -j)}\left(x\right),\hfill & \ell -j\ge 0\hfill \\ 0,\hfill & \ell -j<0\hfill \end{array}\right.$$

for $1\le \ell \le n+1$ and $1\le j\le n$, and the notation $|W_{(n+1)\times (n+1)}\left(x\right)|$ denotes the determinant of the $(n+1)\times (n+1)$ matrix $W_{(n+1)\times (n+1)}\left(x\right)$.

The formula (5) is a reformulation of [4] (p. 40, Exercise 5). See also the paper [5] and those papers collected at the wordpress.com site (https://qifeng618.wordpress.com/2020/03/22/some-papers-authored-by-dr-prof-feng-qi-and-utilizing-a-general-derivative-formula-for-the-ratio-of-two-differentiable-functions, accessed on 10 May 2023).

Lemma 2.

(Monotonicity rule for the ratio of two Maclaurin power series [6]). Let ${\alpha }_k$ and ${\beta }_k$ for $k\in \left\{0\right\}\cup \mathbb{N}$ be real numbers and the power series

$$U\left(x\right)=\sum _{k=0}^{\infty }{\alpha }_k{x}^k\mathit{and}V\left(x\right)=\sum _{k=0}^{\infty }{\beta }_k{x}^k$$

be convergent on $(-R,R)$ for some $R>0$. If ${\beta }_k>0$ and the ratio $\frac{{\alpha }_k}{{\beta }_k}$ is (strictly) increasing for $k\ge 0$, then the function $\frac{U\left(x\right)}{V\left(x\right)}$ is also (strictly) increasing on $(0,R)$.

Lemma 3.

(Monotonicity rule for the ratio of two functions [7] (Theorem 1.25)). For $a,b\in \mathbb{R}$ with $a<b$, let $\lambda \left(x\right)$ and $\mu \left(x\right)$ be continuous on $[a,b]$, differentiable on $(a,b)$, and ${\mu }^{\prime }\left(x\right)\ne 0$ on $(a,b)$. If the ratio $\frac{{\lambda }^{\prime }\left(x\right)}{{\mu }^{\prime }\left(x\right)}$ is increasing on $(a,b)$, then both $\frac{\lambda \left(x\right)-\lambda \left(a\right)}{\mu \left(x\right)-\mu \left(a\right)}$ and $\frac{\lambda \left(x\right)-\lambda \left(b\right)}{\mu \left(x\right)-\mu \left(b\right)}$ are increasing in $x\in (a,b)$.

3. Maclaurin Power Series Expansion

In this section, we solve the first problem posed in the first section of this paper.

Theorem 1.

Let the real numbers

$$e_m={(-1)}^m\frac{\left(2m\right)!}{(2m+3)!},m\ge 0$$

and the determinants

$$E_{2n}=\left|\begin{array}{cc}{A}_{2n-\mathrm{1,1}}& B_{2n-\mathrm{1,2}n-1}\\ \frac{e_{n+1}}{e_0}& C_{\mathrm{1,2}n-1}\end{array}\right|,n\ge 1,$$

where the matrices $A_{2n-1,1}$, $B_{2n-1,2n-1}$, and $C_{1,2n-1}$ for $n\ge 1$ are defined by

$$\begin{array}{cc}\hfill A_{2n-1,1}& =\left(\begin{array}{c}\frac{e_1}{e_0}\\ 0\\ \frac{e_2}{e_0}\\ ⋮\\ \frac{e_{n-1}}{e_0}\\ 0\\ \frac{e_n}{e_0}\end{array}\right)={\left(\begin{array}{c}{a}_{i,j}\end{array}\right)}_{\begin{array}{c}1\le i\le 2n-1\\ j=1\end{array}},\hfill \\ \hfill a_{i,1}& =\{\begin{array}{cc}\frac{e_k}{e_0},\hfill & 1\le i=2k-1\le 2n-1;\hfill \\ 0,\hfill & 2\le i=2k\le 2n-2,\hfill \end{array}\hfill \\ \hfill B_{2n-1,2n-1}& =\left|\begin{array}{cccccc}\left(\genfrac{}{}{0pt}{}{1}{1}\right)& 0& \cdots & 0& 0& 0\\ 0& \left(\genfrac{}{}{0pt}{}{2}{2}\right)& \cdots & 0& 0& 0\\ \left(\genfrac{}{}{0pt}{}{3}{1}\right)\frac{e_1}{e_0}& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ \left(\genfrac{}{}{0pt}{}{2n-3}{1}\right)\frac{e_{n-2}}{e_0}& 0& \cdots & \left(\genfrac{}{}{0pt}{}{2n-3}{2n-3}\right)& 0& 0\\ 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2}\right)\frac{e_{n-2}}{e_0}& \cdots & 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2n-2}\right)& 0\\ \left(\genfrac{}{}{0pt}{}{2n-1}{1}\right)\frac{e_{n-1}}{e_0}& 0& \cdots & \left(\genfrac{}{}{0pt}{}{2n-1}{2n-3}\right)\frac{e_1}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{2n-1}{2n-1}\right)\end{array}\right|\hfill \\ \hfill & ={\left(\begin{array}{c}{b}_{i,j}\end{array}\right)}_{1\le i,j\le 2n-1},\hfill \\ \hfill b_{i,j}& =\{\begin{array}{cc}0,\hfill & 1\le i<j\le 2n-1;\hfill \\ \left(\genfrac{}{}{0pt}{}{i}{j}\right)\frac{e_k}{e_0},\hfill & 0\le i-j=2k\le 2n-2;\hfill \\ 0,\hfill & 1\le i-j=2k-1\le 2n-1,\hfill \end{array}\hfill \\ \hfill C_{1,2n-1}& =\left(\begin{array}{ccccccc}\left(\genfrac{}{}{0pt}{}{2n+1}{1}\right)\frac{e_n}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{2n+1}{3}\right)\frac{e_{n-1}}{e_0}& \cdots & \left(\genfrac{}{}{0pt}{}{2n+1}{2n-3}\right)\frac{e_2}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{2n+1}{2n-1}\right)\frac{e_1}{e_0}\end{array}\right)\hfill \\ \hfill & ={\left(\begin{array}{c}{c}_{i,j}\end{array}\right)}_{\begin{array}{c}i=1\\ 1\le j\le 2n-1\end{array}},\hfill \\ \hfill c_{1,j}& =\{\begin{array}{cc}\left(\genfrac{}{}{0pt}{}{2n+1}{2k-1}\right)\frac{e_{n-k+1}}{e_0},\hfill & 1\le j=2k-1\le 2n-3;\hfill \\ 0,\hfill & 2\le j=2k\le 2n-2.\hfill \end{array}\hfill \end{array}$$

