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A new sequence convergent to Euler–Mascheroni constant

  • Xu You
  • Di-Rong Chen
Open Access
Research

Abstract

In this paper, we provide a new sequence converging to the Euler–Mascheroni constant. Finally, we establish some inequalities for the Euler–Mascheroni constant by the new sequence.

Keywords

Euler–Mascheroni constant Rate of convergence Taylor’s formula Harmonic sequence 

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
$$\begin{aligned} \gamma(n):=\sum_{m=1}^{n}\frac{1}{m} -\ln n. \end{aligned}$$
(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, 6, 7, 8, 9, 10, 11, 12, 13, 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

Letfbe 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).

Notes

Acknowledgements

This work was supported by the National Natural Science Foundation of China (Grant Nos. 11571267, 61403034, and 91538112) and Beijing Municipal Commission of Education Science and Technology Program KM201810017009. Computations made in this paper were performed using Mathematica 9.0.

Authors’ contributions

The authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

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Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  1. 1.Department of Mathematics and PhysicsBeijing Institute of Petrochemical TechnologyBeijingP.R. China
  2. 2.Department of MathematicsWuhan Textile UniversityWuhanP.R. China

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