Absolute Monotonicity of Normalized Tail of Power Series Expansion of Exponential Function ^†

Abstract

In this work, the author reviews the origination of normalized tails of the Maclaurin power series expansions of infinitely differentiable functions, presents that the ratio between two normalized tails of the Maclaurin power series expansion of the exponential function is decreasing on the positive axis, and proves that the normalized tail of the Maclaurin power series expansion of the exponential function is absolutely monotonic on the whole real axis.

1. A Short Review of Normalized Tails

In April 2023, Qi and several mathematicians considered the decreasing property of the ratio ${\displaystyle \frac{F\left(u\right)}{G\left(u\right)}}$ on $(0,{\displaystyle \frac{\pi }{2}})$, where

$$F\left(u\right)=\left\{\begin{array}{cc}ln{\displaystyle \frac{3(tanu-u)}{u^3}},\hfill & 0<\left|u\right|<{\displaystyle \frac{\pi }{2}}\hfill \\ 0,\hfill & u=0\hfill \end{array}\right.$$

(1)

and

$$G(u)=\left\{\begin{array}{cc}\begin{array}{c}ln{\displaystyle \frac{tanu}{u}},\hfill \end{array}\hfill & 0<\left|u\right|<{\displaystyle \frac{\pi }{2}}\hfill \\ 0,\hfill & u=0\hfill \end{array}\right.$$

(2)

are both even functions on $(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$. The reason why we investigated the ratio ${\displaystyle \frac{F\left(u\right)}{G\left(u\right)}}$ and its monotonicity on $(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$ was stated in [1] (Remark 10).

By the study of the ratio ${\displaystyle \frac{F\left(u\right)}{G\left(u\right)}}$ in [1], the authors observed that the functions

$${\displaystyle \frac{tanu}{u}}\mathrm{and}{\displaystyle \frac{3(tanu-u)}{u^3}}={\displaystyle \frac{tanu-u}{u^3/3}}$$

(3)

are closely related to the first two terms in the Maclaurin power series expansion

$$tanu=\sum _{j=1}^{\infty }{\displaystyle \frac{2^{2j}\left(2^{2j}-1\right)}{\left(2j\right)!}}|B_{2j}|u^{2j-1}=u+{\displaystyle \frac{u^3}{3}}+{\displaystyle \frac{2u^5}{15}}+{\displaystyle \frac{17u^7}{315}}+\cdots ,\left|u\right|<{\displaystyle \frac{\pi }{2}},$$

(4)

where the Bernoulli numbers $B_j$ are generated by

$${\displaystyle \frac{u}{e^u-1}}=\sum _{j=0}^{\infty }{B}_j{\displaystyle \frac{u^j}{j!}}=1-{\displaystyle \frac{u}{2}}+\sum _{j=1}^{\infty }{B}_{2j}{\displaystyle \frac{u^{2j}}{\left(2j\right)!}},0<\left|u\right|<2\pi .$$

(5)

Motivated by the above observation, basing on the first two terms in the Maclaurin power series expansions

$$cosu=\sum _{j=0}^{\infty }{(-1)}^j{\displaystyle \frac{u^{2j}}{\left(2j\right)!}}=1-{\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{u^4}{24}}-{\displaystyle \frac{u^6}{720}}+{\displaystyle \frac{u^8}{40320}}-\cdots ,u\in \mathbb{R}$$

(6)

and

$$sinu=\sum _{j=0}^{\infty }{(-1)}^j{\displaystyle \frac{u^{2j+1}}{(2j+1)!}}=u-{\displaystyle \frac{u^3}{6}}+{\displaystyle \frac{u^5}{120}}-{\displaystyle \frac{u^7}{5040}}+{\displaystyle \frac{u^9}{362880}}-\cdots ,u\in \mathbb{R},$$

(7)

Qi and his coauthors further constructed the even functions

$$\left\{\begin{array}{cc}ln{\displaystyle \frac{2(1-cosu)}{u^2}},\hfill & 0<\left|u\right|<2\pi ;\hfill \\ 0,\hfill & u=0,\hfill \end{array}\right.$$

(8)

$$\left\{\begin{array}{cc}\frac{ln{\displaystyle \frac{2(1-cosu)}{u^2}},}{lncosu}\hfill & 0<|u|<\frac{\pi }{2};\hfill \\ \frac{1}{6},\hfill & u=0,\hfill \\ 0,\hfill & u=\pm \frac{\pi }{2},\hfill \end{array}\right.$$

(9)

$$\left\{\begin{array}{cc}ln{\displaystyle \frac{6(u-sinu)}{u^3}},\hfill & 0<\left|u\right|<\infty ;\hfill \\ 0,\hfill & u=0,\hfill \end{array}\right.$$

(10)

and

$$\left\{\begin{array}{cc}{\displaystyle \frac{ln\frac{6(u-sinu)}{u^3}}{ln\frac{sinu}{u}}},\hfill & \left|u\right|\in (0,\pi );\hfill \\ {\displaystyle \frac{3}{10}},\hfill & u=0;\hfill \\ 0,\hfill & u=\pm \pi \hfill \end{array}\right.$$

(11)

in the papers [2,3], respectively.