Then the function $Q\left(x\right)$ defined by (3) can be expanded into the Maclaurin power series expansion

$$\begin{array}{cc}\hfill Q\left(x\right)& =-\frac{1}{20}{x}^2-\sum _{n=1}^{\infty }{E}_{2n}\frac{x^{2n+2}}{(2n+2)!}\hfill \\ \hfill & =-\frac{1}{20}{x}^2-\frac{1}{\mathrm{16,800}}{x}^4+\frac{1}{\mathrm{756,000}}{x}^6+\frac{89}{\mathrm{3,104,640,000}}{x}^8+\cdots \hfill \end{array}$$

(6)

for $x\in \mathbb{R}$.

Proof. 

The first derivative of $Q\left(x\right)$ is

$$Q^{\prime }\left(x\right)=\frac{2x-3sinx+xcosx}{x(sinx-x)}=\frac{\frac{3sin x-2x-xcos x}{x^4}}{\frac{x-sin x}{x^3}},$$

where

$$\frac{x-sinx}{x^3}=\sum _{k=0}^{\infty }{e}_k\frac{x^{2k}}{\left(2k\right)!},0<\left|x\right|<\infty$$

and

$$\frac{3sinx-2x-xcosx}{x^4}=\sum _{k=0}^{\infty }{e}_{k+1}\frac{x^{2k+1}}{(2k+1)!},0<\left|x\right|<\infty .$$

Hence, the functions

$$v\left(x\right)=\{\begin{array}{cc}\hfill \frac{x-sinx}{x^3},& x\ne 0\\ \hfill e_0,& x=0\end{array}\mathrm{and}u\left(x\right)=\{\begin{array}{cc}\hfill \frac{3sinx-2x-xcosx}{x^4},& x\ne 0\\ \hfill 0,& x=0\end{array}$$

satisfy

$$v^{\left(k\right)}\left(0\right)=\{\begin{array}{cc}{e}_m,\hfill & k=2m\hfill \\ 0,\hfill & k=2m+1\hfill \end{array}\mathrm{and}{u}^{\left(k\right)}\left(0\right)=\{\begin{array}{cc}\hfill 0,& k=2m\\ \hfill e_{m+1},& k=2m+1\end{array}$$

