## 1 Introduction

The Euler-Mascheroni constant was first introduced by Leonhard Euler (1707-1783) in 1734 as the limit of the sequence

$\gamma \left(n\right):=\sum _{m=1}^{n}\frac{1}{m}-lnn.$
(1.1)

There are many famous unsolved problems about the nature of this constant (see, e.g., the survey papers or books of Brent and Zimmermann [1], Dence and Dence [2], Havil [3] and Lagarias [4]). For example, it is a long-standing open problem if the Euler-Mascheroni constant is a rational number. A good part of its mystery comes from the fact that the known algorithms converging to γ are not very fast, at least, when they are compared to similar algorithms for π and e.

The sequence ${\left(\gamma \left(n\right)\right)}_{n\in \mathbb{N}}$ converges very slowly toward γ, like ${\left(2n\right)}^{-1}$. Up to now, many authors have been preoccupied with improving its rate of convergence (see, e.g., [2, 522] and the references therein). We list some main results as follows:

$\begin{array}{c}\sum _{m=1}^{n}\frac{1}{m}-ln\left(n+\frac{1}{2}\right)=\gamma +O\left({n}^{-2}\right)\phantom{\rule{1em}{0ex}}\text{(DeTemple [6])},\hfill \\ \sum _{m=1}^{n}\frac{1}{m}-ln\frac{{n}^{3}+\frac{3}{2}{n}^{2}+\frac{227}{240}+\frac{107}{480}}{{n}^{2}+n+\frac{97}{240}}=\gamma +O\left({n}^{-6}\right)\phantom{\rule{1em}{0ex}}\text{(Mortici [13])},\hfill \\ \sum _{m=1}^{n}\frac{1}{m}-ln\left(1+\frac{1}{2n}+\frac{1}{24{n}^{2}}-\frac{1}{48{n}^{3}}+\frac{23}{5,760{n}^{4}}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\gamma +O\left({n}^{-5}\right)\phantom{\rule{1em}{0ex}}\text{(Chen and Mortici [5]).}\hfill \end{array}$

Recently, Mortici and Chen [14] provided a very interesting sequence,

$\begin{array}{rcl}\nu \left(n\right)& =& \sum _{m=1}^{n}\frac{1}{m}-\frac{1}{2}ln\left({n}^{2}+n+\frac{1}{3}\right)\\ -\left(\frac{-\frac{1}{180}}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{2}}+\frac{\frac{8}{2,835}}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{3}}++\frac{\frac{5}{1,512}}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{4}}+\frac{\frac{592}{93,555}}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{5}}\right),\end{array}$

and proved

$\underset{n\to \mathrm{\infty }}{lim}{n}^{12}\left(\nu \left(n\right)-\gamma \right)=-\frac{796,801}{43,783,740}.$
(1.2)

Hence the rate of convergence of the sequence ${\left(\nu \left(n\right)\right)}_{n\in \mathbb{N}}$ is ${n}^{-12}$.

Very recently, by inserting the continued fraction term in (1.1), Lu [9] introduced a class of sequences ${\left({r}_{k}\left(n\right)\right)}_{n\in \mathbb{N}}$ (see Theorem 1) and showed

$\frac{1}{72{\left(n+1\right)}^{3}}<\gamma -{r}_{2}\left(n\right)<\frac{1}{72{n}^{3}},$
(1.3)
$\frac{1}{120{\left(n+1\right)}^{4}}<{r}_{3}\left(n\right)-\gamma <\frac{1}{120{\left(n-1\right)}^{4}}.$
(1.4)

In fact, Lu [9] also found ${a}_{4}$ without proof. In general, the continued fraction method could provide a better approximation than others, and has less numerical computations.

First, we will prove the following theorem.

Theorem 1 For the Euler-Mascheroni constant, we have the following convergent sequence:

$r\left(n\right)=1+\frac{1}{2}+\cdots +\frac{1}{n}-lnn-\frac{{a}_{1}}{n+\frac{{a}_{2}n}{n+\frac{{a}_{3}n}{n+\ddots }}},$

where $\left({a}_{1},{a}_{2},{a}_{4},{a}_{6},{a}_{8},{a}_{10},{a}_{12}\right)=\left(\frac{1}{2},\frac{1}{6},\frac{3}{5},\frac{79}{126},\frac{7,230}{6,241},\frac{4,146,631}{3,833,346},\frac{306,232,774,533}{179,081,182,865}\right)$, and ${a}_{2k+1}=-{a}_{2k}$ for $1\le k\le 6$.

