Continued fraction inequalities for the Euler-Mascheroni constant

Abstract

The aim of this paper is to establish new inequalities for the Euler-Mascheroni constant by the continued fraction method.

MSC:11Y60, 41A25, 41A20.

1 Introduction

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

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

(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 ${(\gamma (n))}_{n\in \mathbb{N}}$ converges very slowly toward γ, like ${(2n)}^{-1}$. Up to now, many authors have been preoccupied with improving its rate of convergence (see, e.g., [2, 5–22] and the references therein). We list some main results as follows:

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

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

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

and proved

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

(1.2)

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

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

$$\frac{1}{72{(n+1)}^3}<\gamma -r_2(n)<\frac{1}{72n^3},$$

(1.3)

$$\frac{1}{120{(n+1)}^4}<r_3(n)-\gamma <\frac{1}{120{(n-1)}^4}.$$

(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:

$$r(n)=1+\frac{1}{2}+\cdots +\frac{1}{n}-lnn-\frac{a_1}{n+\frac{a_{2}n}{n+\frac{a_{3}n}{n+\ddots }}},$$

where $(a_1,a_2,a_4,a_6,a_8,a_{10},a_{12})=(\frac{1}{2},\frac{1}{6},\frac{3}{5},\frac{79}{126},\frac{7,230}{6,241},\frac{4,146,631}{3,833,346},\frac{306,232,774,533}{179,081,182,865})$, and $a_{2k+1}=-a_{2k}$ for $1\le k\le 6$.

Let

$$R_k(n):=\frac{a_1}{n+\frac{a_{2}n}{n+\frac{a_{3}n}{n+\frac{a_{4}n}{\frac{\ddots }{n+a_k}}}}}$$

(see the Appendix for their simple expressions) and

$$r_k(n):=\sum _{m=1}^n\frac{1}{m}-lnn-R_k(n).$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{k+1}(r_k(n)-\gamma )=C_k,$$

(1.5)

where

$$\begin{array}{rcl}(C_1,\dots ,C_{13})& =& (-\frac{1}{12},-\frac{1}{72},\frac{1}{120},\frac{1}{200},-\frac{79}{25,200},-\frac{6,241}{3,175,200},\frac{241}{105,840},\frac{58,081}{22,018,248},\\ -\frac{262,445}{91,974,960},-\frac{2,755,095,121}{892,586,949,408},\frac{20,169,451}{3,821,257,440},\\ \frac{406,806,753,641,401}{45,071,152,103,463,200},-\frac{71,521,421,431}{5,152,068,292,800}).\end{array}$$

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 $r_{10}(n)$, $r_{11}(n)$, $C_{10}$ and $C_{11}$ be defined in Theorem  1, then

$$C_{10}\frac{1}{{(n+1)}^{11}}<\gamma -r_{10}(n)<C_{10}\frac{1}{n^{11}},$$

(1.6)

$$C_{11}\frac{1}{{(n+1)}^{12}}<r_{11}(n)-\gamma <C_{11}\frac{1}{n^{12}}.$$

(1.7)

Remark 1 In fact, Theorem 2 implies that $r_{10}(n)$ is a strictly increasing function of n, whereas $r_{11}(n)$ is a strictly decreasing function of n. Certainly, it has similar inequalities for $r_k(n)$ ($1\le k\le 9$), we leave these for readers to verify. It is also should be noted that (1.4) cannot deduce the monotonicity of $r_3(n)$.