for $k,m\ge 0$. Accordingly, making use of Formula (5) results in

$$\begin{array}{cc}\hfill Q^{(2n+2)}\left(0\right)& =\underset{x\to 0}{lim}{\left[\frac{u\left(x\right)}{v\left(x\right)}\right]}^{(2n+1)}\hfill \\ \hfill & =\frac{{(-1)}^{2n+1}}{v^{2n+2}\left(0\right)}\left|\begin{array}{ccccc}u\left(0\right)& v\left(0\right)& 0& \cdots & 0\\ u^{\prime }\left(0\right)& v^{\prime }\left(0\right)& \left(\genfrac{}{}{0pt}{}{1}{1}\right)v\left(0\right)& \cdots & 0\\ u^{″}\left(0\right)& v^{″}\left(0\right)& \left(\genfrac{}{}{0pt}{}{2}{1}\right)v^{\prime }\left(0\right)& \cdots & 0\\ u^{\left(3\right)}\left(0\right)& v^{\left(3\right)}\left(0\right)& \left(\genfrac{}{}{0pt}{}{3}{1}\right)v^{″}\left(0\right)& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ u^{(2n-2)}\left(0\right)& v^{(2n-2)}\left(0\right)& \left(\genfrac{}{}{0pt}{}{2n-2}{1}\right)v^{(2n-3)}\left(0\right)& \cdots & 0\\ u^{(2n-1)}\left(0\right)& v^{(2n-1)}\left(0\right)& \left(\genfrac{}{}{0pt}{}{2n-1}{1}\right)v^{(2n-2)}\left(0\right)& \cdots & 0\\ u^{\left(2n\right)}\left(0\right)& v^{\left(2n\right)}\left(0\right)& \left(\genfrac{}{}{0pt}{}{2n}{1}\right)v^{(2n-1)}\left(0\right)& \cdots & \left(\genfrac{}{}{0pt}{}{2n}{2n}\right)v\left(0\right)\\ u^{(2n+1)}\left(0\right)& v^{(2n+1)}\left(0\right)& \left(\genfrac{}{}{0pt}{}{2n+1}{1}\right)v^{\left(2n\right)}\left(0\right)& \cdots & \left(\genfrac{}{}{0pt}{}{2n+1}{2n}\right)v^{\prime }\left(0\right)\end{array}\right|\hfill \\ \hfill & =\frac{-1}{e_0^{2n+2}}\left|\begin{array}{cccccc}0& e_0& 0& 0& \cdots & 0\\ e_1& 0& \left(\genfrac{}{}{0pt}{}{1}{1}\right)e_0& 0& \cdots & 0\\ 0& e_1& 0& \left(\genfrac{}{}{0pt}{}{2}{2}\right)e_0& \cdots & 0\\ e_2& 0& \left(\genfrac{}{}{0pt}{}{3}{1}\right)e_1& 0& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& e_{n-1}& 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2}\right)e_{n-2}& \cdots & 0\\ e_n& 0& \left(\genfrac{}{}{0pt}{}{2n-1}{1}\right)e_{n-1}& 0& \cdots & 0\\ 0& e_n& 0& \left(\genfrac{}{}{0pt}{}{2n}{2}\right)e_n& \cdots & \left(\genfrac{}{}{0pt}{}{2n}{2n}\right)e_0\\ e_{n+1}& 0& \left(\genfrac{}{}{0pt}{}{2n+1}{1}\right)e_n& 0& \cdots & 0\end{array}\right|\hfill \\ \hfill & =\frac{1}{e_0^{2n+1}}\left|\begin{array}{ccccccc}{e}_1& \left(\genfrac{}{}{0pt}{}{1}{1}\right)e_0& 0& \cdots & 0& 0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2}{2}\right)e_0& \cdots & 0& 0& 0\\ e_2& \left(\genfrac{}{}{0pt}{}{3}{1}\right)e_1& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2}\right)e_{n-2}& \cdots & \left(\genfrac{}{}{0pt}{}{2n-2}{2n-2}\right)e_0& 0& 0\\ e_n& \left(\genfrac{}{}{0pt}{}{2n-1}{1}\right)e_{n-1}& 0& \cdots & 0& \left(\genfrac{}{}{0pt}{}{2n-1}{2n-1}\right)e_0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2n}{2}\right)e_{n-1}& \cdots & \left(\genfrac{}{}{0pt}{}{2n}{2n-2}\right)e_1& 0& \left(\genfrac{}{}{0pt}{}{2n}{2n}\right)e_0\\ e_{n+1}& \left(\genfrac{}{}{0pt}{}{2n+1}{1}\right)e_n& 0& \cdots & 0& \left(\genfrac{}{}{0pt}{}{2n+1}{2n-1}\right)e_1& 0\end{array}\right|\hfill \\ \hfill & =\left|\begin{array}{ccccccc}\frac{e_1}{e_0}& \left(\genfrac{}{}{0pt}{}{1}{1}\right)& 0& \cdots & 0& 0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2}{2}\right)& \cdots & 0& 0& 0\\ \frac{e_2}{e_0}& \left(\genfrac{}{}{0pt}{}{3}{1}\right)\frac{e_1}{e_0}& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2}\right)\frac{e_{n-2}}{e_0}& \cdots & \left(\genfrac{}{}{0pt}{}{2n-2}{2n-2}\right)& 0& 0\\ \frac{e_n}{e_0}& \left(\genfrac{}{}{0pt}{}{2n-1}{1}\right)\frac{e_{n-1}}{e_0}& 0& \cdots & 0& \left(\genfrac{}{}{0pt}{}{2n-1}{2n-1}\right)& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2n}{2}\right)\frac{e_{n-1}}{e_0}& \cdots & \left(\genfrac{}{}{0pt}{}{2n}{2n-2}\right)\frac{e_1}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{2n}{2n}\right)\\ \frac{e_{n+1}}{e_0}& \left(\genfrac{}{}{0pt}{}{2n+1}{1}\right)\frac{e_n}{e_0}& 0& \cdots & 0& \left(\genfrac{}{}{0pt}{}{2n+1}{2n-1}\right)\frac{e_1}{e_0}& 0\end{array}\right|\hfill \\ \hfill & =-\left|\begin{array}{ccccccc}\frac{e_1}{e_0}& \left(\genfrac{}{}{0pt}{}{1}{1}\right)& 0& \cdots & 0& 0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2}{2}\right)& \cdots & 0& 0& 0\\ \frac{e_2}{e_0}& \left(\genfrac{}{}{0pt}{}{3}{1}\right)\frac{e_1}{e_0}& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ \frac{e_{n-1}}{e_0}& \left(\genfrac{}{}{0pt}{}{2n-3}{1}\right)\frac{e_{n-2}}{e_0}& 0& \cdots & \left(\genfrac{}{}{0pt}{}{2n-3}{2n-3}\right)& 0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2}\right)\frac{e_{n-2}}{e_0}& \cdots & 0& \left(\genfrac{}{}{0pt}{}{2n-2}{2n-2}\right)& 0\\ \frac{e_n}{e_0}& \left(\genfrac{}{}{0pt}{}{2n-1}{1}\right)\frac{e_{n-1}}{e_0}& 0& \cdots & \left(\genfrac{}{}{0pt}{}{2n-1}{2n-3}\right)\frac{e_1}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{2n-1}{2n-1}\right)\\ \frac{e_{n+1}}{e_0}& \left(\genfrac{}{}{0pt}{}{2n+1}{1}\right)\frac{e_n}{e_0}& 0& \cdots & \left(\genfrac{}{}{0pt}{}{2n+1}{2n-3}\right)\frac{e_2}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{2n+1}{2n-1}\right)\frac{e_1}{e_0}\end{array}\right|\hfill \\ \hfill & =-\left|\begin{array}{cc}{A}_{2n-1,1}& B_{2n-1,2n-1}\\ \frac{e_{n+1}}{e_0}& C_{1,2n-1}\end{array}\right|\hfill \\ \hfill & =-E_{2n}\hfill \end{array}$$

