1 Introduction
The Euler–Mascheroni constant was first introduced by Leonhard Euler (1707–1783) in 1734 as the limit of the sequence
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.
$$\begin{aligned} \gamma(n):=\sum_{m=1}^{n}\frac{1}{m} -\ln n. \end{aligned}$$
(1.1)
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:
$$\begin{aligned} &\sum_{m=1}^{n}\frac{1}{m} -\ln \biggl(n+\frac{1}{2} \biggr)=\gamma+O\bigl(n^{-2}\bigr) \quad \mbox{(DeTemple [6]),} \\ &\sum_{m=1}^{n}\frac{1}{m} -\ln \frac{n^{3}+\frac{3}{2}n^{2}+\frac{227}{240}+\frac{107}{480}}{ n^{2}+n+\frac{97}{240}}=\gamma+O\bigl(n^{-6}\bigr) \quad\mbox{(Mortici [13]),} \\ & \begin{aligned} &\sum_{m=1}^{n}\frac{1}{m} -\ln \biggl( 1+\frac{1}{2n}+\frac{1}{24n^{2}}-\frac{1}{48n^{3}} +\frac{23}{5760n^{4}} \biggr)=\gamma+O\bigl(n^{-5}\bigr)\\ & \quad\mbox{(Chen and Mortici [5]).} \end{aligned} \end{aligned}$$
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Recently, Mortici and Chen [14] provided a very interesting sequence
and proved that
Hence the rate of the convergence of the sequence \((\nu(n) )_{n\in\mathbb{N}}\) is \(n^{-12}\).
$$\begin{aligned} \nu(n)&=\sum_{m=1}^{n}\frac{1}{m} - \frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &\quad{}- \biggl(\frac{-\frac{1}{180}}{ (n^{2}+n+\frac{1}{3} )^{2}} +\frac{\frac{8}{2835}}{ (n^{2}+n+\frac{1}{3} )^{3}} +\frac{\frac{5}{1512}}{ (n^{2}+n+\frac{1}{3} )^{4}} + \frac{\frac{592}{93\text{,}555}}{ (n^{2}+n+\frac{1}{3} )^{5}} \biggr) \end{aligned}$$
$$\begin{aligned} \lim_{n\rightarrow\infty}n^{12} \bigl(\nu(n)-\gamma \bigr)=- \frac {796\text{,}801}{43\text{,}783\text{,}740}. \end{aligned}$$
(1.2)
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
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.
$$\begin{aligned} &\frac{1}{72(n+1)^{3}}< \gamma-r_{2}(n)< \frac{1}{72n^{3}}, \end{aligned}$$
(1.3)
$$\begin{aligned} &\frac{1}{120(n+1)^{4}}< r_{3}(n)-\gamma< \frac{1}{120(n-1)^{4}}. \end{aligned}$$
(1.4)
Theorem 1
For the Euler–Mascheroni constant, we have the following convergent sequence:
where
$$\begin{aligned} r(n)=1+\frac{1}{2} +\cdots+\frac{1}{n}-\ln n-\ln \biggl(1+ \frac{a_{1}}{n} \biggr)-\ln \biggl(1+\frac{a_{2}}{n^{2}} \biggr)-\cdots, \end{aligned}$$
$$\begin{aligned} &a_{1}=\frac{1}{2},\qquad a_{2}=\frac{1}{24},\qquad a_{3}=-\frac{1}{24},\qquad a_{4}=\frac {143}{5760}, \\ &a_{5}=-\frac{1}{160}, \qquad a_{6}=-\frac{151}{290\text{,}304},\qquad a_{7}=-\frac{1}{896},\qquad \dots. \end{aligned} $$
Let
$$\begin{aligned} r_{k}(n):=\sum_{m=1}^{n} \frac{1}{m}-\ln n-\sum_{m=1}^{k}\ln \biggl(1+\frac {a_{m}}{n^{m}} \biggr). \end{aligned}$$
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For
\(1\le k\le7\), we have
where
$$\begin{aligned} \lim_{n\rightarrow\infty}n^{k+2} \bigl(r_{k}(n)-\gamma \bigr)=C_{k}, \end{aligned}$$
(1.5)
$$\begin{aligned} &C_{1}=\frac{1}{24},\qquad C_{2}=-\frac{1}{24},\qquad C_{3}=\frac{143}{5760},\qquad C_{4}=-\frac{1}{160},\\ &C_{5}=-\frac{151}{290\text{,}304},\qquad C_{6}=-\frac{1}{896},\qquad C_{7}=\frac{109\text{,}793}{22\text{,}118\text{,}400},\qquad \dots. \end{aligned} $$
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
with
\(s>1\), then 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)
$$\begin{aligned} \lim_{n\rightarrow+\infty}n^{s-1}x_{n}=\frac{l}{s-1}. \end{aligned}$$
(2.2)
We need to find the value \(a_{1}\in\mathbb{R}\) that produces the most accurate approximation of the form
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,