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 ${(x_n)}_{n\in \mathbb{N}}$ is convergent to zero and there exists the limit

$$\underset{n\to +\mathrm{\infty }}{lim}{n}^s(x_n-x_{n+1})=l\in [-\mathrm{\infty },+\mathrm{\infty }]$$

(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

$$r_1(n)=\sum _{m=1}^n\frac{1}{m}-lnn-\frac{a_1}{n},$$

(2.3)

here we note $R_1(n)=a_1/n$. To measure the accuracy of this approximation, we usually say that approximation (2.3) is better as $r_1(n)-\gamma$ faster converges to zero. Clearly,

$$r_1(n)-r_1(n+1)=ln(1+\frac{1}{n})-\frac{1}{n+1}+\frac{a_1}{n+1}-\frac{a_1}{n}.$$

(2.4)

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

$$ln(1+x)=\sum _{m=1}^{\mathrm{\infty }}{(-1)}^{m-1}\frac{x^m}{m}\text{and}\frac{1}{1-x}=\sum _{m=0}^{\mathrm{\infty }}{x}^m.$$

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

$$r_1(n)-r_1(n+1)=(\frac{1}{2}-a_1)\frac{1}{n^2}+(a_1-\frac{2}{3})\frac{1}{n^3}+(\frac{3}{4}-a_1)\frac{1}{n^4}+O\left(\frac{1}{n^5}\right).$$

(2.5)

From Lemma 1, we see that the rate of convergence of the sequence ${(r_1(n)-\gamma )}_{n\in \mathbb{N}}$ is even higher than the value s satisfying (2.1). By Lemma 1, we have

• (i) If $a_1\ne \frac{1}{2}$, then the rate of convergence of ${(r_1(n)-\gamma )}_{n\in \mathbb{N}}$ is $n^{-1}$ since

  $$\underset{n\to \mathrm{\infty }}{lim}n(r_1(n)-\gamma )=\frac{1}{2}-a_1\ne 0.$$

• (ii) If $a_1=\frac{1}{2}$, from (2.5) we have

  $$r_1(n)-r_1(n+1)=-\frac{1}{6}\frac{1}{n^3}+O\left(\frac{1}{n^4}\right).$$

Hence the rate of convergence of ${(r_1(n)-\gamma )}_{n\in \mathbb{N}}$ is $n^{-2}$ since

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2(r_1(n)-\gamma )=-\frac{1}{12}.$$

We also observe that the fastest possible sequence ${(r_1(n))}_{n\in \mathbb{N}}$ 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

$$r_k(n)=\sum _{m=1}^n\frac{1}{m}-lnn-R_k(n),$$

(2.6)

then

$$r_k(n)-r_k(n+1)=ln(1+\frac{1}{n})-\frac{1}{n+1}+R_k(n+1)-R_k(n).$$

(2.7)

It is easy to get the following power series:

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

(2.8)

Hence the key step is to expand $R_k(n+1)-R_k(n)$ 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

$$\begin{array}{rcl}{R}_8(n)& =& \frac{\frac{1}{2}}{n+\frac{\frac{n}{6}}{n+\frac{-\frac{n}{6}}{n+\frac{\frac{3}{5}n}{n+\frac{\frac{-3}{5}n}{n+\frac{\frac{79}{126}n}{n+\frac{\frac{-79}{126}n}{n+a_8}}}}}}}\\ =& \frac{-237+1,405a_8+1,800n+1,740a_{8}n-630n^2+3,780a_8{n}^2+3,780n^3}{6(79a_8+600a_{8}n+600n^2+790a_8{n}^2+1,260a_8{n}^3+1,260n^4)}.\end{array}$$

(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

$$\begin{array}{c}{R}_8(n+1)-R_8(n)\hfill \\ =-\frac{1}{2n^2}+\frac{2}{3n^3}-\frac{3}{4n^4}+\frac{4}{5n^5}-\frac{5}{6n^6}+\frac{6}{7n^7}-\frac{7}{8n^8}\hfill \\ +\frac{360,030-6,241a_8}{396,900n^9}+\frac{-346,440+24,964a_8+6,241a_8^2}{352,800n^{10}}+O\left(\frac{1}{n^{11}}\right).\hfill \end{array}$$

(2.10)

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

$$\begin{array}{rcl}{r}_8(n)-r_8(n+1)& =& (-\frac{8}{9}+\frac{360,030-6,241a_8}{396,900})\frac{1}{n^9}\\ +(\frac{9}{10}+\frac{-346,440+24,964a_8+6,241a_8^2}{352,800})\frac{1}{n^{10}}+O\left(\frac{1}{n^{11}}\right).\end{array}$$

