1 Introduction
Let \(a>0\). Then the generalized Euler–Mascheroni constant \(\gamma (a)\) [1] is given by
$$ \gamma (a)=\lim_{n\rightarrow \infty } \biggl[ \frac{1}{a}+ \frac{1}{a+1}+ \cdots +\frac{1}{a+n-1}-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] . $$
We clearly see that the generalized Euler–Mascheroni constant \(\gamma (a)\) is the natural generalization of the classical Euler–Mascheroni constant [2‐5]
$$ \gamma =\gamma (1)=\lim_{n\rightarrow \infty } \biggl( 1+\frac{1}{2}+ \frac{1}{3}+\cdots +\frac{1}{n}-\log n \biggr) =0.577215664901\ldots \,. $$
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Recently, the two bounds for γ and \(\gamma (a)\) have attracted the attention of many mathematicians. In particular, many remarkable inequalities and asymptotic formulas for γ and \(\gamma (a)\) can be found in the literature [6‐10].
Let
$$\begin{aligned}& \gamma_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n}-\log n, \\& R_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n}-\log \biggl( n+ \frac{1}{2} \biggr) , \\& S_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{2n}- \log n, \\& T_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n}-\log \biggl( n+ \frac{1}{2}+\frac{1}{24n} \biggr) , \\& y_{n}(a)=\frac{1}{a}+\frac{1}{a+1}+\cdots + \frac{1}{a+n-1}-\log \biggl( \frac{a+n-1}{a} \biggr) , \\& \alpha_{n}(a)=\frac{1}{a}+\frac{1}{a+1}+\cdots + \frac{1}{a+n-2}+ \frac{1}{2(a+n-1)}-\log \biggl( \frac{a+n-1}{a} \biggr) , \end{aligned}$$
(1.1)
$$\begin{aligned}& \beta_{n}(a)=\frac{1}{a}+\frac{1}{a+1}+\frac{1}{a+n-1}- \log \biggl( \frac{a+n-1/2}{a} \biggr) , \end{aligned}$$
(1.2)
$$\begin{aligned}& \lambda_{n}(a)=\frac{1}{a}+\frac{1}{a+1}+\frac{1}{a+n-1}- \log \biggl( \frac{a+n-1/2}{a}+ \frac{1}{24a(a+n-1)} \biggr) , \end{aligned}$$
(1.3)
$$\begin{aligned}& \mu_{n}(a)=y_{n}(a)-\frac{1}{2(a+n-1)}+\frac{1}{12(a+n-1)^{2}}- \frac{1}{120(a+n-1)^{4}}. \end{aligned}$$
(1.4)
Negoi [11] proved that the two-sided inequality
is valid for \(n\geq 1\).
$$ \frac{1}{48(n+1)^{3}}\leq \gamma -T_{n}\leq \frac{1}{48n^{3}} $$
(1.5)
Qiu and Vuorinen [12] proved that the two-sided inequality
is valid for \(n\geq 1\) if and only if \(\lambda \geq 1/12\) and \(\mu \leq \gamma -1/2\).
$$ \frac{1}{2n}-\frac{\lambda }{n^{2}}< \gamma_{n}-\gamma \leq \frac{1}{2n}-\frac{\mu }{n^{2}} $$
(1.6)
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In [13], DeTemple proved that the double inequality
holds for all \(n\geq 1\).
$$ \frac{1}{24(n+1)^{2}}\leq R_{n}-\gamma \leq \frac{1}{24n^{2}} $$
(1.7)
Chen [14] proved that \(\alpha =1/\sqrt{12\gamma -6}-1\) and \(\beta =0\) are the best possible constants such that the double inequality
holds for \(n\geq 1\).
$$ \frac{1}{12(n+\alpha )^{2}}\leq \gamma -S_{n}\leq \frac{1}{12(n+ \beta )^{2}} $$
(1.8)
Sîntămărian [15], and Berinde and Mortici [16] proved that the double inequalities
are valid for all \(a>0\) and \(n\geq 1\).
