1 Introduction

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

$$\begin{aligned} \gamma(n):=\sum_{m=1}^{n}\frac{1}{m} -\ln n. \end{aligned}$$
(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 are preoccupied to improve its rate of convergence; see, for example, [2, 5–14] and references therein. We list some main results:

$$\begin{aligned} &\sum_{m=1}^{n}\frac{1}{m} -\ln \biggl(n+\frac{1}{2} \biggr)=\gamma+O\bigl(n^{-2}\bigr) \quad \mbox{(DeTemple [6]),} \\ &\sum_{m=1}^{n}\frac{1}{m} -\ln \frac{n^{3}+\frac{3}{2}n^{2}+\frac{227}{240}+\frac{107}{480}}{ n^{2}+n+\frac{97}{240}}=\gamma+O\bigl(n^{-6}\bigr) \quad\mbox{(Mortici [13]),} \\ & \begin{aligned} &\sum_{m=1}^{n}\frac{1}{m} -\ln \biggl( 1+\frac{1}{2n}+\frac{1}{24n^{2}}-\frac{1}{48n^{3}} +\frac{23}{5760n^{4}} \biggr)=\gamma+O\bigl(n^{-5}\bigr)\\ & \quad\mbox{(Chen and Mortici [5]).} \end{aligned} \end{aligned}$$

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

$$\begin{aligned} \nu(n)&=\sum_{m=1}^{n}\frac{1}{m} - \frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &\quad{}- \biggl(\frac{-\frac{1}{180}}{ (n^{2}+n+\frac{1}{3} )^{2}} +\frac{\frac{8}{2835}}{ (n^{2}+n+\frac{1}{3} )^{3}} +\frac{\frac{5}{1512}}{ (n^{2}+n+\frac{1}{3} )^{4}} + \frac{\frac{592}{93\text{,}555}}{ (n^{2}+n+\frac{1}{3} )^{5}} \biggr) \end{aligned}$$

and proved that

$$\begin{aligned} \lim_{n\rightarrow\infty}n^{12} \bigl(\nu(n)-\gamma \bigr)=- \frac {796\text{,}801}{43\text{,}783\text{,}740}. \end{aligned}$$
(1.2)

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

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

$$\begin{aligned} &\frac{1}{72(n+1)^{3}}< \gamma-r_{2}(n)< \frac{1}{72n^{3}}, \end{aligned}$$
(1.3)
$$\begin{aligned} &\frac{1}{120(n+1)^{4}}< r_{3}(n)-\gamma< \frac{1}{120(n-1)^{4}}. \end{aligned}$$
(1.4)

In fact, Lu [9] also found \(a_{4}\) without proof, and his works motivate our study. In this paper, starting from the well-known sequence \(\gamma_{n}\), based on the early works of Mortici, DeTemple, and Lu, we provide some new classes of convergent sequences for the Euler–Mascheroni constant.

Theorem 1

For the Euler–Mascheroni constant, we have the following convergent sequence:

$$\begin{aligned} r(n)=1+\frac{1}{2} +\cdots+\frac{1}{n}-\ln n-\ln \biggl(1+ \frac{a_{1}}{n} \biggr)-\ln \biggl(1+\frac{a_{2}}{n^{2}} \biggr)-\cdots, \end{aligned}$$

where

$$\begin{aligned} &a_{1}=\frac{1}{2},\qquad a_{2}=\frac{1}{24},\qquad a_{3}=-\frac{1}{24},\qquad a_{4}=\frac {143}{5760}, \\ &a_{5}=-\frac{1}{160}, \qquad a_{6}=-\frac{151}{290\text{,}304},\qquad a_{7}=-\frac{1}{896},\qquad \dots. \end{aligned} $$

Let

$$\begin{aligned} r_{k}(n):=\sum_{m=1}^{n} \frac{1}{m}-\ln n-\sum_{m=1}^{k}\ln \biggl(1+\frac {a_{m}}{n^{m}} \biggr). \end{aligned}$$

For \(1\le k\le7\), we have

$$\begin{aligned} \lim_{n\rightarrow\infty}n^{k+2} \bigl(r_{k}(n)-\gamma \bigr)=C_{k}, \end{aligned}$$
(1.5)

where

$$\begin{aligned} &C_{1}=\frac{1}{24},\qquad C_{2}=-\frac{1}{24},\qquad C_{3}=\frac{143}{5760},\qquad C_{4}=-\frac{1}{160},\\ &C_{5}=-\frac{151}{290\text{,}304},\qquad C_{6}=-\frac{1}{896},\qquad C_{7}=\frac{109\text{,}793}{22\text{,}118\text{,}400},\qquad \dots. \end{aligned} $$

Furthermore, for \(r_{2}(n)\) and \(r_{3}(n)\), we also have the following inequalities.