Developing expression (2.4) into a power series expansion in \(1/n\), we obtain
$$\begin{aligned} r_{1}(n)=\sum_{m=1}^{n} \frac{1}{m} -\ln n-\ln \biggl(1+\frac{a_{1}}{n} \biggr). \end{aligned}$$
(2.3)
$$\begin{aligned} r_{1}(n)-r_{1}(n+1)=\ln \biggl(1+\frac{1}{n} \biggr)-\frac{1}{n+1}+\ln \biggl(1+\frac{a_{1}}{n+1} \biggr)-\ln \biggl(1+ \frac{a_{1}}{n} \biggr). \end{aligned}$$
(2.4)
$$\begin{aligned} r_{1}(n)-r_{1}(n+1)= \biggl(\frac{1}{2} -a_{1} \biggr)\frac{1}{n^{2}}+ \biggl(-\frac{2}{3}+a_{1}+{a_{1}}^{2} \biggr)\frac{1}{n^{3}}+O \biggl(\frac{1}{n^{4}} \biggr). \end{aligned}$$
(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
Then
$$\begin{aligned} r_{k}(n):=\sum_{m=1}^{n} \frac{1}{m}-\ln n-\sum_{m=1}^{k}\ln \biggl(1+\frac {a_{m}}{n^{m}} \biggr). \end{aligned}$$
(2.6)
$$\begin{aligned} &r_{k}(n)-r_{k}(n+1) \\ &\quad =\ln \biggl(1+\frac{1}{n} \biggr)-\frac{1}{n+1}+\sum_{m=1}^{k}\ln \biggl(1+\frac{a_{m}}{(n+1)^{m}} \biggr)-\sum_{m=1}^{k} \ln \biggl(1+\frac{a_{m}}{n^{m}} \biggr). \end{aligned}$$
(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
The rate of convergence of \((r_{8}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-7}\), since
$$\begin{aligned} r_{5}(n)-r_{5}(n+1)=-\frac{151}{48\text{,}384}\frac{1}{n^{7}} +O \biggl(\frac {1}{n^{8}} \biggr). \end{aligned}$$
(2.9)
$$\begin{aligned} \lim_{n\rightarrow\infty}n^{7} \bigl(r_{5}(n)-\gamma \bigr)=-\frac{151}{290\text{,}304}:=C_{5}. \end{aligned}$$
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
The rate of convergence of \((r_{6}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-8}\), since
$$\begin{aligned} r_{6}(n)-r_{6}(n+1)=-\frac{1}{128}\frac{1}{n^{8}}+O \biggl(\frac{1}{n^{9}} \biggr). \end{aligned}$$
(2.10)
$$\begin{aligned} \lim_{n\rightarrow\infty}n^{8} \bigl(r_{6}(n)-\gamma \bigr)=-\frac{1}{896}:=C_{6}. \end{aligned}$$
Finally, we check that \(a_{7}=-\frac{1}{896}\):
Since \(a_{7}=-\frac{1}{896}\) and
the rate of convergence of the \((r_{7}(n)-\gamma )_{n\in\mathbb{N}}\) is \(n^{-9}\).
$$\begin{aligned} r_{7}(n)-r_{7}(n+1)= \biggl(-\frac{1}{128}-7a_{7} \biggr)\frac{1}{n^{8}} + \biggl(\frac{196\text{,}193}{2\text{,}764\text{,}800}+28a_{7} \biggr) \frac{1}{n^{9}}+O \biggl(\frac {1}{n^{10}} \biggr). \end{aligned}$$
(2.11)
$$\begin{aligned} \lim_{n\rightarrow\infty}n^{9} \bigl(r_{7}(n)-\gamma \bigr)= \frac{109\text{,}793}{22\text{,}118\text{,}400}:=C_{7}, \end{aligned}$$
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
where
Let \(D_{1}=1/2\). By using the Mathematica software we have
and
$$\begin{aligned} \gamma-r_{2}(n)=\sum_{m=n}^{\infty} \bigl(r_{2}(m+1)-r_{2}(m) \bigr) =\sum _{m=n}^{\infty}f(m), \end{aligned}$$
(3.2)
$$\begin{aligned} f(m)&=\frac{1}{m+1}-\ln \biggl(1+\frac{1}{m} \biggr)-\ln \biggl(1+ \frac {a_{1}}{m+1} \biggr)-\ln \biggl(1+\frac{a_{2}}{(m+1)^{2}} \biggr)\\ &\quad {}+\ln \biggl(1+ \frac{a_{1}}{m} \biggr)+\ln \biggl(1+\frac{a_{2}}{m^{2}} \biggr). \end{aligned} $$
$$\begin{aligned} &-f'(x)-D_{1}\frac{1}{(x+1)^{5}}\\ &\quad =\frac {300+2739x+11\text{,}434x^{2}+24\text{,}870x^{3}+28\text{,}314x^{4}+15\text{,}936x^{5}+3472x^{6}}{ 2x(1+x)^{5}(1+2x)(3+2x)(1+24x^{2})(25+48x+24x^{2})}>0 \end{aligned}$$
$$\begin{aligned} & -f'(x)-D_{1}\frac{1}{(x+\frac{1}{2})^{5}}\\ &\quad =-\frac{P_{6}(x)(x-1)+151\text{,}085}{2x^{5} (1 +x)^{2} (1 + 2x) (3 + 2x) (1 + 24x^{2}) (25 + 48x + 24 x^{2})}< 0. \end{aligned}$$
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
and
Combining (3.5) and (3.6) completes the proof of (1.6).