$$\begin{aligned}& \frac{1}{2(n+a)}\leq y_{n}(a)-\gamma (a)\leq \frac{1}{2(n+a-1)}, \end{aligned}$$
(1.9)
$$\begin{aligned}& \frac{1}{24(n+a)^{2}}\leq \beta_{n}(a)-\gamma (a)\leq \frac{1}{24(n+a-1)^{2}} \end{aligned}$$
(1.10)
The main purpose of this article is to find the best possible constants \(\alpha_{1}\), \(\alpha_{2}\), \(\alpha_{3}\), \(\alpha_{4}\), \(\beta_{1}\), \(\beta_{2}\), \(\beta_{3}\) and \(\beta_{4}\) such that the double inequalities
hold for all \(a>0\) and \(n\geq n_{0}\) and improve the bounds for the Euler–Mascheroni constant.
$$\begin{aligned}& \frac{1}{12(a+n-\alpha_{1})^{2}}\leq \gamma (a)-\alpha_{n}(a)< \frac{1}{12(a+n- \beta_{1})^{2}}, \\& \frac{1}{24(a+n-\alpha_{2})^{2}}\leq \beta_{n}(a)-\gamma (a)< \frac{1}{24(a+n- \beta_{2})^{2}}, \\& \frac{1}{48(a+n-\alpha_{3})^{3}}\leq \gamma (a)-\lambda_{n}(a)< \frac{1}{48(a+n- \beta_{3})^{3}}, \\& \frac{\alpha_{4}}{(a+n-1)^{6}}\leq \gamma (a)-\mu_{n}(a)< \frac{\beta _{4}}{(a+n-1)^{6}} \end{aligned}$$
2 Main results
In order to prove our main results, we need several formulas and lemmas which we present in this section.
For \(x>0\), the classical gamma function Γ and its logarithmic derivative, the so-called psi function ψ are defined [17‐24] as
respectively.
$$ \Gamma (x)= \int_{0}^{\infty }t^{x-1}e^{-t}dt, \qquad \psi (x)=\frac{ \Gamma^{\prime }(x)}{\Gamma (x)}, $$
The psi function ψ has the recurrence and asymptotic formulas [25] as follows:
$$\begin{aligned}& \psi (x+1)=\psi (x)+\frac{1}{x}, \end{aligned}$$
(2.1)
$$\begin{aligned}& \psi (x)\sim \log x-\frac{1}{2x}-\frac{1}{12x^{2}}+\frac{1}{120x^{4}}- \frac{1}{252x ^{6}}+\cdots \quad (x\rightarrow \infty ). \end{aligned}$$
(2.2)
Lemma 2.1
(See [14, Proof of Theorem 1])
The function
is strictly decreasing on
\([2, \infty )\)
with
\(f_{1}(\infty )=0\).
$$ f_{1}(x)=\frac{1}{ \sqrt{12 ( \log x-\psi (x+1)+\frac{1}{2x} ) }}-x $$
(2.3)
Lemma 2.2
The function
is strictly decreasing on
\([2, \infty )\)
with
\(f_{2}(\infty )=1/2\).
$$ f_{2}(x)=\frac{1}{\sqrt{24 ( \psi (x+1)-\log (x+1/2) ) }}-x $$
(2.4)
Lemma 2.3
(See [28, Proof of Theorem 2])
The function
is strictly decreasing on
\([5, \infty )\)
with
\(f_{3}(\infty )=83/360\).
$$ f_{3}(x)=\frac{1}{\sqrt[3]{48 [ \log ( x+\frac{1}{2}+ \frac{1}{24x} ) -\psi (x+1) ] }}-x $$
(2.5)
Lemma 2.4
(See [29, Theorem 1.2(2)])
The function
is strictly increasing from
\((0, \infty )\)
onto
\((0, 1/252)\).