Theorem 2

Let \(r_{2}(n)\) and \(r_{3}(n)\) be as in Theorem 1. Then

$$\begin{aligned} &\frac{1}{24}\frac{1}{(n+1)^{3}}< \gamma-r_{2}(n)< \frac{1}{24}\frac {1}{n^{3}}, \end{aligned}$$
(1.6)
$$\begin{aligned} &\frac{143}{5760}\frac{1}{(n+1)^{4}}< r_{3}(n)-\gamma< \frac{143}{5760}\frac {1}{n^{4}}. \end{aligned}$$
(1.7)

Remark 1

Certainly, there are similar inequalities for \(r_{k}(n)\) (\(1\le k\le7\)); we omit the details.

2 Proof of Theorem 1

The following lemma gives a method for measuring the rate of convergence. This lemma was first used by Mortici [15, 16] for constructing asymptotic expansions or accelerating some convergences. For a proof and other details, see, for example, [16].

Lemma 1

If the sequence \((x_{n})_{n\in\mathbb{N}}\) converges to zero and there exists the limit

$$\begin{aligned} \lim_{n\rightarrow+\infty}n^{s}(x_{n}-x_{n+1})=l \in[-\infty,+\infty] \end{aligned}$$
(2.1)

with \(s>1\), then there exists the limit

$$\begin{aligned} \lim_{n\rightarrow+\infty}n^{s-1}x_{n}=\frac{l}{s-1}. \end{aligned}$$
(2.2)

We need to find the value \(a_{1}\in\mathbb{R}\) that produces the most accurate approximation of the form

$$\begin{aligned} r_{1}(n)=\sum_{m=1}^{n} \frac{1}{m} -\ln n-\ln \biggl(1+\frac{a_{1}}{n} \biggr). \end{aligned}$$
(2.3)

To measure the accuracy of this approximation, we usually say that an approximation (2.3) is better as \(r_{1}(n) -\gamma\) faster converges to zero. Clearly,

$$\begin{aligned} r_{1}(n)-r_{1}(n+1)=\ln \biggl(1+\frac{1}{n} \biggr)-\frac{1}{n+1}+\ln \biggl(1+\frac{a_{1}}{n+1} \biggr)-\ln \biggl(1+ \frac{a_{1}}{n} \biggr). \end{aligned}$$
(2.4)

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

$$\begin{aligned} r_{1}(n)-r_{1}(n+1)= \biggl(\frac{1}{2} -a_{1} \biggr)\frac{1}{n^{2}}+ \biggl(-\frac{2}{3}+a_{1}+{a_{1}}^{2} \biggr)\frac{1}{n^{3}}+O \biggl(\frac{1}{n^{4}} \biggr). \end{aligned}$$
(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 as the value s satisfies (2.1). By Lemma 1 we have

(i) If \(a_{1}\neq1/2\), then the rate of convergence of the \((r_{1}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-2}\), since

$$\begin{aligned} \lim_{n\rightarrow\infty}n \bigl(r_{1}(n)-\gamma \bigr)= \frac{1}{2}-a_{1}\neq0. \end{aligned}$$

(ii) If \(a_{1}= 1/2\), then from (2.5) we have

$$\begin{aligned} r_{1}(n)-r_{1}(n+1)=\frac{1}{12}\frac{1}{n^{3}}+O \biggl(\frac{1}{n^{4}} \biggr). \end{aligned}$$

Hence the rate of convergence of the \((r_{1}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-3}\), since

$$\begin{aligned} \lim_{n\rightarrow\infty}n^{3} \bigl(r_{1}(n)-\gamma \bigr)=\frac{1}{24}:=C_{1}. \end{aligned}$$

We also observe that the fastest possible sequence \((r_{1}(n) )_{n\in\mathbb{N}}\) is obtained only for \(a_{1}=1/2\).