Let

${R}_{k}\left(n\right):=\frac{{a}_{1}}{n+\frac{{a}_{2}n}{n+\frac{{a}_{3}n}{n+\frac{{a}_{4}n}{\frac{\ddots }{n+{a}_{k}}}}}}$

(see the Appendix for their simple expressions) and

${r}_{k}\left(n\right):=\sum _{m=1}^{n}\frac{1}{m}-lnn-{R}_{k}\left(n\right).$

For $1\le k\le 13$, we have

$\underset{n\to \mathrm{\infty }}{lim}{n}^{k+1}\left({r}_{k}\left(n\right)-\gamma \right)={C}_{k},$
(1.5)

where

$\begin{array}{rcl}\left({C}_{1},\dots ,{C}_{13}\right)& =& \left(-\frac{1}{12},-\frac{1}{72},\frac{1}{120},\frac{1}{200},-\frac{79}{25,200},-\frac{6,241}{3,175,200},\frac{241}{105,840},\frac{58,081}{22,018,248},\\ -\frac{262,445}{91,974,960},-\frac{2,755,095,121}{892,586,949,408},\frac{20,169,451}{3,821,257,440},\\ \frac{406,806,753,641,401}{45,071,152,103,463,200},-\frac{71,521,421,431}{5,152,068,292,800}\right).\end{array}$

Open problem For every $k\ge 1$, we have ${a}_{2k+1}=-{a}_{2k}$ .

The main aim of this paper is to improve (1.3) and (1.4). We establish the following more precise inequalities.

Theorem 2 Let ${r}_{10}\left(n\right)$, ${r}_{11}\left(n\right)$, ${C}_{10}$ and ${C}_{11}$ be defined in Theorem  1, then

${C}_{10}\frac{1}{{\left(n+1\right)}^{11}}<\gamma -{r}_{10}\left(n\right)<{C}_{10}\frac{1}{{n}^{11}},$
(1.6)
${C}_{11}\frac{1}{{\left(n+1\right)}^{12}}<{r}_{11}\left(n\right)-\gamma <{C}_{11}\frac{1}{{n}^{12}}.$
(1.7)

Remark 1 In fact, Theorem 2 implies that ${r}_{10}\left(n\right)$ is a strictly increasing function of n, whereas ${r}_{11}\left(n\right)$ is a strictly decreasing function of n. Certainly, it has similar inequalities for ${r}_{k}\left(n\right)$ ($1\le k\le 9$), we leave these for readers to verify. It is also should be noted that (1.4) cannot deduce the monotonicity of ${r}_{3}\left(n\right)$.

Remark 2 It is worth to point out that Theorem 2 provides sharp bounds for a harmonic sequence which are superior to Theorems 3 and 4 of Mortici and Chen [14].

## 2 The proof of Theorem 1

The following lemma gives a method for measuring the rate of convergence. This lemma was first used by Mortici [23, 24] for constructing asymptotic expansions or to accelerate some convergences. For proof and other details, see, e.g., [24].

Lemma 1 If the sequence ${\left({x}_{n}\right)}_{n\in \mathbb{N}}$ is convergent to zero and there exists the limit

$\underset{n\to +\mathrm{\infty }}{lim}{n}^{s}\left({x}_{n}-{x}_{n+1}\right)=l\in \left[-\mathrm{\infty },+\mathrm{\infty }\right]$
(2.1)

with $s>1$, then there exists the limit

$\underset{n\to +\mathrm{\infty }}{lim}{n}^{s-1}{x}_{n}=\frac{l}{s-1}.$
(2.2)

In the sequel, we always assume $n\ge 2$.