$$ f_{4}(x)=\frac{x^{2}}{120}- \biggl( \psi (x)-\log x+ \frac{1}{2x}+\frac{1}{12x ^{2}} \biggr) x^{6} $$
(2.6)
Theorem 2.5
Let
\(\alpha_{n}(a)\)
and
\(f_{1}(x)\)
be, respectively, defined by (1.1) and (2.3). Then
\(\alpha_{1}=1-f_{1}(a+2)\)
and
\(\beta_{1}=1\)
are the best possible constants such that the double inequality
holds for all
\(a>0\)
and
\(n\geq 3\).
$$ \frac{1}{12(a+n-\alpha_{1})^{2}}\leq \gamma (a)-\alpha_{n}(a)< \frac{1}{12(a+n- \beta_{1})^{2}} $$
(2.7)
Proof
It follows from (1.1), (2.1) and (2.2) that
$$\begin{aligned} \gamma (a)-\alpha_{n}(a)&=\lim_{n\rightarrow \infty } \biggl[ \psi (n+a)- \psi (a)-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] \\ &\quad {}- \biggl[ \psi (n+a)-\psi (a)-\frac{1}{2(a+n-1)}-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] \\ &=\lim_{n\rightarrow \infty }\bigl[\psi (n+a)-\log (a+n-1)\bigr] \\ &\quad {}-\psi (n+a)+\frac{1}{2(a+n-1)}+\log (a+n-1) \\ &=\log (a+n-1)-\psi (n+a)+\frac{1}{2(a+n-1)}. \end{aligned}$$
(2.8)
Theorem 2.6
Let
\(\beta_{n}(a)\)
and
\(f_{2}(x)\)
be, respectively, defined by (1.2) and (2.4). Then
\(\alpha_{2}=1-f_{2}(a+2)\)
and
\(\beta_{2}=1/2\)
are the best possible constants such that the double inequality
holds for all
\(a>0\)
and
\(n\geq 3\).
$$ \frac{1}{24(a+n-\alpha_{2})^{2}}\leq \beta_{n}(a)-\gamma (a)< \frac{1}{24(a+n- \beta_{2})^{2}} $$
(2.10)
Proof
It follows from (1.2), (2.1) and (2.2) that
$$ \beta_{n}(a)-\gamma (a)=\psi (n+a)-\log \biggl( a+n-\frac{1}{2} \biggr) . $$
(2.11)
Remark 2.1
Theorem 2.7
Let
\(\lambda_{n}(a)\)
and
\(f_{3}(x)\)
be, respectively, defined by (1.3) and (2.5). Then
\(\alpha_{3}=1-f_{3}(a+5)\)
and
\(\beta_{3}=277/360\)
are the best possible constants such that the double inequality
holds for all
\(a>0\)
and
\(n\geq 6\).
$$ \frac{1}{48(a+n-\alpha_{3})^{3}}\leq \gamma (a)-\lambda_{n}(a)< \frac{1}{48(a+n- \beta_{3})^{3}} $$
(2.13)
Proof
From (1.3), (2.1) and (2.2) we have
$$ \gamma (a)-\lambda_{n}(a)=\log \biggl( a+n- \frac{1}{2}+ \frac{1}{24(a+n-1)} \biggr) -\psi (a+n). $$
(2.14)
Theorem 2.8
Let
\(\mu_{n}(a)\)
and
\(f_{4}(x)\)
be, respectively, defined by (1.4) and (2.6). Then
\(\alpha_{4}=f_{4}(a)\)
and
\(\beta_{4}=1/252\)
are the best possible constants such that the double inequality
holds for all
\(a>0\)
and
\(n\geq 1\).