We repeat our approach to determine \(a_{1}\) to \(a_{7}\) step by step. In fact, we can easily compute \(a_{k}\), \(k\leq15\), by the Mathematica software. In this paper, we use the Mathematica software to manipulate symbolic computations.

Let

$$\begin{aligned} r_{k}(n):=\sum_{m=1}^{n} \frac{1}{m}-\ln n-\sum_{m=1}^{k}\ln \biggl(1+\frac {a_{m}}{n^{m}} \biggr). \end{aligned}$$
(2.6)

Then

$$\begin{aligned} &r_{k}(n)-r_{k}(n+1) \\ &\quad =\ln \biggl(1+\frac{1}{n} \biggr)-\frac{1}{n+1}+\sum_{m=1}^{k}\ln \biggl(1+\frac{a_{m}}{(n+1)^{m}} \biggr)-\sum_{m=1}^{k} \ln \biggl(1+\frac{a_{m}}{n^{m}} \biggr). \end{aligned}$$
(2.7)

Hence the key step is to expand \(r_{k}(n)-r_{k}(n+1)\) into power series in \(1/n\). Here we use some examples to explain our method.

Step 1: For example, given \(a_{1}\) to \(a_{4}\), find \(a_{5}\). Define

$$\begin{aligned} r_{5}(n):=\sum_{m=1}^{n} \frac{1}{m}-\ln n-\sum_{m=1}^{5}\ln \biggl(1+\frac {a_{m}}{n^{m}} \biggr). \end{aligned}$$

By using the Mathematica software (the Mathematica Program is very similar to that given further in Remark 2; however, it has the parameter \(a_{8}\)) we obtain

$$\begin{aligned} r_{5}(n)-r_{5}(n+1)= \biggl(-\frac{1}{32}-5a_{5} \biggr)\frac{1}{n^{6}}+ \biggl(\frac{4385}{48\text{,}384}+15a_{5} \biggr) \frac{1}{n^{7}}+O \biggl(\frac {1}{n^{8}} \biggr). \end{aligned}$$
(2.8)

The fastest possible sequence \((r_{5}(n) )_{n\in\mathbb{N}}\) is obtained only for \(a_{5}=-\frac{1}{160}\). At the same time, it follows from (2.8) that

$$\begin{aligned} r_{5}(n)-r_{5}(n+1)=-\frac{151}{48\text{,}384}\frac{1}{n^{7}} +O \biggl(\frac {1}{n^{8}} \biggr). \end{aligned}$$
(2.9)

The rate of convergence of \((r_{8}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-7}\), since

$$\begin{aligned} \lim_{n\rightarrow\infty}n^{7} \bigl(r_{5}(n)-\gamma \bigr)=-\frac{151}{290\text{,}304}:=C_{5}. \end{aligned}$$

We can use this approach to find \(a_{k}\) (\(1\le k\le15\)). From the computations we may the conjecture \(a_{n+1}=C_{n}\), \(n\geq1\). Now, let us check it carefully.

Step 2: Check \(a_{6}=-\frac{151}{290\text{,}304}\) to \(a_{7}=-\frac{1}{896}\).

Let \(a_{1},\dots, a_{6}\), and \(r_{6}(n)\) be defined as in Theorem 1. Applying the Mathematica software, we obtain

$$\begin{aligned} r_{6}(n)-r_{6}(n+1)=-\frac{1}{128}\frac{1}{n^{8}}+O \biggl(\frac{1}{n^{9}} \biggr). \end{aligned}$$
(2.10)

The rate of convergence of \((r_{6}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-8}\), since

$$\begin{aligned} \lim_{n\rightarrow\infty}n^{8} \bigl(r_{6}(n)-\gamma \bigr)=-\frac{1}{896}:=C_{6}. \end{aligned}$$

Finally, we check that \(a_{7}=-\frac{1}{896}\):

$$\begin{aligned} r_{7}(n)-r_{7}(n+1)= \biggl(-\frac{1}{128}-7a_{7} \biggr)\frac{1}{n^{8}} + \biggl(\frac{196\text{,}193}{2\text{,}764\text{,}800}+28a_{7} \biggr) \frac{1}{n^{9}}+O \biggl(\frac {1}{n^{10}} \biggr). \end{aligned}$$
(2.11)

Since \(a_{7}=-\frac{1}{896}\) and

$$\begin{aligned} \lim_{n\rightarrow\infty}n^{9} \bigl(r_{7}(n)-\gamma \bigr)= \frac{109\text{,}793}{22\text{,}118\text{,}400}:=C_{7}, \end{aligned}$$

the rate of convergence of the \((r_{7}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-9}\).