We need to find the value ${a}_{1}\in \mathbb{R}$ which produces the most accurate approximation of the form

${r}_{1}\left(n\right)=\sum _{m=1}^{n}\frac{1}{m}-lnn-\frac{{a}_{1}}{n},$
(2.3)

here we note ${R}_{1}\left(n\right)={a}_{1}/n$. To measure the accuracy of this approximation, we usually say that approximation (2.3) is better as ${r}_{1}\left(n\right)-\gamma$ faster converges to zero. Clearly,

${r}_{1}\left(n\right)-{r}_{1}\left(n+1\right)=ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}+\frac{{a}_{1}}{n+1}-\frac{{a}_{1}}{n}.$
(2.4)

It is well known that for $|x|<1$,

$ln\left(1+x\right)=\sum _{m=1}^{\mathrm{\infty }}{\left(-1\right)}^{m-1}\frac{{x}^{m}}{m}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{1}{1-x}=\sum _{m=0}^{\mathrm{\infty }}{x}^{m}.$

Developing expression (2.4) into power series expansion in $1/n$, we obtain

${r}_{1}\left(n\right)-{r}_{1}\left(n+1\right)=\left(\frac{1}{2}-{a}_{1}\right)\frac{1}{{n}^{2}}+\left({a}_{1}-\frac{2}{3}\right)\frac{1}{{n}^{3}}+\left(\frac{3}{4}-{a}_{1}\right)\frac{1}{{n}^{4}}+O\left(\frac{1}{{n}^{5}}\right).$
(2.5)

From Lemma 1, we see that the rate of convergence of the sequence ${\left({r}_{1}\left(n\right)-\gamma \right)}_{n\in \mathbb{N}}$ is even higher than the value s satisfying (2.1). By Lemma 1, we have

1. (i)

If ${a}_{1}\ne \frac{1}{2}$, then the rate of convergence of ${\left({r}_{1}\left(n\right)-\gamma \right)}_{n\in \mathbb{N}}$ is ${n}^{-1}$ since

$\underset{n\to \mathrm{\infty }}{lim}n\left({r}_{1}\left(n\right)-\gamma \right)=\frac{1}{2}-{a}_{1}\ne 0.$
2. (ii)

If ${a}_{1}=\frac{1}{2}$, from (2.5) we have

${r}_{1}\left(n\right)-{r}_{1}\left(n+1\right)=-\frac{1}{6}\frac{1}{{n}^{3}}+O\left(\frac{1}{{n}^{4}}\right).$

Hence the rate of convergence of ${\left({r}_{1}\left(n\right)-\gamma \right)}_{n\in \mathbb{N}}$ is ${n}^{-2}$ since

$\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\left({r}_{1}\left(n\right)-\gamma \right)=-\frac{1}{12}.$

We also observe that the fastest possible sequence ${\left({r}_{1}\left(n\right)\right)}_{n\in \mathbb{N}}$ is obtained only for ${a}_{1}=\frac{1}{2}$.

Just as Lu [9] did, we may repeat the above approach to determine ${a}_{1}$ to ${a}_{4}$ step by step. However, the computations become very difficult when $k\ge 5$. In this paper we use Mathematica software to manipulate symbolic computations.

Let

${r}_{k}\left(n\right)=\sum _{m=1}^{n}\frac{1}{m}-lnn-{R}_{k}\left(n\right),$
(2.6)

then

${r}_{k}\left(n\right)-{r}_{k}\left(n+1\right)=ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}+{R}_{k}\left(n+1\right)-{R}_{k}\left(n\right).$
(2.7)

It is easy to get the following power series:

$ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}=\sum _{m=2}^{\mathrm{\infty }}{\left(-1\right)}^{m}\frac{m-1}{m}\frac{1}{{n}^{m}}.$
(2.8)

Hence the key step is to expand ${R}_{k}\left(n+1\right)-{R}_{k}\left(n\right)$ into power series in $\frac{1}{n}$. Here we use some examples to explain our method.