for $n\ge 1$ and, due to the evenness of $Q\left(x\right)$ on $\mathbb{R}$, $Q^{(2n+1)}\left(0\right)=0$ for $n\ge 0$. Consequently, we obtain

$$\begin{array}{c}Q\left(x\right)=\sum _{n=0}^{\infty }\frac{Q^{\left(n\right)}\left(0\right)}{n!}{x}^n=\sum _{n=1}^{\infty }\frac{Q^{\left(2n\right)}\left(0\right)}{\left(2n\right)!}{x}^{2n}\hfill \\ \hfill =\frac{Q^{″}\left(0\right)}{2!}{x}^2+\sum _{n=1}^{\infty }\frac{Q^{(2n+2)}\left(0\right)}{(2n+2)!}{x}^{2n+2}=-\frac{1}{20}{x}^2-\sum _{n=1}^{\infty }\frac{E_{2n}\left(0\right)}{(2n+2)!}{x}^{2n+2}.\end{array}$$

The required proof is thus complete. □

Remark 1.

For $n=3$, the determinant $E_6$ is

$$\begin{array}{cc}\hfill E_6& =\left|\begin{array}{cccccc}\frac{e_1}{e_0}& \left(\genfrac{}{}{0pt}{}{1}{1}\right)& 0& 0& 0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{2}{2}\right)& 0& 0& 0\\ \frac{e_2}{e_0}& \left(\genfrac{}{}{0pt}{}{3}{1}\right)\frac{e_1}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{3}{3}\right)& 0& 0\\ 0& 0& \left(\genfrac{}{}{0pt}{}{4}{2}\right)\frac{e_1}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{4}{4}\right)& 0\\ \frac{e_3}{e_0}& \left(\genfrac{}{}{0pt}{}{5}{1}\right)\frac{e_2}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{5}{3}\right)\frac{e_1}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{5}{5}\right)\\ \frac{e_4}{e_0}& \left(\genfrac{}{}{0pt}{}{7}{1}\right)\frac{e_3}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{7}{3}\right)\frac{e_2}{e_0}& 0& \left(\genfrac{}{}{0pt}{}{7}{5}\right)\frac{e_1}{e_0}\end{array}\right|\hfill \\ \hfill & =\left|\begin{array}{cccccc}-\frac{1}{10}& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ \frac{1}{35}& -\frac{3}{10}& 0& 1& 0& 0\\ 0& 0& -\frac{3}{5}& 0& 1& 0\\ -\frac{1}{84}& \frac{1}{7}& 0& -1& 0& 1\\ \frac{1}{165}& -\frac{1}{12}& 0& 1& 0& -\frac{21}{10}\end{array}\right|\hfill \\ \hfill & =-\frac{89}{\mathrm{77,000}}\hfill \end{array}$$

and $-\frac{E_6}{8!}=\frac{89}{3,104,640,000}$. This coincides with the coefficient of the term $x^8$ in the Maclaurin power series expansion (6).

4. Decreasing Property

In this section, we solve the second problem posed in the first section of this paper.

Theorem 2.

The function $R\left(x\right)$ defined by (4) decreasingly maps $[0,\pi ]$ onto $\left[0,\frac{3}{10}\right]$.

First proof. 

Directly differentiating gives

$$\frac{Q^{\prime }\left(x\right)}{{\left(ln\frac{sinx}{x}\right)}^{\prime }}=\frac{(2x-3sinx+xcosx)sinx}{(sinx-x)(xcosx-sinx)}$$

and

$${\left[\frac{Q^{\prime }\left(x\right)}{{\left(ln\frac{sinx}{x}\right)}^{\prime }}\right]}^{\prime }=-\frac{h\left(x\right)}{{(x-sinx)}^2{(xcosx-sinx)}^2}.$$

(7)

where

$$\begin{array}{cc}\hfill h\left(x\right)& =2x^3+x^3{cos}^{3}x-4x^2{sin}^{3}x-2x^{2}sinxcosx-{sin}^{3}x+2x{sin}^{2}x\hfill \\ \hfill & -2cosx{sin}^{3}x+7xcosx{sin}^{2}x-7x^{2}sinx{cos}^{2}x.\hfill \end{array}$$

In [1] (p. 43), we find

$$\begin{array}{cc}\hfill {sin}^{2}x& =\sum _{k=1}^{\infty }{(-1)}^{k+1}\frac{2^{2k-1}}{\left(2k\right)!}{x}^{2k},\left|x\right|<\infty ,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill {sin}^{3}x& =\frac{1}{4}\sum _{k=1}^{\infty }{(-1)}^{k+1}\frac{3^{2k+1}-3}{(2k+1)!}{x}^{2k+1},\left|x\right|<\infty ,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill {cos}^{3}x& =\frac{1}{4}\sum _{k=0}^{\infty }{(-1)}^k\frac{3^{2k}+3}{\left(2k\right)!}{x}^{2k},\left|x\right|<\infty .\hfill \end{array}$$

(10)

Differentiating results in

$$\begin{array}{cc}\hfill sinxcosx& =\frac{1}{2}\sum _{k=0}^{\infty }{(-1)}^k\frac{2^{2k+1}}{(2k+1)!}{x}^{2k+1},\left|x\right|<\infty ,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill cosx{sin}^{2}x& =\frac{1}{12}\sum _{k=1}^{\infty }{(-1)}^{k+1}\frac{3^{2k+1}-3}{\left(2k\right)!}{x}^{2k},\left|x\right|<\infty ,\hfill \end{array}$$

(12)

and

$$sinx{cos}^{2}x=\frac{1}{12}\sum _{k=0}^{\infty }{(-1)}^k\frac{3^{2k+2}+3}{(2k+1)!}{x}^{2k+1},\left|x\right|<\infty .$$