This completes the proof of Theorem 1.

Remark 2

From the computations we can guess that \(a_{n+1}=C_{n}\), \(n\geq1\). It is a very interesting problem to prove this. However, it seems impossible by the provided method.

3 Proof of Theorem 2

Before we prove Theorem 2, let us give a simple inequality, which follows from the Hermite–Hadamard inequality and plays an important role in the proof.

Lemma 2

Let f be a twice continuously differentiable function. If \(f''(x)>0\), then

$$\begin{aligned} \int_{a}^{a+1}f(x) \,dx > f(a+1/2). \end{aligned}$$
(3.1)

By \(P_{k}(x)\) we denote polynomials of degree k in x such that all its nonzero coefficients are positive; it may be different at each occurrence.

Let us prove Theorem 2. Noting that \(r_{2}(\infty)=0\), we easily see that

$$\begin{aligned} \gamma-r_{2}(n)=\sum_{m=n}^{\infty} \bigl(r_{2}(m+1)-r_{2}(m) \bigr) =\sum _{m=n}^{\infty}f(m), \end{aligned}$$
(3.2)

where

$$\begin{aligned} f(m)&=\frac{1}{m+1}-\ln \biggl(1+\frac{1}{m} \biggr)-\ln \biggl(1+ \frac {a_{1}}{m+1} \biggr)-\ln \biggl(1+\frac{a_{2}}{(m+1)^{2}} \biggr)\\ &\quad {}+\ln \biggl(1+ \frac{a_{1}}{m} \biggr)+\ln \biggl(1+\frac{a_{2}}{m^{2}} \biggr). \end{aligned} $$

Let \(D_{1}=1/2\). By using the Mathematica software we have

$$\begin{aligned} &-f'(x)-D_{1}\frac{1}{(x+1)^{5}}\\ &\quad =\frac {300+2739x+11\text{,}434x^{2}+24\text{,}870x^{3}+28\text{,}314x^{4}+15\text{,}936x^{5}+3472x^{6}}{ 2x(1+x)^{5}(1+2x)(3+2x)(1+24x^{2})(25+48x+24x^{2})}>0 \end{aligned}$$

and

$$\begin{aligned} & -f'(x)-D_{1}\frac{1}{(x+\frac{1}{2})^{5}}\\ &\quad =-\frac{P_{6}(x)(x-1)+151\text{,}085}{2x^{5} (1 +x)^{2} (1 + 2x) (3 + 2x) (1 + 24x^{2}) (25 + 48x + 24 x^{2})}< 0. \end{aligned}$$

Hence, we get the following inequalities for \(x\ge1\):

$$\begin{aligned} D_{1}\frac{1}{(x+1)^{5}}< -f'(x)< D_{1} \frac{1}{(x+\frac{1}{2})^{5}}. \end{aligned}$$
(3.3)

Since \(f(\infty)=0\), from the right-hand side of (3.3) and Lemma 2 we get

$$\begin{aligned} f(m)&= - \int_{m}^{\infty}f'(x) \,dx \le D_{1} \int_{m}^{\infty}\biggl( x+\frac{1}{2} \biggr)^{-5} \,dx \\ &=\frac{D_{1}}{4}\biggl(m+\frac{1}{2}\biggr)^{-4} \le \frac{D_{1}}{4} \int _{m}^{m+1}x^{-4} \,dx. \end{aligned}$$
(3.4)

From (3.1) and (3.4) we obtain

$$\begin{aligned} \gamma-r_{2}(n)&\le \sum_{m=n}^{\infty} \frac{D_{1}}{4} \int_{m}^{m+1}x^{-4} \,dx \\ &=\frac{D_{1}}{4} \int_{n}^{\infty}x^{-4} \,dx=\frac{D_{1}}{12} \frac {1}{n^{3}}. \end{aligned}$$
(3.5)