For generalizing the above observations, in the papers [4,5,6,7] and [8] (Remark 7), Qi and his coauthors introduced the concept of the normalized tails (also known as the normalized remainders) of the Maclaurin power series expansions (6) and (7) by

$${CosR}_n\left(u\right)=\left\{\begin{array}{cc}{(-1)}^n{\displaystyle \frac{\left(2n\right)!}{u^{2n}}}\left[cosu-{\displaystyle \sum _{j=0}^{n-1}}{(-1)}^j{\displaystyle \frac{u^{2j}}{\left(2j\right)!}}\right],\hfill & u\ne 0\hfill \\ 1,\hfill & u=0\hfill \end{array}\right.$$

and

$${SinR}_n(u)=\left\{\begin{array}{cc}{(-1)}^n{\displaystyle \frac{(2n+1)!}{u^{2n+1}}}\left[sinu-{\displaystyle \sum _{j=0}^{n-1}}{(-1)}^j{\displaystyle \frac{u^{2j+1}}{(2j+1)!}}\right],\hfill & u\ne 0\hfill \\ 1,\hfill & u=0\hfill \end{array}\right.$$

for $n\in \mathbb{N}$ and $u\in \mathbb{R}$. These two normalized tails are generalizations of the functions

$$cosu={\displaystyle \frac{cosu}{1}},{\displaystyle \frac{2(1-cosu)}{u^2}}={\displaystyle \frac{cosu-1}{-u^2/2}},{\displaystyle \frac{sinu}{u}},{\displaystyle \frac{6(u-sinu)}{u^3}}={\displaystyle \frac{sinu-u}{-u^3/6}}$$

which appeared in (8), (9), (10), and (11), respectively.

In [9], basing on the Maclaurin power series expansion (5), Qi and his joint authors invented the normalized tail:

$$\left\{\begin{array}{cc}{\displaystyle \frac{1}{B_{2n+2}}}{\displaystyle \frac{(2n+2)!}{u^{2n+2}}}\left[{\displaystyle \frac{u}{e^u-1}}-1+{\displaystyle \frac{u}{2}}-{\displaystyle \sum _{j=1}^n}{B}_{2j}{\displaystyle \frac{u^{2j}}{\left(2j\right)!}}\right],\hfill & u\ne 0;\\ 1,\hfill & u=0.\end{array}\right.$$

Through studying this normalized tail, some new knowledge about the Bernoulli polynomials was obtained in [9] (Proposition 1) and the arXiv preprint at https://doi.org/10.48550/arxiv.2405.05280.

On the basis of the Maclaurin power series expansion (4) and utilizing the idea and thought mentioned above, Qi and his coauthors introduced the following normalized tail

$$\left\{\begin{array}{cc}{\displaystyle \frac{\left(2n\right)!}{2^{2n}\left(2^{2n}-1\right)}}{\displaystyle \frac{1}{|B_{2n}|u^{2n-1}}}\left[tanu-{\displaystyle \sum _{j=1}^{n-1}}{\displaystyle \frac{2^{2j}(2^{2j}-1)}{\left(2j\right)!}}\left|B_{2j}\right|u^{2j-1}\right],\hfill & 0<\left|u\right|<{\displaystyle \frac{\pi }{2}};\\ 1,\hfill & u=0.\end{array}\right.$$

This normalized tail is a generalization of the functions in (3), which appeared in (1) and (2). Qi and his coauthors have investigated this normalized tail in a forthcoming paper.