(2.11)

The fastest possible sequence ${(r_8(n))}_{n\in \mathbb{N}}$ is obtained only for $a_8=\frac{7,230}{6,241}$. At the same time, it follows from (2.11) that

$$r_8(n)-r_8(n+1)=\frac{58,081}{2,446,472}\frac{1}{n^{10}}+O\left(\frac{1}{n^{11}}\right),$$

(2.12)

the rate of convergence of ${(r_8(n)-\gamma )}_{n\in \mathbb{N}}$ is $n^{-9}$ since

$$\underset{n\to \mathrm{\infty }}{lim}{n}^9(r_8(n)-\gamma )=-\frac{58,081}{22,018,248}.$$

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 $R_9(n)$ be defined in Theorem 1. Applying Mathematica software, we obtain

$$\begin{array}{rcl}{R}_9(n+1)-R_9(n)& =& -\frac{1}{2n^2}+\frac{2}{3n^3}-\frac{3}{4n^4}+\frac{4}{5n^5}-\frac{5}{6n^6}+\frac{6}{7n^7}-\frac{7}{8n^8}+\frac{8}{9}\frac{1}{n^9}\\ -\frac{9}{10}\frac{1}{n^{10}}+\frac{736,265}{836,136}\frac{1}{n^{11}}+O\left(\frac{1}{n^{12}}\right),\end{array}$$

(2.13)

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

$$r_9(n)-r_9(n+1)=-\frac{262,445}{9,197,496}\frac{1}{n^{11}}+O\left(\frac{1}{n^{12}}\right),$$

(2.14)

the rate of convergence of ${(r_9(n)-\gamma )}_{n\in \mathbb{N}}$ is $n^{-10}$ since

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{10}(r_9(n)-\gamma )=-\frac{262,445}{91,974,960}.$$

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 $a_{13}=-\frac{306,232,774,533}{179,081,182,865}$.

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

(2.15)

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

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

(2.16)

Since

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

thus the rate of convergence of ${(r_{13}(n)-\gamma )}_{n\in \mathbb{N}}$ 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 $R_{13}(n+1)-R_{13}(n)$ into power series in $\frac{1}{n}$ with order 16: $\mathrm{Normal}[\mathrm{Series}[(R_{13}[n+1]-R_{13}[n])/.n\to 1/x,\{x,0,16\}]]/$. $x\to 1/n$.

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 $f^{″}(x)>0$, then

$${\int }_a^{a+1}f(x)dx>f(a+1/2).$$

(3.1)

In the sequel, the notation $P_k(x)$ 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 $r_{10}(\mathrm{\infty })=0$, it is easy to see

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

(3.2)

where

$$f(m)=\frac{1}{m+1}-ln(1+\frac{1}{m})-R_{10}(m+1)+R_{10}(m).$$

Let $D_1=\frac{2,755,095,121}{6,762,022,344}$. By using Mathematica software, we have

$$f^{\prime }(x)+D_1\frac{1}{{(x+1)}^{13}}=-\frac{P_{19}(x)(x-1)+1,619,906,998,377\cdots 5,270,931}{33,810,111,720x{(1+x)}^{13}{P}_{10}^{(1)}(x)P_{10}^{(2)}(x)}<0,$$

and

$$f^{\prime }(x)+D_1\frac{1}{{(x+\frac{1}{2})}^{13}}=\frac{P_{22}(x)}{4,226,263,965x{(1+x)}^2{(1+2x)}^{13}{P}_{10}^{(3)}(x)P_{10}^{(4)}(x)}>0.$$

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

$$D_1\frac{1}{{(x+1)}^{13}}<-f^{\prime }(x)<D_1\frac{1}{{(x+\frac{1}{2})}^{13}}.$$

(3.3)