$$\begin{aligned} f(m)&= - \int_{m}^{\infty}f'(x) \,dx \ge D_{1} \int_{m}^{\infty}(x+1)^{-5} \,dx \\ &=\frac{D_{1}}{4}(m+1)^{-4}\ge\frac{D_{1}}{4} \int_{m+1}^{m+2}x^{-4} \,dx \end{aligned}$$
$$ \begin{aligned}[b] \gamma-r_{2}(n)&\ge \sum_{m=n}^{\infty} \frac{D_{1}}{4} \int _{m+1}^{m+2}x^{-4} \,dx \\ &=\frac{D_{1}}{4} \int_{n+1}^{\infty}x^{-4} \,dx=\frac{D_{1}}{12} \frac {1}{(n+1)^{3}}. \end{aligned} $$
(3.6)
Noting that \(r_{3}(\infty)=0\), we easily deduce
where
Let \(D_{2}=\frac{143}{288}\). By using the Mathematica software we have
and
Hence, for \(x\ge4\),
$$\begin{aligned} r_{3}(n)-\gamma=\sum_{m=n}^{\infty} \bigl(r_{3}(m)-r_{3}(m+1) \bigr)=\sum _{m=n}^{\infty}g(m), \end{aligned}$$
(3.7)
$$g(m)=r_{3}(m)-r_{3}(m+1). $$
$$\begin{aligned} &-g'(x)-D_{2}\frac{1}{(x+1)^{6}}\\ &\quad =\frac{P_{12}(x)}{288 n (1 + n)^{6} (1 + 2 n) (3 + 2 n) (1 + 24 n^{2}) (25 + 48 n +24 n^{2}) (-1 + 24 n^{3})P_{3}(x)}>0 \end{aligned}$$
$$\begin{aligned} &-g'(x)-D_{2}\frac{1}{(x+\frac{1}{2})^{6}}\\ &\quad =-\frac{P_{12}(x)(x-4)+2\text{,}052\text{,}948\text{,}001\text{,}087\text{,}775 }{9 x (1 + x)^{2} (1 + 2 x)^{6} (3 + 2 x) (1 + 24 x^{2}) (25 + 48 x + 24 x^{2}) (-1 + 24 x^{3})P_{3}(x)}< 0. \end{aligned}$$
$$\begin{aligned} D_{2}\frac{1}{(n+1)^{6}}< -g'(x)< D_{2} \frac{1}{(x+\frac{1}{2})^{6}}. \end{aligned}$$
(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
Finally,
and
Combining (3.10) and (3.11) completes the proof of (1.7).
$$\begin{aligned} r_{3}(n)-\gamma&\le \sum_{m=n}^{\infty} \frac{D_{2}}{5} \int _{m}^{m+1}x^{-5} \,dx \\ &=\frac{D_{2}}{5} \int_{n}^{\infty}x^{-5} \,dx=\frac{D_{2}}{20} \frac {1}{n^{4}}. \end{aligned}$$
(3.10)
$$\begin{aligned} g(m)&= - \int_{m}^{\infty}g'(x) \,dx \ge D_{2} \int_{m}^{\infty}(x+1)^{-6} \,dx \\ &=\frac{D_{2}}{5}(m+1)^{-5}\ge\frac{D_{2}}{5} \int_{m+1}^{m+2}x^{-5} \,dx \end{aligned}$$
$$\begin{aligned} r_{3}(n)-\gamma&\ge \sum_{m=n}^{\infty} \frac{D_{2}}{5} \int _{m+1}^{m+2}x^{-5} \,dx \\ &=\frac{D_{2}}{5} \int_{n+1}^{\infty}x^{-5} \,dx =\frac{D_{2}}{20} \frac{1}{(n+1)^{4}}. \end{aligned}$$
(3.11)
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.
Competing interests
The authors declare that they have no competing interests.
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