In the paper [10] (p. 798) and the handbook [11] (pp. 42 and 55), we can find the Maclaurin power series expansion

$${tan}^{2}u=\sum _{j=1}^{\infty }{\displaystyle \frac{2^{2j+2}\left(2^{2j+2}-1\right)(2j+1)}{(2j+2)!}}\left|B_{2j+2}\right|u^{2j}$$

(12)

for $\left|u\right|<{\displaystyle \frac{\pi }{2}}$. Basing on the series expansion (12), imitating the above observations, and employing the above initiating idea and thought to define the normalized tails, Zhang and Qi built the normalized tail

$$h_n\left(u\right)=\left\{\begin{array}{cc}{\displaystyle \frac{{\displaystyle (2n+2)!\left[{tan}^{2}u-\sum _{\ell =1}^{n-1}\frac{2^{2\ell +2}\left(2^{2\ell +2}-1\right)(2\ell +1)}{(2\ell +2)!}\left|B_{2\ell +2}\right|u^{2\ell }\right]}}{2^{2n+2}(2^{2n+2}-1)(2n+1)\left|B_{2n+2}\right|u^{2n}}},\hfill & u\ne 0\\ 1,\hfill & u=0\end{array}\right.$$

for $n\in \mathbb{N}$ and $\left|u\right|<{\displaystyle \frac{\pi }{2}}$ in the paper [12].

Considering the relation ${sec}^{2}u=1+{tan}^{2}u$, we reformulate the normalized tail $h_n\left(u\right)$ as

$$h_n\left(u\right)=\left\{\begin{array}{cc}{\displaystyle \frac{{\displaystyle (2n+2)!\left[{sec}^{2}u-\sum _{\ell =0}^{n-1}\frac{2^{2\ell +2}\left(2^{2\ell +2}-1\right)(2\ell +1)}{(2\ell +2)!}\left|B_{2\ell +2}\right|u^{2\ell }\right]}}{2^{2n+2}(2^{2n+2}-1)(2n+1)\left|B_{2n+2}\right|u^{2n}}},\hfill & u\ne 0\\ 1,\hfill & u=0\end{array}\right.$$

for $n\in \mathbb{N}$ and $\left|u\right|<{\displaystyle \frac{\pi }{2}}$. Hence, the quantity $h_n\left(u\right)$ is also the normalized tail of the Maclaurin power series expansion of the square ${sec}^{2}u$ about $u=0$.

In [13] (Section 5), the authors summed up the above ideas and thoughts to design normalized tails as follows:

Suppose that a real function $f\left(u\right)$ has a formal Maclaurin power series expansion

$$f\left(u\right)=\sum _{j=0}^{\infty }{f}^{\left(j\right)}\left(0\right){\displaystyle \frac{u^j}{j!}}.$$

(13)

If $f^{(n+1)}\left(0\right)\ne 0$ for some $n\in {\mathbb{N}}_0$, then we call the function

$$\left\{\begin{array}{cc}{\displaystyle \frac{1}{f^{(n+1)}\left(0\right)}}{\displaystyle \frac{(n+1)!}{u^{n+1}}}\left[f\left(u\right)-{\displaystyle \sum _{j=0}^n}{f}^{\left(j\right)}\left(0\right){\displaystyle \frac{u^j}{j!}}\right],\hfill & u\ne 0\\ 1,\hfill & u=0\end{array}\right.$$

the normalized tail of the Maclaurin power series expansion (13).

2. Motivations of This Paper

It is well known that

$$e^u=\sum _{j=0}^{\infty }{\displaystyle \frac{u^j}{j!}},\left|u\right|\in \mathbb{R}$$

(14)

and that, for $n\in \mathbb{N}$ and $u\in \mathbb{R}$, the quantity

$$R_n\left(u\right)=e^u-\sum _{k=0}^{n-1}{\displaystyle \frac{u^k}{k!}}$$

is called the nth tail of the Maclaurin power series expansion (14).

In the paper [13], the authors designed the normalized tail

$$f_n\left(u\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n!}{u^n}}{R}_n\left(u\right),\hfill & u\ne 0\\ 1,\hfill & u=0\end{array}\right.$$

(15)

for $n\in \mathbb{N}$ and $u\in \mathbb{R}$. The main results of the paper [13] include the following information:

• The normalized tail $f_n\left(u\right)$ for $n\in \mathbb{N}$ is an increasing and logarithmically convex function of $u\in \mathbb{R}$;
• The logarithm $ln{f}_n\left(u\right)$ for $n\in \mathbb{N}$ was expanded into a Maclaurin power series;
• The function

  $$\left\{\begin{array}{cc}{\displaystyle \frac{ln{f}_n\left(u\right)}{u}},\hfill & u\ne 0\\ {\displaystyle \frac{1}{n+1}},\hfill & u=0\end{array}\right.$$

  for $n\in \mathbb{N}$ is increasing in $u\in \mathbb{R}$;
• The inequality

  $$\left[\sum _{j=0}^{\infty }{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+2}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\left[\sum _{j=0}^{\infty }{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\ge {\left[\sum _{j=0}^{\infty }{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]}^2$$

  (16)

  is sound for $n\in \mathbb{N}$ and $u\in \mathbb{R}$.