(13)

Theorem 2.1 in the paper [8] reads that

$${\left(\frac{sinz}{z}\right)}^{\ell }=1+\sum _{j=1}^{\infty }{(-1)}^j\frac{T(\ell +2j,\ell )}{\left(\genfrac{}{}{0pt}{}{\ell +2j}{\ell }\right)}\frac{{\left(2z\right)}^{2j}}{\left(2j\right)!}$$

(14)

for $\ell \ge 0$ and $z\in \mathbb{C}$, where

$$T(n,\ell )=\frac{1}{\ell !}\sum _{m=0}^{\ell }{(-1)}^m\left(\genfrac{}{}{0pt}{}{\ell }{m}\right){\left(\frac{\ell }{2}-m\right)}^n.$$

(15)

Taking $n=4+2j$ and $\ell =4$ in (15) gives

$$T(4+2j,4)=\frac{1}{4!}\sum _{m=0}^4{(-1)}^m\left(\genfrac{}{}{0pt}{}{4}{m}\right){(2-m)}^{4+2j}=\frac{4^{j+1}-1}{3}.$$

Setting $\ell =4$ in (14) leads to

$$\begin{array}{cc}\hfill {\left(\frac{sinz}{z}\right)}^4& =1+\sum _{j=1}^{\infty }{(-1)}^j\frac{T(4+2j,4)}{\left(\genfrac{}{}{0pt}{}{4+2j}{4}\right)}\frac{{\left(2z\right)}^{2j}}{\left(2j\right)!}=\sum _{j=0}^{\infty }{(-1)}^j\frac{2^{2j+3}(4^{j+1}-1)}{(2j+4)!}{z}^{2j},\hfill \end{array}$$

which can be rearranged as

$${sin}^{4}z=\sum _{j=0}^{\infty }{(-1)}^j\frac{2^{2j+3}(2^{2j+2}-1)}{(2j+4)!}{z}^{2j+4},\left|z\right|<\infty .$$

Differentiating gives

$$cosz{sin}^{3}z=\sum _{j=0}^{\infty }{(-1)}^j\frac{2^{2j+1}(2^{2j+2}-1)}{(2j+3)!}{z}^{2j+3},\left|z\right|<\infty .$$

(16)

Making use of the power series expansions (8)–(13) and (16), we can expand the function $h\left(x\right)$ into

$$\begin{array}{cc}\hfill h\left(x\right)& =2x^3+\frac{1}{4}\sum _{k=0}^{\infty }{(-1)}^k\frac{3^{2k}+3}{\left(2k\right)!}{x}^{2k+3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{3^{2k+1}-3}{(2k+1)!}{x}^{2k+3}\hfill \\ \hfill & -\sum _{k=0}^{\infty }{(-1)}^k\frac{2^{2k+1}}{(2k+1)!}{x}^{2k+3}-\frac{1}{4}\sum _{k=0}^{\infty }{(-1)}^k\frac{3^{2k+3}-3}{(2k+3)!}{x}^{2k+3}\hfill \\ \hfill & +2\sum _{k=0}^{\infty }{(-1)}^k\frac{2^{2k+1}}{(2k+2)!}{x}^{2k+3}-2\sum _{k=0}^{\infty }{(-1)}^k\frac{2^{2k+1}(2^{2k+2}-1)}{(2k+3)!}{x}^{2k+3}\hfill \\ \hfill & +\frac{7}{12}\sum _{k=0}^{\infty }{(-1)}^k\frac{3^{2k+3}-3}{(2k+2)!}{x}^{2k+3}-\frac{7}{12}\sum _{k=0}^{\infty }{(-1)}^k\frac{3^{2k+2}+3}{(2k+1)!}{x}^{2k+3}\hfill \\ \hfill & =\frac{1}{2}\sum _{k=5}^{\infty }{(-1)}^k\frac{\left[\begin{array}{c}\left(4k^3-6k^2+29k+57\right)3^{2k}\\ -\left(2k^2+3k-1\right)2^{2k+3}-2^{4k+5}\\ +12k^3-2k^2-69k-57\end{array}\right]}{(2k+3)!}{x}^{2k+3}\hfill \\ \hfill & =\frac{x^{13}}{2}\sum _{k=0}^{\infty }{(-1)}^{k+1}\frac{\left[\begin{array}{c}\left(4k^3+54k^2+269k+552\right)3^{2k+10}\\ -\left(2k^2+23k+64\right)2^{2k+13}-2^{4k+25}\\ +12k^3+178k^2+811k+1048\end{array}\right]}{(2k+13)!}{x}^{2k}\hfill \\ \hfill & \triangleq \frac{x^{13}}{2}\sum _{k=0}^{\infty }{(-1)}^k{\mathsf{\Theta }}_k{x}^{2k}.\hfill \end{array}$$

We now consider the functions

$${\mathsf{\Phi }}_k\left(x\right)={(-1)}^k{\mathsf{\Theta }}_k{x}^{2k}+{(-1)}^{k+1}{\mathsf{\Theta }}_{k+1}{x}^{2k+2}={(-1)}^k{x}^{2k}\left({\mathsf{\Theta }}_k-{\mathsf{\Theta }}_{k+1}{x}^2\right)$$

for $k\ge 0$ and $x\in [0,\pi ]$. Then the function $h\left(x\right)$ can be expressed as

$$h\left(x\right)=\frac{x^{13}}{2}\sum _{k=0}^{\infty }{\mathsf{\Phi }}_{2k}\left(x\right)=\frac{x^{13}}{2}\sum _{k=0}^{\infty }{\mathsf{\Theta }}_{2k+1}\left(\frac{{\mathsf{\Theta }}_{2k}}{{\mathsf{\Theta }}_{2k+1}}-x^2\right)x^{4k},x\in [0,\pi ].$$