Step 1: For example, given ${a}_{1}$ to ${a}_{7}$, find ${a}_{8}$. Define

$\begin{array}{rcl}{R}_{8}\left(n\right)& =& \frac{\frac{1}{2}}{n+\frac{\frac{n}{6}}{n+\frac{-\frac{n}{6}}{n+\frac{\frac{3}{5}n}{n+\frac{\frac{-3}{5}n}{n+\frac{\frac{79}{126}n}{n+\frac{\frac{-79}{126}n}{n+{a}_{8}}}}}}}}\\ =& \frac{-237+1,405{a}_{8}+1,800n+1,740{a}_{8}n-630{n}^{2}+3,780{a}_{8}{n}^{2}+3,780{n}^{3}}{6\left(79{a}_{8}+600{a}_{8}n+600{n}^{2}+790{a}_{8}{n}^{2}+1,260{a}_{8}{n}^{3}+1,260{n}^{4}\right)}.\end{array}$
(2.9)

By using Mathematica software (Mathematica Program is very similar to the one given in Remark 3; however, it has a parameter ${a}_{8}$), we obtain

$\begin{array}{c}{R}_{8}\left(n+1\right)-{R}_{8}\left(n\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{1}{2{n}^{2}}+\frac{2}{3{n}^{3}}-\frac{3}{4{n}^{4}}+\frac{4}{5{n}^{5}}-\frac{5}{6{n}^{6}}+\frac{6}{7{n}^{7}}-\frac{7}{8{n}^{8}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{360,030-6,241{a}_{8}}{396,900{n}^{9}}+\frac{-346,440+24,964{a}_{8}+6,241{a}_{8}^{2}}{352,800{n}^{10}}+O\left(\frac{1}{{n}^{11}}\right).\hfill \end{array}$
(2.10)

Substituting (2.8) and (2.10) into (2.7), we get

$\begin{array}{rcl}{r}_{8}\left(n\right)-{r}_{8}\left(n+1\right)& =& \left(-\frac{8}{9}+\frac{360,030-6,241{a}_{8}}{396,900}\right)\frac{1}{{n}^{9}}\\ +\left(\frac{9}{10}+\frac{-346,440+24,964{a}_{8}+6,241{a}_{8}^{2}}{352,800}\right)\frac{1}{{n}^{10}}+O\left(\frac{1}{{n}^{11}}\right).\end{array}$
(2.11)

The fastest possible sequence ${\left({r}_{8}\left(n\right)\right)}_{n\in \mathbb{N}}$ is obtained only for ${a}_{8}=\frac{7,230}{6,241}$. At the same time, it follows from (2.11) that

${r}_{8}\left(n\right)-{r}_{8}\left(n+1\right)=\frac{58,081}{2,446,472}\frac{1}{{n}^{10}}+O\left(\frac{1}{{n}^{11}}\right),$
(2.12)

the rate of convergence of ${\left({r}_{8}\left(n\right)-\gamma \right)}_{n\in \mathbb{N}}$ is ${n}^{-9}$ since

$\underset{n\to \mathrm{\infty }}{lim}{n}^{9}\left({r}_{8}\left(n\right)-\gamma \right)=-\frac{58,081}{22,018,248}.$

We can use the above approach to find ${a}_{k}$ ($3\le k\le 8$). Unfortunately, it does not work well for ${a}_{9}$. Since ${a}_{3}=-{a}_{2}$, ${a}_{5}=-{a}_{4}$ and ${a}_{7}=-{a}_{6}$. So, we may conjecture ${a}_{9}=-{a}_{8}$. Now let us check it carefully.

Step 2: Check ${a}_{9}=-\frac{7,230}{6,241}$ to ${a}_{13}=-\frac{306,232,774,533}{179,081,182,865}$.

Let ${a}_{1},\dots ,{a}_{9}$ and ${R}_{9}\left(n\right)$ be defined in Theorem 1. Applying Mathematica software, we obtain

$\begin{array}{rcl}{R}_{9}\left(n+1\right)-{R}_{9}\left(n\right)& =& -\frac{1}{2{n}^{2}}+\frac{2}{3{n}^{3}}-\frac{3}{4{n}^{4}}+\frac{4}{5{n}^{5}}-\frac{5}{6{n}^{6}}+\frac{6}{7{n}^{7}}-\frac{7}{8{n}^{8}}+\frac{8}{9}\frac{1}{{n}^{9}}\\ -\frac{9}{10}\frac{1}{{n}^{10}}+\frac{736,265}{836,136}\frac{1}{{n}^{11}}+O\left(\frac{1}{{n}^{12}}\right),\end{array}$
(2.13)

which is the desired result. Substituting (2.8) and (2.13) into (2.7), we get

${r}_{9}\left(n\right)-{r}_{9}\left(n+1\right)=-\frac{262,445}{9,197,496}\frac{1}{{n}^{11}}+O\left(\frac{1}{{n}^{12}}\right),$
(2.14)

the rate of convergence of ${\left({r}_{9}\left(n\right)-\gamma \right)}_{n\in \mathbb{N}}$ is ${n}^{-10}$ since