The inequality (16) is equivalent to

$${\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}\frac{u^j}{j!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}}}\le {\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+k+1}{n}\right)}\frac{u^j}{j!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+k}{n}\right)}\frac{u^j}{j!}}}$$

(17)

for $k,n\in \mathbb{N}$ and $u\in \mathbb{R}$. What is the limit of the ratio between two series on the right hand side of the inequality (17) as $k\to \infty$? See the second problem at the website https://math.stackexchange.com/q/4956563 (accessed on 11 August 2024).

In the paper [13], the authors did not mention the positivity of the normalized tail $f_n\left(u\right)$ for $n\in \mathbb{N}$ and $u\in \mathbb{R}$. When $u>0$, it is immediate that $R_n\left(u\right)$ and $f_n\left(u\right)$ are both positive for $n\in \mathbb{N}$. When $u<0$, the inequality $e^u>{\sum }_{k=0}^n{\displaystyle \frac{u^k}{k!}}$ is valid for odd $n\in \mathbb{N}$ and reverses for even $n\in \mathbb{N}$; see the paper [14] and the closely related references therein. As a result, the normalized tail $f_n\left(u\right)$ is positive for $n\in \mathbb{N}$ and $u\in \mathbb{R}$.

Let $I\subseteq \mathbb{R}$ be an interval. A real infinitely-differentiable function $f\left(u\right)$ defined on I is said to be absolutely monotonic in $u\in I$ if and only if all of its derivatives satisfy $f^{\left(k\right)}\left(u\right)\ge 0$ for $k\in {\mathbb{N}}_0$ and $u\in I$. A real infinitely-differentiable function $f\left(u\right)$ defined on I is said to be completely monotonic in $u\in I$ if and only if all of its derivatives satisfy ${(-1)}^k{f}^{\left(k\right)}\left(u\right)\ge 0$ for $k\in {\mathbb{N}}_0$ and $u\in I$. When $I=(0,\infty )$ or $I=[0,\infty )$, one can refer to plenty of classical investigations on absolutely (or completely, respectively) monotonic functions in the chapters [15] (Chapter XIII) and [16] (Chapter IV) and in the monograph [17]. In the papers [18,19,20], the authors invented the notions of logarithmically absolutely (or logarithmically completely, respectively) monotonic functions.

In this paper, we aim to discuss the decreasing property of the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ in $u\in (0,\infty )$ and the absolute monotonicity of the normalized tail $f_n\left(u\right)$ in $u\in \mathbb{R}$ for $n\in \mathbb{N}$. This paper is a continuation of the paper [13].

3. Lemmas

For attaining our aims, we need the following lemmas, including two monotonicity rules for the ratio between two Maclaurin power series and for the ratio between two definite integrals involving one parameter.

Lemma 1

(Monotonicity rule; see [21], [22] (Theorem 4.3), and [23] (Lemma 2.1)). Let $a_j,b_j\in \mathbb{R}$ for $j\in {\mathbb{N}}_0$ be two real sequences and let the Maclaurin power series

$$U\left(t\right)=\sum _{j=0}^{\infty }{a}_j{t}^{j}andV\left(t\right)=\sum _{j=0}^{\infty }{b}_j{t}^j$$

be convergent on $(-r,r)$ for some positive number $r>0$. If $b_j>0$ and the quotient ${\displaystyle \frac{a_j}{b_j}}$ is increasing for $j\in {\mathbb{N}}_0$, then the quotient ${\displaystyle \frac{U\left(t\right)}{V\left(t\right)}}$ is also increasing on $(0,r)$.

Lemma 2

([24] (p. 502)). For $m\in {\mathbb{N}}_0$ and $t\in \mathbb{R}$, we have

$$R_{m+1}\left(t\right)=e^t-\sum _{j=0}^m{\displaystyle \frac{t^j}{j!}}={\displaystyle \frac{t^{m+1}}{(m+1)!}}\left[1+t{\int }_0^1{s}^{m+1}{e}^{t(1-s)}\mathrm{d}s\right].$$

Lemma 3

(Monotonicity rule; see [25] (Lemma 9)). Let the functions $U\left(s\right)$, $V\left(s\right)>0$, and $W(s,t)>0$ be integrable in $s\in (\alpha ,\beta )$.