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

$$\begin{array}{rcl}f(m)& =& -{\int }_m^{\mathrm{\infty }}{f}^{\prime }(x)dx\le D_1{\int }_m^{\mathrm{\infty }}{(x+\frac{1}{2})}^{-13}dx\\ =& \frac{D_1}{12}{(m+\frac{1}{2})}^{-12}\le \frac{D_1}{12}{\int }_m^{m+1}{x}^{-12}dx.\end{array}$$

(3.4)

From (3.1) and (3.4) we obtain

$$\begin{array}{rcl}\gamma -r_{10}(n)& \le & \sum _{m=n}^{\mathrm{\infty }}\frac{D_1}{12}{\int }_m^{m+1}{x}^{-12}dx\\ =& \frac{D_1}{12}{\int }_n^{\mathrm{\infty }}{x}^{-12}dx=\frac{D_1}{132}\frac{1}{n^{11}}.\end{array}$$

(3.5)

Similarly, we also have

$$\begin{array}{rcl}f(m)& =& -{\int }_m^{\mathrm{\infty }}{f}^{\prime }(x)dx\ge D_1{\int }_m^{\mathrm{\infty }}{(x+1)}^{-13}dx\\ =& \frac{D_1}{12}{(m+1)}^{-12}\ge \frac{D_1}{12}{\int }_{m+1}^{m+2}{x}^{-12}dx\end{array}$$

and

$$\begin{array}{rcl}\gamma -r_{10}(n)& \ge & \sum _{m=n}^{\mathrm{\infty }}\frac{D_1}{12}{\int }_{m+1}^{m+2}{x}^{-12}dx\\ =& \frac{D_1}{12}{\int }_{n+1}^{\mathrm{\infty }}{x}^{-12}dx=\frac{D_1}{132}\frac{1}{{(n+1)}^{11}}.\end{array}$$

(3.6)

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

Note $r_{11}(\mathrm{\infty })=0$, it is easy to deduce

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

(3.7)

where

$$g(m)=ln(1+\frac{1}{m})-\frac{1}{m+1}-R_{11}(m)+R_{11}(m+1).$$

We write $D_2=\frac{20,169,451}{24,495,240}$. By using Mathematica software, we have

$$-g^{\prime }(x)-D_2\frac{1}{{(x+1)}^{14}}=\frac{P_{18}(x)}{24,495,240x^3{(1+x)}^{14}{P}_8^{(1)}(x)P_8^{(2)}(x)}>0$$

and

$$\begin{array}{rcl}-g^{\prime }(x)-D_2\frac{1}{{(x+\frac{1}{2})}^{14}}& =& -\frac{P_{19}(x)(x-1)+4,622,005,677,839,353,997,724,676,307,741}{6,123,810x^3{(1+x)}^3{(1+2x)}^{14}{P}_8^{(3)}(x)P_8^{(4)}(x)}\\ <& 0.\end{array}$$

Hence, for $x\ge 1$,

$$D_2\frac{1}{{(x+1)}^{14}}<-g^{\prime }(x)<D_2\frac{1}{{(x+\frac{1}{2})}^{14}}.$$

(3.8)

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

$$\begin{array}{rcl}g(m)& =& -{\int }_m^{\mathrm{\infty }}{g}^{\prime }(x)dx\le D_2{\int }_m^{\mathrm{\infty }}{(x+\frac{1}{2})}^{-14}dx\\ =& \frac{D_2}{13}{(m+\frac{1}{2})}^{-13}\le \frac{D_2}{13}{\int }_m^{m+1}{x}^{-13}dx.\end{array}$$

(3.9)

It follows from (3.7) and (3.9) that

$$\begin{array}{rcl}{r}_{11}(n)-\gamma & \le & \sum _{m=n}^{\mathrm{\infty }}\frac{D_2}{13}{\int }_m^{m+1}{x}^{-13}dx\\ =& \frac{D_2}{13}{\int }_n^{\mathrm{\infty }}{x}^{-13}dx=\frac{D_2}{156}\frac{1}{n^{12}}.\end{array}$$

(3.10)