$\underset{n\to \mathrm{\infty }}{lim}{n}^{10}\left({r}_{9}\left(n\right)-\gamma \right)=-\frac{262,445}{91,974,960}.$

Next, we can use Step 1 to find ${a}_{10}$, and Step 2 to check ${a}_{11}$ and ${a}_{12}$. It should be noted that Theorem 2 will provide the other proofs for ${a}_{10}$ and ${a}_{11}$. So we omit the details here.

Finally, we check ${a}_{13}=-\frac{306,232,774,533}{179,081,182,865}$.

$\begin{array}{c}{R}_{13}\left(n+1\right)-{R}_{13}\left(n\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{1}{2{n}^{2}}+\frac{2}{3{n}^{3}}-\frac{3}{4{n}^{4}}+\frac{4}{5{n}^{5}}-\frac{5}{6{n}^{6}}+\frac{6}{7{n}^{7}}-\frac{7}{8{n}^{8}}+\frac{8}{9}\frac{1}{{n}^{9}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\frac{9}{10}\frac{1}{{n}^{10}}+\frac{10}{11}\frac{1}{{n}^{11}}-\frac{11}{12}\frac{1}{{n}^{12}}+\frac{12}{13}\frac{1}{{n}^{13}}-\frac{13}{14}\frac{1}{{n}^{14}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{1,903,648,586,623}{2,576,034,146,400}\frac{1}{{n}^{15}}+O\left(\frac{1}{{n}^{16}}\right).\hfill \end{array}$
(2.15)

Substituting (2.8) and (2.15) into (2.7), one has

${r}_{13}\left(n\right)-{r}_{13}\left(n+1\right)=-\frac{500,649,950,017}{2,576,034,146,400}\frac{1}{{n}^{15}}+O\left(\frac{1}{{n}^{16}}\right).$
(2.16)

Since

$\underset{n\to \mathrm{\infty }}{lim}{n}^{14}\left({r}_{13}\left(n\right)-\gamma \right)=-\frac{71,521,421,431}{5,152,068,292,800},$

thus the rate of convergence of ${\left({r}_{13}\left(n\right)-\gamma \right)}_{n\in \mathbb{N}}$ is ${n}^{-14}$.

This completes the proof of Theorem 1.

Remark 3 In fact, if the assertion ${a}_{13}=-\frac{306,232,774,533}{179,081,182,865}$ holds, then the other values ${a}_{j}$ ($1\le j\le 12$) must be true. The following Mathematica Program will generate ${R}_{13}\left(n+1\right)-{R}_{13}\left(n\right)$ into power series in $\frac{1}{n}$ with order 16: $\mathrm{Normal}\left[\mathrm{Series}\left[\left({R}_{13}\left[n+1\right]-{R}_{13}\left[n\right]\right)/.\phantom{\rule{0.25em}{0ex}}n\to 1/x,\left\{x,0,16\right\}\right]\right]/$. $x\to 1/n$.

Remark 4 It is a very interesting question to find ${a}_{k}$ for $k\ge 14$. However, it seems impossible by the above method.

## 3 The proof of Theorem 2

Before we prove Theorem 2, let us give a simple inequality by the Hermite-Hadamard inequality, which plays an important role in the proof.

Lemma 2 Let f be twice derivable with ${f}^{″}$ continuous. If ${f}^{″}\left(x\right)>0$, then

${\int }_{a}^{a+1}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx>f\left(a+1/2\right).$
(3.1)

In the sequel, the notation ${P}_{k}\left(x\right)$ means a polynomial of degree k in x with all of its non-zero coefficients positive, which may be different at each occurrence.