$$ \frac{\alpha_{4}}{(a+n-1)^{6}}\leq \gamma (a)-\mu_{n}(a)< \frac{\beta _{4}}{(a+n-1)^{6}} $$
(2.16)
Proof
It follows from (1.4), (2.1) and (2.2) that
$$\begin{aligned} &\gamma (a)-\mu_{n}(a) \\ &\quad =\frac{1}{120(n+a-1)^{4}} \\ &\quad \quad {}- \biggl[ \psi (n+a-1)-\log (n+a-1)+ \frac{1}{2(n+a-1)}+ \frac{1}{12(n+a-1)^{2}} \biggr] . \end{aligned}$$
(2.17)
Remark 2.2
Note that
$$ \alpha_{n}(a)=y_{n}(a)-\frac{1}{2(a+n-1)}. $$
(2.19)
It follows from (1.4), Theorem 2.5, Theorem 2.8 and (2.19) that \(\alpha_{1}=1-f_{1}(a+2)\), \(\beta_{1}=1\), \(\alpha_{4}=f_{4}(a)\) and \(\beta_{4}=1/252\) are the best possible constants such that the double inequalities
hold for all \(a>0\) and \(n\geq 3\).
$$\begin{aligned}& \frac{1}{2(a+n-1)}-\frac{1}{12(a+n-\beta_{1})^{2}}< y_{n}(a)-\gamma (a) \\& \hphantom{\frac{1}{2(a+n-1)}-\frac{1}{12(a+n-\beta_{1})^{2}}}\leq \frac{1}{2(a+n-1)}-\frac{1}{12(a+n-\alpha_{1})^{2}}, \end{aligned}$$
(2.20)
$$\begin{aligned}& \frac{1}{2(a+n-1)}-\frac{1}{12(a+n-1)^{2}}+\frac{1}{120(a+n-1)^{4}}-\frac{ \beta_{4}}{(a+n-1)^{6}} \\& \quad < y_{n}(a)-\gamma (a) \\& \quad \leq \frac{1}{2(a+n-1)}-\frac{1}{12(a+n-1)^{2}}+ \frac{1}{120(a+n-1)^{4}}-\frac{\alpha_{4}}{(a+n-1)^{6}}, \end{aligned}$$
(2.21)
We clearly see that the two inequalities (2.20) and (2.21) are the improvements of the inequality (1.9) for \(n\geq 3\).
Let \(a=1\) and
and
Then
$$\begin{aligned}& c_{1}=f_{1}(3)=1/\sqrt{12(\gamma +\log 3)-20}-3=0.015998 \ldots \,, \\& c_{2}=f_{2}(3)=1/ \sqrt{44-24(\gamma +\log 7-\log 2)}-3=0.5242567\ldots \,, \\& c_{3}=f_{3}(6)=-6+1/\sqrt[3]{48(\gamma -49/20+\log 937-\log 144)}=0.242347\ldots \end{aligned}$$
$$\begin{aligned}& c_{4}=f_{4}(1)=\gamma -23/40=0.00221566\ldots \,. \end{aligned}$$
$$\begin{aligned}& \gamma (1)=\gamma , \qquad \alpha_{n}(1)=\gamma_{n}- \frac{1}{2n}=S_{n}, \qquad \beta_{n}(1)=R_{n}, \\& \lambda_{n}(1)=T_{n}, \qquad \mu_{n}(1)= \gamma_{n}-\frac{1}{2n}+\frac{1}{12n ^{2}}-\frac{1}{120n^{4}}. \end{aligned}$$
Corollary 2.1
The double inequality
holds for all
\(n\geq 3\).
$$ \frac{1}{2n}-\frac{1}{12n^{2}}< \gamma_{n}-\gamma \leq \frac{1}{2n}-\frac{1}{12(n+c _{1})^{2}} $$
(2.22)
Corollary 2.2
The double inequality
holds for all
\(n\geq 3\).
$$ \frac{1}{12(n+c_{1})^{2}}\leq \gamma -S_{n}< \frac{1}{12n^{2}} $$
(2.23)
Corollary 2.3
The double inequality
holds for all
\(n\geq 3\).
$$ \frac{1}{24(n+c_{2})^{2}}\leq R_{n}-\gamma < \frac{1}{24(n+1/2)^{2}} $$
(2.24)
Corollary 2.4
The double inequality
holds for all
\(n\geq 6\).