1. 

If the quotients ${\displaystyle \frac{\partial W(s,t)/\partial t}{W(s,t)}}$ and ${\displaystyle \frac{U\left(s\right)}{V\left(s\right)}}$ are both increasing or both decreasing in $s\in (\alpha ,\beta )$, then the quotient

$$R\left(t\right)={\displaystyle \frac{{\int }_{\alpha }^{\beta }W(s,t)U\left(s\right)\mathrm{d}s}{{\int }_{\alpha }^{\beta }W(s,t)V\left(s\right)\mathrm{d}s}}$$

is increasing in t.

2. 

If one of the quotients ${\displaystyle \frac{\partial W(s,t)/\partial t}{W(s,t)}}$ and ${\displaystyle \frac{U\left(s\right)}{V\left(s\right)}}$ is increasing and another one of them is decreasing in $s\in (\alpha ,\beta )$, then the quotient $R\left(t\right)$ is decreasing in t.

Lemma 4

([26] (p. 22, Problem 94)). Let $a_n$ and $b_n>0$ for $n\in {\mathbb{N}}_0$ be two sequences such that the infinite series ${\sum }_{n=0}^{\infty }{b}_n{t}^n$ converges for $t\in \mathbb{R}$ and ${lim}_{n\to \infty }{\displaystyle \frac{a_n}{b_n}}=s$. Then, the infinite series ${\sum }_{n=0}^{\infty }{a}_n{t}^n$ converges too for $t\in \mathbb{R}$; in addition,

$$\underset{t\to \infty }{lim}{\displaystyle \frac{{\sum }_{n=0}^{\infty }{a}_n{t}^n}{{\sum }_{n=0}^{\infty }{b}_n{t}^n}}=s.$$

Remark 1.

In the paper [27], motivated by [28], Yang and Chu employed Lemma 4 to bind the modified Bessel function of the first kind ${\sum }_{n=0}^{\infty }{\displaystyle \frac{1}{{(n!)}^2}}{\left({\displaystyle \frac{t}{2}}\right)}^{2n}$ and the Toader–Qi mean

$${\displaystyle \frac{2}{\pi }}{\int }_0^{\pi /2}{a}^{{cos}^2\theta }{b}^{{sin}^2\theta }\mathrm{d}\theta .$$

4. Decreasing Property on Positive Half-Axis

In this section, we verify that the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ is decreasing in $u\in (0,\infty )$.

Theorem 1.

For $n\in \mathbb{N}$, the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ is decreasing in $u\in (0,\infty )$.

Proof. 

Making use of the Maclaurin power series expansion (14), we can write the normalized tail $f_n\left(u\right)$ as defined by (15) for $n\in \mathbb{N}$ as

$$f_n\left(u\right)=\sum _{j=0}^{\infty }{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}}{\displaystyle \frac{u^j}{j!}},u\in \mathbb{R}.$$

(18)

Then, for $u\in \mathbb{R}$ and $n\in \mathbb{N}$, we have

$${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}={\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}\frac{u^j}{j!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}}}.$$

(19)

The ratio between coefficients of $u^j$ of two series in the above fraction is

$${\displaystyle \frac{\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}\frac{1}{j!}}{\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{1}{j!}}}={\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}}={\displaystyle \frac{(j+n)!}{n!j!}}{\displaystyle \frac{(n+1)!j!}{(j+n+1)!}}={\displaystyle \frac{n+1}{j+n+1}}$$

and it is decreasing in $j\in {\mathbb{N}}_0$ for given $n\in \mathbb{N}$. Considering Lemma 1, we conclude that the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ is decreasing in $u\in (0,\infty )$ for a given $n\in \mathbb{N}$. The proof of Theorem 1 is complete. □

5. Absolute Monotonicity

In this section, we discover the absolute monotonicity of the normalized tail $f_n\left(u\right)$ in $u\in \mathbb{R}$ for $n\in \mathbb{N}$.

Theorem 2.

For $n\in \mathbb{N}$, the normalized tail $f_n\left(u\right)$ is an absolutely monotonic function in $u\in \mathbb{R}$.

Proof. 

When $u>0$, from the Maclaurin power series expansion (18), it follows immediately that the normalized tail $f_n\left(u\right)$ for $n\in \mathbb{N}$ is an absolutely monotonic function in $u>0$.