Let us begin to prove Theorem 2. Note ${r}_{10}\left(\mathrm{\infty }\right)=0$, it is easy to see

$\gamma -{r}_{10}\left(n\right)=\sum _{m=n}^{\mathrm{\infty }}\left({r}_{10}\left(m+1\right)-{r}_{10}\left(m\right)\right)=\sum _{m=n}^{\mathrm{\infty }}f\left(m\right),$
(3.2)

where

$f\left(m\right)=\frac{1}{m+1}-ln\left(1+\frac{1}{m}\right)-{R}_{10}\left(m+1\right)+{R}_{10}\left(m\right).$

Let ${D}_{1}=\frac{2,755,095,121}{6,762,022,344}$. By using Mathematica software, we have

${f}^{\prime }\left(x\right)+{D}_{1}\frac{1}{{\left(x+1\right)}^{13}}=-\frac{{P}_{19}\left(x\right)\left(x-1\right)+1,619,906,998,377\cdots 5,270,931}{33,810,111,720x{\left(1+x\right)}^{13}{P}_{10}^{\left(1\right)}\left(x\right){P}_{10}^{\left(2\right)}\left(x\right)}<0,$

and

${f}^{\prime }\left(x\right)+{D}_{1}\frac{1}{{\left(x+\frac{1}{2}\right)}^{13}}=\frac{{P}_{22}\left(x\right)}{4,226,263,965x{\left(1+x\right)}^{2}{\left(1+2x\right)}^{13}{P}_{10}^{\left(3\right)}\left(x\right){P}_{10}^{\left(4\right)}\left(x\right)}>0.$

Hence, we get the following inequalities for $x\ge 1$:

${D}_{1}\frac{1}{{\left(x+1\right)}^{13}}<-{f}^{\prime }\left(x\right)<{D}_{1}\frac{1}{{\left(x+\frac{1}{2}\right)}^{13}}.$
(3.3)

Applying $f\left(\mathrm{\infty }\right)=0$, (3.3) and Lemma 2, we get

$\begin{array}{rcl}f\left(m\right)& =& -{\int }_{m}^{\mathrm{\infty }}{f}^{\prime }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le {D}_{1}{\int }_{m}^{\mathrm{\infty }}{\left(x+\frac{1}{2}\right)}^{-13}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{1}}{12}{\left(m+\frac{1}{2}\right)}^{-12}\le \frac{{D}_{1}}{12}{\int }_{m}^{m+1}{x}^{-12}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(3.4)

From (3.1) and (3.4) we obtain

$\begin{array}{rcl}\gamma -{r}_{10}\left(n\right)& \le & \sum _{m=n}^{\mathrm{\infty }}\frac{{D}_{1}}{12}{\int }_{m}^{m+1}{x}^{-12}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{1}}{12}{\int }_{n}^{\mathrm{\infty }}{x}^{-12}\phantom{\rule{0.2em}{0ex}}dx=\frac{{D}_{1}}{132}\frac{1}{{n}^{11}}.\end{array}$
(3.5)

Similarly, we also have

$\begin{array}{rcl}f\left(m\right)& =& -{\int }_{m}^{\mathrm{\infty }}{f}^{\prime }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\ge {D}_{1}{\int }_{m}^{\mathrm{\infty }}{\left(x+1\right)}^{-13}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{1}}{12}{\left(m+1\right)}^{-12}\ge \frac{{D}_{1}}{12}{\int }_{m+1}^{m+2}{x}^{-12}\phantom{\rule{0.2em}{0ex}}dx\end{array}$

and

$\begin{array}{rcl}\gamma -{r}_{10}\left(n\right)& \ge & \sum _{m=n}^{\mathrm{\infty }}\frac{{D}_{1}}{12}{\int }_{m+1}^{m+2}{x}^{-12}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{1}}{12}{\int }_{n+1}^{\mathrm{\infty }}{x}^{-12}\phantom{\rule{0.2em}{0ex}}dx=\frac{{D}_{1}}{132}\frac{1}{{\left(n+1\right)}^{11}}.\end{array}$
(3.6)

Combining (3.5) and (3.6) completes the proof of (1.6).