By induction, we can verify that

$$\begin{array}{c}{2}^{4k+25}-\left(4k^3+54k^2+269k+552\right)3^{2k+10}\hfill \\ \hfill =32\times 3^{2k+10}\left[{\left(\frac{4}{3}\right)}^{2k+10}-\frac{4k^3+54k^2+269k+552}{32}\right]>0\end{array}$$

and

$$\begin{array}{c}\left(2k^2+23k+64\right)2^{2k+13}-(12k^3+178k^2+811k+1048)\hfill \\ \hfill =\left(2k^2+23k+64\right)\left(2^{2k+13}-\frac{12k^3+178k^2+811k+1048}{2k^2+23k+64}\right)>0\end{array}$$

for $k\ge 0$. Therefore, we acquire that ${\mathsf{\Theta }}_k>0$ for $k\ge 0$.

The partial sum

$$\begin{array}{c}P\left(x\right)=\sum _{k=0}^2{\mathsf{\Theta }}_{2k+1}\left(\frac{{\mathsf{\Theta }}_{2k}}{{\mathsf{\Theta }}_{2k+1}}-x^2\right)x^{4k}\hfill \\ \hfill =\frac{1}{4200}-\frac{x^2}{\mathrm{18,000}}+\frac{361x^4}{\mathrm{58,212,000}}-\frac{19x^6}{\mathrm{43,243,200}}+\frac{\mathrm{338,857}{x}^8}{\mathrm{15,256,200,960,000}}-\frac{\mathrm{1,331,567}{x}^{10}}{\mathrm{1,556,132,497,920,000}}\end{array}$$

satisfies

$$\begin{array}{cc}\hfill P^{\prime }\left(x\right)& =-\frac{\mathrm{1,331,567}{x}^4}{\mathrm{311,226,499,584,000}}+\frac{\mathrm{338,857}{x}^3}{\mathrm{3,814,050,240,000}}-\frac{19x^2}{\mathrm{14,414,400}}+\frac{361x}{\mathrm{29,106,000}}-\frac{1}{\mathrm{18,000}}\hfill \\ \hfill & \to -\frac{1}{\mathrm{18,000}}+\frac{361\pi }{\mathrm{29,106,000}}-\frac{19{\pi }^2}{\mathrm{14,414,400}}+\frac{33\mathrm{8,857}{\pi }^3}{\mathrm{3,814,050,240,00}0}\hfill \\ \hfill & -\frac{1,\mathrm{331,567}{\pi }^4}{3\mathrm{11,226,499,584,000}},x\to \pi \hfill \\ \hfill & <0,\hfill \\ \hfill P^{″}\left(x\right)& =-\frac{\mathrm{1,331,567}{x}^3}{\mathrm{77,806,624,896,0}00}+\frac{338,857x^2}{\mathrm{1,271,350,080,000}}-\frac{19x}{\mathrm{7,207,200}}+\frac{361}{\mathrm{29,106,000}}\hfill \\ \hfill & \to \frac{361}{\mathrm{29,106,000}}-\frac{19\pi }{\mathrm{7,207,200}}+\frac{\mathrm{338,857}{\pi }^2}{1,\mathrm{271,350,080},000}-\frac{\mathrm{1,331,567}{\pi }^3}{7\mathrm{7,806,624,896,00}0},x\to \pi \hfill \\ \hfill & >0,\hfill \\ \hfill P^{‴}\left(x\right)& =-\frac{1,\mathrm{331,567}{x}^2}{\mathrm{25,935,541,632,000}}+\frac{\mathrm{338,857}x}{63\mathrm{5,675,040,000}}-\frac{19}{\mathrm{7,207,20}0}\hfill \\ \hfill & \to -\frac{19}{\mathrm{7,207,20}0}+\frac{3\mathrm{38,85}7\pi }{\mathrm{635,675,040,0}00}-\frac{\mathrm{1,331,56}7{\pi }^2}{\mathrm{25,935,541,632,00}0},x\to \pi \hfill \\ \hfill & <0,\hfill \\ \hfill P^{\left(4\right)}\left(x\right)& =\frac{\mathrm{338,857}}{6\mathrm{35,675,040,0}00}-\frac{\mathrm{1,331,56}7x}{\mathrm{12,967,770,816,000}}\hfill \\ \hfill & \to \frac{3\mathrm{38,857}}{\mathrm{635,675,040,000}}-\frac{\mathrm{1,331,567}\pi }{\mathrm{12,967,770,816,000}},x\to \pi \hfill \\ \hfill & >0.\hfill \end{array}$$

Accordingly, the function $P\left(x\right)$ is decreasing on $[0,\pi ]$. From

$$\begin{array}{cc}\hfill P\left(\pi \right)& =\frac{1}{4200}-\frac{\pi }{\mathrm{18,000}}+\frac{361{\pi }^2}{\mathrm{58,212,000}}-\frac{19{\pi }^3}{\mathrm{43,243,200}}\hfill \\ \hfill & +\frac{\mathrm{338,857}{\pi }^4}{\mathrm{15,256,200,960,00}0}-\frac{\mathrm{1,331,567}{\pi }^5}{\mathrm{1,556,132,497,920,000}}\hfill \\ \hfill & =0.000113\cdots ,\hfill \end{array}$$

it follows that $P\left(x\right)>0$ on $[0,\pi ]$.