Utilizing Lemma 2, we arrive at

$$f_n\left(u\right)=1+u{\int }_0^1{v}^n{e}^{u(1-v)}\mathrm{d}v,u\in \mathbb{R}$$

(20)

and $f_n\left(0\right)=1$ for $n\in \mathbb{N}$. For $m\in \mathbb{N}$, a direct differentiation gives

$$\begin{array}{cc}{f}_n^{\left(m\right)}\left(u\right)\hfill & =e^u\left[u{\int }_0^1{v}^n{(1-v)}^m{e}^{-uv}\mathrm{d}v+m{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v\right]\hfill \\ \hfill & \to m{\int }_0^1{v}^n{(1-v)}^{m-1}\mathrm{d}v,u\to 0\hfill \\ \hfill & ={\displaystyle \frac{m!n!}{(m+n)!}}\hfill \\ \hfill & >0,m,n\in \mathbb{N}.\hfill \end{array}$$

Accordingly, in order to prove the absolute monotonicity of $f_n\left(u\right)$ in $u<0$ for $n\in \mathbb{N}$, we need to show

$${\displaystyle \frac{u{\int }_0^1{v}^n{(1-v)}^m{e}^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}\ge -m$$

(21)

for $m,n\in \mathbb{N}$ and $u<0$.

It is easy to see that, via integration in part,

$$u{\int }_0^1{v}^n{(1-v)}^m{e}^{-uv}\mathrm{d}v=-{\int }_0^1{v}^n{(1-v)}^m{\displaystyle \frac{\mathrm{d}{e}^{-uv}}{\mathrm{d}v}}\mathrm{d}v={\int }_0^1{\left[v^n{(1-v)}^m\right]}^{\prime }{e}^{-uv}\mathrm{d}v$$

for $m,n\in \mathbb{N}$ and $u<0$. Applying Lemma 3 to

$$W(u,v)=e^{-uv}>0,U\left(v\right)={\left[v^n{(1-v)}^m\right]}^{\prime },V\left(v\right)=v^n{(1-v)}^{m-1}>0$$

for $v\in (0,1)$ and $m,n\in \mathbb{N}$, which satisfy that both

$${\displaystyle \frac{\partial [lnW(v,u\left)\right]}{\partial u}}={\displaystyle \frac{\partial W(v,u)/\partial u}{W(v,u)}}=-v$$

and

$${\displaystyle \frac{U\left(v\right)}{V\left(v\right)}}={\displaystyle \frac{{\left[v^n{(1-v)}^m\right]}^{\prime }}{v^n{(1-v)}^{m-1}}}={\displaystyle \frac{n}{v}}-m-n$$

are decreasing in $v\in (0,1)$, reveals that the ratio

$${\displaystyle \frac{u{\int }_0^1{v}^n{(1-v)}^m{e}^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}={\displaystyle \frac{{\int }_0^1{\left[v^n{(1-v)}^m\right]}^{\prime }{e}^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}$$

(22)

is increasing in $u<0$ for $m,n\in \mathbb{N}$.

In light of Equation (22), we acquire

$$\begin{array}{cc}{\displaystyle \frac{u{\int }_0^1{v}^n{(1-v)}^m{e}^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}\hfill & ={\displaystyle \frac{{\int }_0^1[n{v}^{n-1}{(1-v)}^m-m{v}^n{(1-v)}^{m-1}]e^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}\hfill \\ \hfill & =n{\displaystyle \frac{{\int }_0^1{v}^{n-1}{(1-v)}^m{e}^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}-m\hfill \end{array}$$

for $m,n\in \mathbb{N}$ and $u\in \mathbb{R}$. When $u\to -\infty$, making use of Lemma 4, we obtain

$$\begin{array}{cc}\underset{u\to -\infty }{lim}{\displaystyle \frac{{\int }_0^1{v}^{n-1}{(1-v)}^m{e}^{-uv}\mathrm{d}v}{{\int }_0^1{v}^n{(1-v)}^{m-1}{e}^{-uv}\mathrm{d}v}}\hfill & =\underset{u\to -\infty }{lim}{\displaystyle \frac{{\sum }_{k=0}^{\infty }{(-1)}^k\frac{u^k}{k!}{\int }_0^1{v}^{n+k-1}{(1-v)}^m\mathrm{d}v}{{\sum }_{k=0}^{\infty }{(-1)}^k\frac{u^k}{k!}{\int }_0^1{v}^{n+k}{(1-v)}^{m-1}\mathrm{d}v}}\hfill \\ \hfill & =\underset{u\to \infty }{lim}{\displaystyle \frac{{\sum }_{k=0}^{\infty }\frac{u^k}{k!}\frac{m!(n+k-1)!}{(m+n+k)!}}{{\sum }_{k=0}^{\infty }\frac{u^k}{k!}\frac{(n+k)!(m-1)!}{(m+n+k)!}}}\hfill \\ \hfill & =\underset{k\to \infty }{lim}{\displaystyle \frac{\frac{1}{k!}\frac{m!(n+k-1)!}{(m+n+k)!}}{\frac{1}{k!}\frac{(n+k)!(m-1)!}{(m+n+k)!}}}\hfill \\ \hfill & =\underset{k\to \infty }{lim}{\displaystyle \frac{m}{n+k}}\hfill \\ \hfill & =0\hfill \end{array}$$