Finally,

$$\begin{array}{rcl}g(m)& =& -{\int }_m^{\mathrm{\infty }}{g}^{\prime }(x)dx\ge D_2{\int }_m^{\mathrm{\infty }}{(x+1)}^{-14}dx\\ =& \frac{D_2}{13}{(m+1)}^{-13}\ge \frac{D_2}{13}{\int }_{m+1}^{m+2}{x}^{-13}dx\end{array}$$

and

$$\begin{array}{rcl}{r}_{11}(n)-\gamma & \ge & \sum _{m=n}^{\mathrm{\infty }}\frac{D_2}{13}{\int }_{m+1}^{m+2}{x}^{-13}dx\\ =& \frac{D_2}{13}{\int }_{n+1}^{\mathrm{\infty }}{x}^{-13}dx=\frac{D_2}{156}\frac{1}{{(n+1)}^{12}}.\end{array}$$

(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):

• (i) Together $[D[f[x],\{x,1\}]+D_1{(x+1)}^{13}]$;

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

Appendix

For the reader’s convenience, we rewrite $R_k(n)$ ($k\le 13$) with minimal denominators as follows.

$$\begin{array}{c}{R}_1(n)=\frac{1}{2n},\hfill \\ R_3(n)=\frac{1}{2n}-\frac{1}{12}\frac{1}{n^2},\hfill \\ R_5(n)=\frac{1}{2n}-\frac{5}{6(1+10n^2)},\hfill \\ R_7(n)=\frac{1}{2n}-\frac{79}{1,200}\frac{1}{n^2}-\frac{147}{400(10+21n^2)},\hfill \\ R_9(n)=\frac{1}{2n}-\frac{7(871+790n^2)}{20(241+3,990n^2+3,318n^4)},\hfill \\ R_{11}(n)=\frac{1}{2n}-\frac{52,489}{894,348}\frac{1}{n^2}-\frac{1,237,227,621+584,280,400n^2}{4,471,740(3,549+13,020n^2+5,302n^4)},\hfill \\ R_{13}(n)=\frac{1}{2n}-\frac{39,577,260,671+66,288,226,620n^2+15,762,446,700n^4}{1,260(20,169,451+434,410,620n^2+646,328,298n^4+150,118,540n^6)},\hfill \\ R_2(n)=\frac{3}{6n+1},\hfill \\ R_4(n)=\frac{13+30n}{6(1+6n+10n^2)},\hfill \\ R_6(n)=\frac{5(281+348n+756n^2)}{6(79+600n+790n^2+1,260n^3)},\hfill \\ R_8(n)=\frac{964,337+2,646,000n+2,599,730n^2+2,621,220n^3}{20(19,039+144,600n+315,210n^2+303,660n^3+262,122n^4)},\hfill \\ R_{10}(n)=\left(7\right(108,237,701+208,886,046n+523,341,290n^2\hfill \\ \phantom{R_{10}(n)=}+210,464,400n^3+230,000,760n^4\left)\right)\hfill \\ \phantom{R_{10}(n)=}/\left(20\right(12,649,849+107,768,934n+209,431,110n^2\hfill \\ \phantom{R_{10}(n)=}+395,365,320n^3+174,158,502n^4+161,000,532n^5\left)\right),\hfill \\ R_{12}(n)=(3,604,759,235,968,501+11,032,319,618,513,046n\hfill \\ \phantom{R_{12}(n)=}+17,366,281,558,290,420n^2+19,958,033,982,902,400n^3\hfill \\ \phantom{R_{12}(n)=}+7,661,417,445,218,460n^4+4,964,130,389,017,800n^5)\hfill \\ \phantom{R_{12}(n)=}/(1,260(1,058,674,313,539+9,019,254,081,474n\hfill \\ \phantom{R_{12}(n)=}+22,801,779,033,180n^2+33,088,387,754,520n^3+33,925,126,033,722n^4\hfill \\ \phantom{R_{12}(n)=}+13,474,242,079,452n^5+7,879,572,046,060n^6\left)\right).\hfill \end{array}$$