Similarly, we also have

$$\begin{aligned} f(m)&= - \int_{m}^{\infty}f'(x) \,dx \ge D_{1} \int_{m}^{\infty}(x+1)^{-5} \,dx \\ &=\frac{D_{1}}{4}(m+1)^{-4}\ge\frac{D_{1}}{4} \int_{m+1}^{m+2}x^{-4} \,dx \end{aligned}$$

and

$$ \begin{aligned}[b] \gamma-r_{2}(n)&\ge \sum_{m=n}^{\infty} \frac{D_{1}}{4} \int _{m+1}^{m+2}x^{-4} \,dx \\ &=\frac{D_{1}}{4} \int_{n+1}^{\infty}x^{-4} \,dx=\frac{D_{1}}{12} \frac {1}{(n+1)^{3}}. \end{aligned} $$
(3.6)

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

Noting that \(r_{3}(\infty)=0\), we easily deduce

$$\begin{aligned} r_{3}(n)-\gamma=\sum_{m=n}^{\infty} \bigl(r_{3}(m)-r_{3}(m+1) \bigr)=\sum _{m=n}^{\infty}g(m), \end{aligned}$$
(3.7)

where

$$g(m)=r_{3}(m)-r_{3}(m+1). $$

Let \(D_{2}=\frac{143}{288}\). By using the Mathematica software we have

$$\begin{aligned} &-g'(x)-D_{2}\frac{1}{(x+1)^{6}}\\ &\quad =\frac{P_{12}(x)}{288 n (1 + n)^{6} (1 + 2 n) (3 + 2 n) (1 + 24 n^{2}) (25 + 48 n +24 n^{2}) (-1 + 24 n^{3})P_{3}(x)}>0 \end{aligned}$$

and

$$\begin{aligned} &-g'(x)-D_{2}\frac{1}{(x+\frac{1}{2})^{6}}\\ &\quad =-\frac{P_{12}(x)(x-4)+2\text{,}052\text{,}948\text{,}001\text{,}087\text{,}775 }{9 x (1 + x)^{2} (1 + 2 x)^{6} (3 + 2 x) (1 + 24 x^{2}) (25 + 48 x + 24 x^{2}) (-1 + 24 x^{3})P_{3}(x)}< 0. \end{aligned}$$

Hence, for \(x\ge4\),

$$\begin{aligned} D_{2}\frac{1}{(n+1)^{6}}< -g'(x)< D_{2} \frac{1}{(x+\frac{1}{2})^{6}}. \end{aligned}$$
(3.8)

Since \(g(\infty)=0\), by (3.8) we get

$$\begin{aligned} g(m)&= - \int_{m}^{\infty}g'(x) \,dx \le D_{2} \int_{m}^{\infty} \biggl(x+\frac{1}{2} \biggr)^{-6} \,dx \\ &=\frac{D_{2}}{5} \biggl(m+\frac{1}{2} \biggr)^{-5} \le \frac{D_{2}}{5} \int_{m}^{m+1}x^{-5} \,dx. \end{aligned}$$
(3.9)

It follows from (3.7), (3.9), and Lemma 2 that

$$\begin{aligned} r_{3}(n)-\gamma&\le \sum_{m=n}^{\infty} \frac{D_{2}}{5} \int _{m}^{m+1}x^{-5} \,dx \\ &=\frac{D_{2}}{5} \int_{n}^{\infty}x^{-5} \,dx=\frac{D_{2}}{20} \frac {1}{n^{4}}. \end{aligned}$$
(3.10)

Finally,

$$\begin{aligned} g(m)&= - \int_{m}^{\infty}g'(x) \,dx \ge D_{2} \int_{m}^{\infty}(x+1)^{-6} \,dx \\ &=\frac{D_{2}}{5}(m+1)^{-5}\ge\frac{D_{2}}{5} \int_{m+1}^{m+2}x^{-5} \,dx \end{aligned}$$

and

$$\begin{aligned} r_{3}(n)-\gamma&\ge \sum_{m=n}^{\infty} \frac{D_{2}}{5} \int _{m+1}^{m+2}x^{-5} \,dx \\ &=\frac{D_{2}}{5} \int_{n+1}^{\infty}x^{-5} \,dx =\frac{D_{2}}{20} \frac{1}{(n+1)^{4}}. \end{aligned}$$
(3.11)

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