for $m,n\in \mathbb{N}$. Consequently, the inequality (21) is sound. The proof of Theorem 2 is complete. □

6. More Remarks

In this section, we give some remarks about our main results.

Remark 2.

By the definition (15), we acquire

$${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}={\displaystyle \frac{\frac{(n+1)!}{u^{n+1}}\left(e^u-{\sum }_{j=0}^n\frac{u^j}{j!}\right)}{\frac{n!}{u^n}\left(e^u-{\sum }_{j=0}^{n-1}\frac{u^j}{j!}\right)}}=(n+1){\displaystyle \frac{e^u-{\sum }_{j=0}^n\frac{u^j}{j!}}{u\left(e^u-{\sum }_{j=0}^{n-1}\frac{u^j}{j!}\right)}}\to \left\{\begin{array}{cc}0,\hfill & u\to \infty \\ 1,\hfill & u\to 0\\ {\displaystyle \frac{n+1}{n}},\hfill & u\to -\infty \end{array}\right.$$

for $n\in \mathbb{N}$. Accordingly, Theorem 1 implies the inequality $f_{n+1}\left(u\right)<f_n\left(u\right)$ for $u\in (0,\infty )$ and $n\in \mathbb{N}$. In particular, when $n=1$, we acquire

$${\displaystyle \frac{2(e^u-1-u)}{u^2}}<{\displaystyle \frac{e^u-1}{u}},u\in (0,\infty ),$$

which can be further reformulated as

$$e^u<{\displaystyle \frac{2+u}{2-u}},u\in (0,2).$$

Generally, utilizing Theorem 1, for $n\in \mathbb{N}$, we have

$$e^u<\sum _{k=0}^{n-1}{\displaystyle \frac{u^k}{k!}}+{\displaystyle \frac{n+1}{n+1-u}}{\displaystyle \frac{u^n}{n!}},u\in (0,n+1).$$

(23)

This inequality extends and refines an inequality

$$e^u<\sum _{k=0}^{n-1}{\displaystyle \frac{u^k}{k!}}+{\displaystyle \frac{n}{n-u}}{\displaystyle \frac{u^n}{n!}},u\in (0,n)$$

(24)

collected in [29] (p. 269, Entry 3.6.6). This demonstrates that Theorem 1 and the notion of the normalized tail $f_n\left(u\right)$ of the Maclaurin power series expansion (14) of the exponential function $e^u$ are significant in mathematics.

In the paper [14], Qi constructed many inequalities similar to (23) and (24).

Remark 3.

Taking $n=1,2,3$ in Theorem 2 gives that the functions

$$f_1\left(u\right)={\displaystyle \frac{e^u-1}{u}},f_2\left(u\right)={\displaystyle \frac{2(e^u-1-u)}{u^2}},f_2\left(u\right)={\displaystyle \frac{6\left(e^u-1-u-\frac{u^2}{2}\right)}{u^3}}$$

are absolutely monotonic on $\mathbb{R}$. These special cases are not trivial for $u<0$.

Remark 4.

In [15] (p. 369), it was stated that if f is a completely monotonic function such that $f^{\left(k\right)}\left(t\right)\ne 0$ for $k\in {\mathbb{N}}_0$, then

$$a_i=ln\left[{(-1)}^{i-1}{f}^{(i-1)}\left(t\right)\right],i\in \mathbb{N}$$

is a convex sequence. By virtue of this conclusion or [16] (p. 167, Corollary 16), we see that the absolute monotonicity of $f_n\left(u\right)$ in $u\in \mathbb{R}$ for $n\in \mathbb{N}$ in Theorem 2 is stronger than the logarithmic convexity of $f_n\left(u\right)$ in $u\in \mathbb{R}$ for $n\in \mathbb{N}$ in [13] (Corollary 1).

Remark 5.

We guess that the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ is decreasing for $n\in \mathbb{N}$ in $u\in (-\infty ,0)$.