Note ${r}_{11}\left(\mathrm{\infty }\right)=0$, it is easy to deduce

${r}_{11}\left(n\right)-\gamma =\sum _{m=n}^{\mathrm{\infty }}\left({r}_{11}\left(m\right)-{r}_{11}\left(m+1\right)\right)=\sum _{m=n}^{\mathrm{\infty }}g\left(m\right),$
(3.7)

where

$g\left(m\right)=ln\left(1+\frac{1}{m}\right)-\frac{1}{m+1}-{R}_{11}\left(m\right)+{R}_{11}\left(m+1\right).$

We write ${D}_{2}=\frac{20,169,451}{24,495,240}$. By using Mathematica software, we have

$-{g}^{\prime }\left(x\right)-{D}_{2}\frac{1}{{\left(x+1\right)}^{14}}=\frac{{P}_{18}\left(x\right)}{24,495,240{x}^{3}{\left(1+x\right)}^{14}{P}_{8}^{\left(1\right)}\left(x\right){P}_{8}^{\left(2\right)}\left(x\right)}>0$

and

$\begin{array}{rcl}-{g}^{\prime }\left(x\right)-{D}_{2}\frac{1}{{\left(x+\frac{1}{2}\right)}^{14}}& =& -\frac{{P}_{19}\left(x\right)\left(x-1\right)+4,622,005,677,839,353,997,724,676,307,741}{6,123,810{x}^{3}{\left(1+x\right)}^{3}{\left(1+2x\right)}^{14}{P}_{8}^{\left(3\right)}\left(x\right){P}_{8}^{\left(4\right)}\left(x\right)}\\ <& 0.\end{array}$

Hence, for $x\ge 1$,

${D}_{2}\frac{1}{{\left(x+1\right)}^{14}}<-{g}^{\prime }\left(x\right)<{D}_{2}\frac{1}{{\left(x+\frac{1}{2}\right)}^{14}}.$
(3.8)

Applying $g\left(\mathrm{\infty }\right)=0$, (3.8) and (3.1), we get

$\begin{array}{rcl}g\left(m\right)& =& -{\int }_{m}^{\mathrm{\infty }}{g}^{\prime }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le {D}_{2}{\int }_{m}^{\mathrm{\infty }}{\left(x+\frac{1}{2}\right)}^{-14}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{2}}{13}{\left(m+\frac{1}{2}\right)}^{-13}\le \frac{{D}_{2}}{13}{\int }_{m}^{m+1}{x}^{-13}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(3.9)

It follows from (3.7) and (3.9) that

$\begin{array}{rcl}{r}_{11}\left(n\right)-\gamma & \le & \sum _{m=n}^{\mathrm{\infty }}\frac{{D}_{2}}{13}{\int }_{m}^{m+1}{x}^{-13}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{2}}{13}{\int }_{n}^{\mathrm{\infty }}{x}^{-13}\phantom{\rule{0.2em}{0ex}}dx=\frac{{D}_{2}}{156}\frac{1}{{n}^{12}}.\end{array}$
(3.10)

Finally,

$\begin{array}{rcl}g\left(m\right)& =& -{\int }_{m}^{\mathrm{\infty }}{g}^{\prime }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\ge {D}_{2}{\int }_{m}^{\mathrm{\infty }}{\left(x+1\right)}^{-14}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{2}}{13}{\left(m+1\right)}^{-13}\ge \frac{{D}_{2}}{13}{\int }_{m+1}^{m+2}{x}^{-13}\phantom{\rule{0.2em}{0ex}}dx\end{array}$

and

$\begin{array}{rcl}{r}_{11}\left(n\right)-\gamma & \ge & \sum _{m=n}^{\mathrm{\infty }}\frac{{D}_{2}}{13}{\int }_{m+1}^{m+2}{x}^{-13}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{D}_{2}}{13}{\int }_{n+1}^{\mathrm{\infty }}{x}^{-13}\phantom{\rule{0.2em}{0ex}}dx=\frac{{D}_{2}}{156}\frac{1}{{\left(n+1\right)}^{12}}.\end{array}$
(3.11)

Combining (3.10) and (3.11) completes the proof of (1.7).