The inequality

$$\frac{{\mathsf{\Theta }}_{2k}}{{\mathsf{\Theta }}_{2k+1}}>10>{\pi }^2,k\ge 1$$

(17)

is equivalent to

$${\mathsf{\Theta }}_{2k}>10{\mathsf{\Theta }}_{2k+1},k\ge 1.$$

In other words, this inequality can be formulated as

$$\begin{array}{cc}\hfill & \left(8k^2+58k+25\right)2^{8k+25}-(256k^5+3584k^4+18,752k^3+46,420k^2+53,469k\hfill \\ \hfill & +18,405)3^{4k+10}+\left(32k^4+416k^3+1930k^2+3731k+2470\right)2^{4k+14}\hfill \\ \hfill & -768k^5-11,264k^4-63,872k^3-172,940k^2-219,064k-99,795\hfill \\ \hfill & =32\left(8k^2+58k+25\right)3^{4k+10}[{\left(\frac{4}{3}\right)}^{4k+10}\hfill \\ \hfill & -\frac{256k^5+3584k^4+\mathrm{18,752}{k}^3+\mathrm{46,420}{k}^2+\mathrm{53,469}k+\mathrm{18,405}}{32(8k^2+58k+25)}]\hfill \\ \hfill & +\left(32k^4+416k^3+1930k^2+3731k+2470\right)\hfill \\ \hfill & \times \left(4^{2k+7}-\frac{768k^5+11,264k^4+63,872k^3+172,940k^2+219,064k+99,795}{32k^4+416k^3+1930k^2+3731k+2470}\right)\hfill \\ \hfill & >0\hfill \end{array}$$

for $k\ge 1$. By induction, we reveal that

$${\left(\frac{4}{3}\right)}^{4k+10}-\frac{256k^5+3584k^4+18,752k^3+46,420k^2+53,469k+18,405}{32(8k^2+58k+25)}>0$$

and

$$4^{2k+7}-\frac{768k^5+11,264k^4+63,872k^3+172,940k^2+219,064k+99,795}{32k^4+416k^3+1930k^2+3731k+2470}>0$$

for $k\ge 1$. Consequently, the inequality (17) holds for $k\ge 1$.

Basing on the above-derived results, we conclude that

$$h\left(x\right)=\frac{x^{13}}{2}\left[P\left(x\right)+\sum _{k=3}^{\infty }{\mathsf{\Theta }}_{2k+1}\left(\frac{{\mathsf{\Theta }}_{2k}}{{\mathsf{\Theta }}_{2k+1}}-x^2\right)x^{4k}\right]>0,x\in [0,\pi ].$$

From (7), it follows that the derivative ratio, $\frac{Q^{\prime }\left(x\right)}{{(ln\frac{sinx}{x})}^{\prime }}$ is decreasing on $(0,\pi )$. Using Lemma 3, we derive that the function $R\left(x\right)$ defined by (4) is decreasing on $[0,\pi ]$. The proof of Theorem 2 is complete. □

Second proof. 

We just state the outline and sketch of this proof because the method and approach are similar to the first proof in this paper.

Differentiating gives

$$\begin{array}{cc}\hfill h^{\prime }\left(x\right)& =\frac{1}{2}(15xcos\frac{5x}{2}-3x^{3}cos\frac{5x}{2}+15xcos\frac{3x}{2}-3x^{3}cos\frac{3x}{2}\hfill \\ \hfill & -30xcos\frac{x}{2}-6x^{3}cos\frac{x}{2}+24x^{2}sin\frac{x}{2}+4sin\frac{x}{2}+8sin\frac{3x}{2}\hfill \\ \hfill & +14x^{2}sin\frac{3x}{2}+6x^{2}sin\frac{5x}{2}-4sin\frac{7x}{2})sin\frac{x}{2}\hfill \\ \hfill & =\frac{x^{11}sin\frac{x}{2}}{150\times 2^{2k}}\sum _{k=5}^{\infty }{(-1)}^{k+1}\frac{\left[\begin{array}{c}150\times 7^{2k+1}\\ -25\left(32k^3-112k^2+26k+81\right)3^{2k}\\ -9(2k+1)\left(16k^2-48k+125\right)5^{2k}\\ -300\left(48k^3-48k^2-51k-7\right)\end{array}\right]}{(2k+1)!}{x}^{2(k-5)}\hfill \\ \hfill & \triangleq \frac{x^{11}sin\frac{x}{2}}{2^{2k+1}}\sum _{k=0}^{\infty }{(-1)}^k{W}_k{x}^{2k}\hfill \\ \hfill & =\frac{x^{11}sin\frac{x}{2}}{2^{2k+1}}\sum _{k=0}^{\infty }{W}_{2k+1}\left(\frac{W_{2k}}{W_{2k+1}}-x^2\right)x^{4k}\hfill \end{array}$$

for $x\in [0,\pi ]$ where

$$W_k=\frac{\left[\begin{array}{c}2\times 7^{2k+11}-3(2k+11)\left(16k^2+112k+285\right)5^{2k+8}\\ -\left(32k^3+368k^2+1306k+1411\right)3^{2k+9}\\ -4\left(48k^3+672k^2+3069k+4538\right)\end{array}\right]}{(2k+11)!},k\ge 0.$$

By induction, we can verify that

$$\begin{array}{c}{3}^{2k+9}-\frac{4\left(48k^3+672k^2+3069k+4538\right)}{32k^3+368k^2+1306k+1411}>0,k\ge 0,\\ \begin{array}{cc}& 7^{2k+11}-2\left(32k^3+368k^2+1306k+1411\right)3^{2k+9}\hfill \\ & =49\times 3^{2k+9}\left[{\left(\frac{7}{3}\right)}^{2k+9}-\frac{2\left(32k^3+368k^2+1306k+1411\right)}{49}\right]\hfill \\ & >0,k\ge 0,\hfill \end{array}\end{array}$$

and

$$\begin{array}{cc}\hfill & 7^{2k+11}-3(2k+11)\left(16k^2+112k+285\right)5^{2k+8}\hfill \\ \hfill & =343\times 5^{2k+8}\left[{\left(\frac{7}{5}\right)}^{2k+8}-\frac{3(2k+11)\left(16k^2+112k+285\right)}{343}\right]\hfill \\ \hfill & >0,k\ge 4.\hfill \end{array}$$

Moreover, it is easy to calculate

$$W_0=\frac{3328}{7},W_1=\frac{9088}{21},W_2=\frac{1,479,584}{8085},W_3=\frac{5,030,768}{105,105}.$$

Therefore, the sequence $W_k$ is positive for $k\ge 0$.