A direct differentiation on both sides of (19) yields

$$\begin{array}{cc}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}u}}\left[{\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}\right]\hfill & ={\displaystyle \frac{\left[{\displaystyle \sum _{j=1}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}\frac{u^{j-1}}{(j-1)!}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}\right]-\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}\frac{u^j}{j!}\right]\left[{\displaystyle \sum _{j=1}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^{j-1}}{(j-1)!}\right]}{{\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}\right]}^2}}\hfill \\ \hfill & ={\displaystyle \frac{\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+2}{n+1}\right)}\frac{u^j}{j!}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}\right]-\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}\frac{u^j}{j!}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}\frac{u^j}{j!}\right]}{{\left[{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}\right]}^2}}\hfill \end{array}$$

(25)

for $u\in \mathbb{R}$ and $n\in \mathbb{N}$. Accordingly, in order to prove the decreasing property of the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ for $n\in \mathbb{N}$ in $u\in (-\infty ,0)$, it suffices to show that the inequality

$$\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+2}{n+1}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\le \left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}}{\displaystyle \frac{u^j}{j!}}\right]{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}}{\displaystyle \frac{u^j}{j!}}]$$

(26)

is sound for $u\in \mathbb{R}$ and $n\in \mathbb{N}$. See the first problem at the website https://math.stackexchange.com/q/4956563 (accessed on 11 August 2024).

The case $u>0$ of the inequality (26) can be easily deduced from the combination of Theorem 1 with the derivative in (25).

The inequality (26) can be rearranged as

$${\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+2}{n+1}\right)}\frac{u^j}{j!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}\frac{u^j}{j!}}}\le {\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}\frac{u^j}{j!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}}}$$

for $u\in \mathbb{R}$ and $n\in \mathbb{N}$. See the third problem at the website https://math.stackexchange.com/q/4956563 (accessed on 11 August 2024). Accordingly, in order to prove the inequality (26), it is sufficient to show that the sequence

$${\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}\frac{u^j}{j!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}\frac{u^j}{j!}}}={\displaystyle \frac{{\displaystyle \sum _{j=0}^{\infty }}\frac{(j+1)u^j}{(j+n+1)!}}{{\displaystyle \sum _{j=0}^{\infty }}\frac{u^j}{(j+n)!}}}$$

is decreasing in $n\in \mathbb{N}$ for $u<0$.

Remark 6.

If the guess in Remark 5 is true, making use of the limits in Remark 2, we conclude the inequality ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}<{\displaystyle \frac{n+1}{n}}$ for $n\in \mathbb{N}$ and $u\in \mathbb{R}$. Equivalently, the inequality

$${\displaystyle \frac{R_{n+1}\left(u\right)}{u^{n+1}}}<{\displaystyle \frac{1}{n}}{\displaystyle \frac{R_n\left(u\right)}{u^n}}$$

is sound for $n\in \mathbb{N}$ and $u\in \mathbb{R}$. Therefore, for $n\in \mathbb{N}$, the inequality

$$(n-u){\displaystyle \frac{R_n\left(u\right)}{u^n}}<{\displaystyle \frac{1}{(n-1)!}}$$

is valid in $u>0$ and reverses in $u<0$.

Remark 7.

For $n\in \mathbb{N}$ and $u\in \mathbb{R}$, let

$$G_n\left(u\right)=\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n+1}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]-\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+2}{n+1}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]$$

and

$$H_n\left(u\right)=\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+2}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]-{\left[{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+n+1}{n}\right)}}{\displaystyle \frac{u^j}{j!}}\right]}^2.$$

We guess that the functions $G_n\left(u\right)$ and $H_n\left(u\right)$ for $n\in \mathbb{N}$ are absolutely monotonic in $u\in \mathbb{R}$. This guess is stronger than the inequalities (16) and (26).

7. Conclusions

In this paper, we reviewed the origination of the notion of normalized tails of infinitely differentiable functions’ Maclaurin power series expansions, presenting that the ratio ${\displaystyle \frac{f_{n+1}\left(u\right)}{f_n\left(u\right)}}$ between the normalized tails $f_{n+1}\left(u\right)$ and $f_n\left(u\right)$ for $n\in \mathbb{N}$ is decreasing in $u\in (0,\infty )$, as well as proving that the normalized tail $f_n\left(u\right)$ for $n\in \mathbb{N}$ is absolutely monotonic in $u\in \mathbb{R}$.

Moreover, the guesses posed in Remarks 5 and 7 are interesting in mathematics.