$$ \frac{1}{48(n+c_{3})^{2}}\leq \gamma -T_{n}< \frac{1}{48(n+83/360)^{2}} $$
(2.25)
Corollary 2.5
The double inequality
holds for all
\(n\geq 1\).
$$ \frac{1}{2n}-\frac{1}{12n^{2}}+\frac{1}{120n^{4}}-\frac{1}{252n^{6}} < \gamma_{n}-\gamma \leq \frac{1}{2n}-\frac{1}{12n^{2}}+ \frac{1}{120n ^{4}}-\frac{c_{4}}{n^{6}} $$
(2.26)
Remark 2.3
We clearly see that the upper bound given in (2.22) is better than that given in (1.6) for \(n\geq 3\) due to \(n>\sqrt{12(\gamma -1/2)}c_{1}/(1-\sqrt{12(\gamma -1/2)})=0.4117\ldots \) is the solution of the inequality \(1/[12(n+c_{1})^{2}]>( \gamma -1/2)/n^{2}\), the lower bound given in (2.23) is better than that given in (1.8) for \(n\geq 3\) due to \(c_{1}<1\sqrt{12\gamma -6}-1=0.03885914\ldots\) , both the upper and the lower bounds given in (2.24) are improvements of that given in (1.7) for \(n\geq 3\), inequality (2.25) is stronger than inequality (1.5) for \(n\geq 6\), the lower bound given in (2.26) is better than that given in (1.6) for \(n\geq 1\), and the upper bound given in (2.26) is stronger than that given in (1.6) for \(n\geq 2\) due to
being the solution of the inequality
$$ n> \biggl( \frac{1+\sqrt{1-4800[1-12(\gamma -1/2)]c_{4}}}{20[1-12( \gamma -1/2)]} \biggr) ^{1/2}=1.00000000006823\ldots $$
$$ \frac{1}{2n}-\frac{1}{12n^{2}}+\frac{1}{120n^{4}}-\frac{c_{4}}{n^{6}}< \frac{1}{2n}-\frac{\gamma -1/2}{n^{2}}. $$
3 Results and discussion
As the natural generalization of the Euler–Mascheroni constant
the generalized Euler–Mascheroni constant is defined by
for \(a>0\).
$$ \gamma =\lim_{n\rightarrow \infty } \biggl( 1+\frac{1}{2}+ \frac{1}{3}+ \cdots +\frac{1}{n}-\log n \biggr) =0.5772156649\ldots\, , $$
$$ \gamma (a)=\lim_{n\rightarrow \infty } \biggl[ \frac{1}{a}+ \frac{1}{a+1}+ \cdots +\frac{1}{a+n-1}-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] $$
Recently, the evaluations for γ and \(\gamma (a)\) have been the subject of intensive research. In the article, we provide several sharp upper and lower bounds for the generalized Euler–Mascheroni constant \(\gamma (a)\). As applications, we improve some previously results on the Euler–Mascheroni constant γ. The idea presented may stimulate further research in the theory of special function.
4 Conclusion
In this paper, we present several best possible approximations for the generalized Euler–Mascheroni constant
and improve some well-known bounds for the Euler–Mascheroni constant,
$$ \gamma (a)=\lim_{n\rightarrow \infty } \biggl[ \frac{1}{a}+ \frac{1}{a+1}+ \cdots +\frac{1}{a+n-1}-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] $$
$$ \gamma =\lim_{n\rightarrow \infty } \biggl( 1+\frac{1}{2}+ \frac{1}{3}+ \cdots +\frac{1}{n}-\log n \biggr) =0.5772156649\ldots\,. $$
Acknowledgements
The research was supported by the Natural Science Foundation of China (Grants Nos. 61673169, 11401531, 11601485), the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11626101), the Natural Science Foundation of Zhejiang Province (Grant No. LQ17A010010) and the Science Foundation of Zhejiang Sci-Tech University (Grant No. 14062093-Y).
Competing interests
The authors declare that they have no competing interests.
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