Remark 5 As an example, we give Mathematica Program for the proof of the left-hand side of (3.3):

1. (i)

Together $\left[D\left[f\left[x\right],\left\{x,1\right\}\right]+{D}_{1}{\left(x+1\right)}^{13}\right]$;

2. (ii)

Take out the numerator $P\left[x\right]$ of the above rational function, then manipulate the program: Apart $\left[P\left[x\right]/\left(x-1\right)\right]$.

## Appendix

For the reader’s convenience, we rewrite ${R}_{k}\left(n\right)$ ($k\le 13$) with minimal denominators as follows.

$\begin{array}{c}{R}_{1}\left(n\right)=\frac{1}{2n},\hfill \\ {R}_{3}\left(n\right)=\frac{1}{2n}-\frac{1}{12}\frac{1}{{n}^{2}},\hfill \\ {R}_{5}\left(n\right)=\frac{1}{2n}-\frac{5}{6\left(1+10{n}^{2}\right)},\hfill \\ {R}_{7}\left(n\right)=\frac{1}{2n}-\frac{79}{1,200}\frac{1}{{n}^{2}}-\frac{147}{400\left(10+21{n}^{2}\right)},\hfill \\ {R}_{9}\left(n\right)=\frac{1}{2n}-\frac{7\left(871+790{n}^{2}\right)}{20\left(241+3,990{n}^{2}+3,318{n}^{4}\right)},\hfill \\ {R}_{11}\left(n\right)=\frac{1}{2n}-\frac{52,489}{894,348}\frac{1}{{n}^{2}}-\frac{1,237,227,621+584,280,400{n}^{2}}{4,471,740\left(3,549+13,020{n}^{2}+5,302{n}^{4}\right)},\hfill \\ {R}_{13}\left(n\right)=\frac{1}{2n}-\frac{39,577,260,671+66,288,226,620{n}^{2}+15,762,446,700{n}^{4}}{1,260\left(20,169,451+434,410,620{n}^{2}+646,328,298{n}^{4}+150,118,540{n}^{6}\right)},\hfill \\ {R}_{2}\left(n\right)=\frac{3}{6n+1},\hfill \\ {R}_{4}\left(n\right)=\frac{13+30n}{6\left(1+6n+10{n}^{2}\right)},\hfill \\ {R}_{6}\left(n\right)=\frac{5\left(281+348n+756{n}^{2}\right)}{6\left(79+600n+790{n}^{2}+1,260{n}^{3}\right)},\hfill \\ {R}_{8}\left(n\right)=\frac{964,337+2,646,000n+2,599,730{n}^{2}+2,621,220{n}^{3}}{20\left(19,039+144,600n+315,210{n}^{2}+303,660{n}^{3}+262,122{n}^{4}\right)},\hfill \\ {R}_{10}\left(n\right)=\left(7\left(108,237,701+208,886,046n+523,341,290{n}^{2}\hfill \\ \phantom{{R}_{10}\left(n\right)=}+210,464,400{n}^{3}+230,000,760{n}^{4}\right)\right)\hfill \\ \phantom{{R}_{10}\left(n\right)=}/\left(20\left(12,649,849+107,768,934n+209,431,110{n}^{2}\hfill \\ \phantom{{R}_{10}\left(n\right)=}+395,365,320{n}^{3}+174,158,502{n}^{4}+161,000,532{n}^{5}\right)\right),\hfill \\ {R}_{12}\left(n\right)=\left(3,604,759,235,968,501+11,032,319,618,513,046n\hfill \\ \phantom{{R}_{12}\left(n\right)=}+17,366,281,558,290,420{n}^{2}+19,958,033,982,902,400{n}^{3}\hfill \\ \phantom{{R}_{12}\left(n\right)=}+7,661,417,445,218,460{n}^{4}+4,964,130,389,017,800{n}^{5}\right)\hfill \\ \phantom{{R}_{12}\left(n\right)=}/\left(1,260\left(1,058,674,313,539+9,019,254,081,474n\hfill \\ \phantom{{R}_{12}\left(n\right)=}+22,801,779,033,180{n}^{2}+33,088,387,754,520{n}^{3}+33,925,126,033,722{n}^{4}\hfill \\ \phantom{{R}_{12}\left(n\right)=}+13,474,242,079,452{n}^{5}+7,879,572,046,060{n}^{6}\right)\right).\hfill \end{array}$