The partial sum

$$\sum _{k=0}^9{W}_{2k+1}\left(\frac{W_{2k}}{W_{2k+1}}-x^2\right)x^{4k}>0,x\in [0,\pi ].$$

For $k\ge 3$, the inequality

$$\frac{W_{2k}}{W_{2k+1}}>10>{\pi }^2$$

is valid. Hence, the derivative $h^{\prime }\left(x\right)$ is positive on $[0,\pi ]$. Then the function $h\left(x\right)$ is increasing on $[0,\pi ]$. Since $h\left(0\right)=0$, the function $h\left(x\right)$ is positive on $(0,\pi ]$. From the derivative in (7), it follows that the derivative ratio $\frac{Q^{\prime }\left(x\right)}{{(ln\frac{sinx}{x})}^{\prime }}$ is decreasing on $(0,\pi )$. Employing Lemma 3, we deduce that the function $R\left(x\right)$ defined by (4) is decreasing on $[0,\pi ]$. The second proof of Theorem 2 is complete. □

Remark 2.

Directly differentiating, carefully rearranging, deliberately expanding, and making use of the series expansion (1) give

$$\begin{array}{cc}\hfill \frac{Q^{\prime }\left(x\right)}{{\left(ln\frac{sinx}{x}\right)}^{\prime }}& =\frac{(2x-3sinx+xcosx)sinx}{(sinx-x)(xcosx-sinx)}\hfill \\ \hfill & =\frac{2xsinx+\frac{x}{2}sin\left(2x\right)+\frac{3}{2}cos\left(2x\right)-\frac{3}{2}}{xsinx+\frac{x}{2}sin\left(2x\right)+\frac{1}{2}cos\left(2x\right)-x^{2}cosx-\frac{1}{2}}\hfill \\ \hfill & =\frac{{\sum }_{k=3}^{\infty }{(-1)}^{k+1}\frac{(k-3)2^{2k-1}+4k}{\left(2k\right)!}{x}^{2k}}{{\sum }_{k=3}^{\infty }{(-1)}^{k+1}\frac{(k-1)(2^{2k-1}-4k)}{\left(2k\right)!}{x}^{2k}}\hfill \\ \hfill & =\frac{{\sum }_{k=0}^{\infty }\frac{k{2}^{2k+5}+4k+12}{(2k+6)!}{(-x^2)}^k}{{\sum }_{k=0}^{\infty }\frac{(k+2)(2^{2k+5}-4k-12)}{(2k+6)!}{(-x^2)}^k}.\hfill \end{array}$$

We now consider the sequence

$$S\left(k\right)=\frac{\frac{k{2}^{2k+5}+4k+12}{(2k+6)!}}{\frac{(k+2)(2^{2k+5}-4k-12)}{(2k+6)!}}=\frac{k{2}^{2k+5}+4k+12}{(k+2)(2^{2k+5}-4k-12)},k\ge 0.$$

Because the inequality

$$S\left(k\right)=\frac{k{2}^{2k+5}+4k+12}{(k+2)(2^{2k+5}-4k-12)}<\frac{(k+1)2^{2k+7}+4k+16}{(k+3)(2^{2k+7}-4k-16)}=S(k+1)$$

for $k\ge 0$ can be reformulated as

$$\left[2^{2k+6}-\left(3k^3+20k^2+50k+52\right)\right]2^{2k+3}+k^2+7k+12>0$$

and the inequality

$$2^{2k+6}>3k^3+20k^2+50k+52,k\ge 0$$

can be verified by induction, we derive that the inequality $S\left(k\right)<S(k+1)$ is true for $k\ge 0$.

Finally, employing Lemma 2 and increasing property of the sequence $S\left(k\right)$ in $k\ge 0$, we acquire that the function

$$\begin{array}{c}\frac{{\sum }_{k=0}^{\infty }\frac{k{2}^{2k+5}+4k+12}{(2k+6)!}{t}^k}{{\sum }_{k=0}^{\infty }\frac{(k+2)(2^{2k+5}-4k-12)}{(2k+6)!}{t}^k}\hfill \\ \hfill =\frac{2sinh\left(\sqrt{t}\right)\left[2\sqrt{t}-3sinh\left(\sqrt{t}\right)+\sqrt{t}cosh\left(\sqrt{t}\right)\right]}{2\sqrt{t}sinh\left(\sqrt{t}\right)+\sqrt{t}sinh\left(2\sqrt{t}\right)-2tcosh\left(\sqrt{t}\right)-cosh\left(2\sqrt{t}\right)+1}\end{array}$$

is increasing in $t>0$.

5. Conclusions

In this paper, we expanded the function $Q\left(x\right)$ defined by (3) into the Maclaurin power series (6) around $x=0$ in terms of specific determinants $E_{2n}$ and, with the aid of two monotonicity rules, proved the decreasing property of the function $R\left(x\right)$ defined by (4).