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]

$$ \gamma =\gamma (1)=\lim_{n\rightarrow \infty } \biggl( 1+\frac{1}{2}+ \frac{1}{3}+\cdots +\frac{1}{n}-\log n \biggr) =0.577215664901\ldots \,. $$

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

$$ \frac{1}{48(n+1)^{3}}\leq \gamma -T_{n}\leq \frac{1}{48n^{3}} $$
(1.5)

is valid for \(n\geq 1\).

Qiu and Vuorinen [12] proved that the two-sided inequality

$$ \frac{1}{2n}-\frac{\lambda }{n^{2}}< \gamma_{n}-\gamma \leq \frac{1}{2n}-\frac{\mu }{n^{2}} $$
(1.6)

is valid for \(n\geq 1\) if and only if \(\lambda \geq 1/12\) and \(\mu \leq \gamma -1/2\).

In [13], DeTemple proved that the double inequality

$$ \frac{1}{24(n+1)^{2}}\leq R_{n}-\gamma \leq \frac{1}{24n^{2}} $$
(1.7)

holds for all \(n\geq 1\).

Chen [14] proved that \(\alpha =1/\sqrt{12\gamma -6}-1\) and \(\beta =0\) are the best possible constants such that the double inequality

$$ \frac{1}{12(n+\alpha )^{2}}\leq \gamma -S_{n}\leq \frac{1}{12(n+ \beta )^{2}} $$
(1.8)

holds for \(n\geq 1\).

Sîntămărian [15], and Berinde and Mortici [16] proved that the double inequalities

$$\begin{aligned}& \frac{1}{2(n+a)}\leq y_{n}(a)-\gamma (a)\leq \frac{1}{2(n+a-1)}, \end{aligned}$$
(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)

are valid for all \(a>0\) and \(n\geq 1\).

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

$$\begin{aligned}& \frac{1}{12(a+n-\alpha_{1})^{2}}\leq \gamma (a)-\alpha_{n}(a)< \frac{1}{12(a+n- \beta_{1})^{2}}, \\& \frac{1}{24(a+n-\alpha_{2})^{2}}\leq \beta_{n}(a)-\gamma (a)< \frac{1}{24(a+n- \beta_{2})^{2}}, \\& \frac{1}{48(a+n-\alpha_{3})^{3}}\leq \gamma (a)-\lambda_{n}(a)< \frac{1}{48(a+n- \beta_{3})^{3}}, \\& \frac{\alpha_{4}}{(a+n-1)^{6}}\leq \gamma (a)-\mu_{n}(a)< \frac{\beta _{4}}{(a+n-1)^{6}} \end{aligned}$$

hold for all \(a>0\) and \(n\geq n_{0}\) and improve the bounds for the Euler–Mascheroni constant.

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

$$ \Gamma (x)= \int_{0}^{\infty }t^{x-1}e^{-t}dt, \qquad \psi (x)=\frac{ \Gamma^{\prime }(x)}{\Gamma (x)}, $$

respectively.

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

$$ f_{1}(x)=\frac{1}{ \sqrt{12 ( \log x-\psi (x+1)+\frac{1}{2x} ) }}-x $$
(2.3)

is strictly decreasing on \([2, \infty )\) with \(f_{1}(\infty )=0\).

Lemma 2.2

(See [26, Proof of Theorem 1], [27, Remark 4])

The function

$$ f_{2}(x)=\frac{1}{\sqrt{24 ( \psi (x+1)-\log (x+1/2) ) }}-x $$
(2.4)

is strictly decreasing on \([2, \infty )\) with \(f_{2}(\infty )=1/2\).

Lemma 2.3

(See [28, Proof of Theorem 2])

The function

$$ f_{3}(x)=\frac{1}{\sqrt[3]{48 [ \log ( x+\frac{1}{2}+ \frac{1}{24x} ) -\psi (x+1) ] }}-x $$
(2.5)

is strictly decreasing on \([5, \infty )\) with \(f_{3}(\infty )=83/360\).

Lemma 2.4

(See [29, Theorem 1.2(2)])

The function

$$ f_{4}(x)=\frac{x^{2}}{120}- \biggl( \psi (x)-\log x+ \frac{1}{2x}+\frac{1}{12x ^{2}} \biggr) x^{6} $$
(2.6)

is strictly increasing from \((0, \infty )\) onto \((0, 1/252)\).

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

$$ \frac{1}{12(a+n-\alpha_{1})^{2}}\leq \gamma (a)-\alpha_{n}(a)< \frac{1}{12(a+n- \beta_{1})^{2}} $$
(2.7)

holds for all \(a>0\) and \(n\geq 3\).

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)

From (2.3) and (2.8) we clearly see that inequality (2.7) is equivalent to

$$ \alpha_{1}\leq 1-f_{1}(n+a-1)< \beta_{1}. $$
(2.9)

Therefore, Theorem 2.5 follows easily from Lemma 2.1 and (2.19). □

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

$$ \frac{1}{24(a+n-\alpha_{2})^{2}}\leq \beta_{n}(a)-\gamma (a)< \frac{1}{24(a+n- \beta_{2})^{2}} $$
(2.10)

holds for all \(a>0\) and \(n\geq 3\).

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)

From (2.4) and (2.11) we clearly see that inequality (2.10) can be rewritten as

$$ \alpha_{2}\leq 1-f_{2}(n+a-1)< \beta_{2}. $$
(2.12)

Therefore, Theorem 2.6 follows easily from Lemma 2.2 and (2.12). □

Remark 2.1

We clearly see that both the upper and the lower bounds given in (2.10) for \(\beta_{n}(a)-\gamma (a)\) are better than that given in (1.10) for \(n\geq 3\) due to \(1-f_{2}(2)=3-1/\sqrt{36-24( \gamma +\log 5-\log 2)}=0.466904841516\ldots \) .

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

$$ \frac{1}{48(a+n-\alpha_{3})^{3}}\leq \gamma (a)-\lambda_{n}(a)< \frac{1}{48(a+n- \beta_{3})^{3}} $$
(2.13)

holds for all \(a>0\) and \(n\geq 6\).

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)

It follows from (2.5) and (2.14) that inequality (2.13) can be rewritten as

$$ \alpha_{3}\leq 1-f_{3}(a+n-1)< \beta_{3}. $$
(2.15)

Therefore, Theorem 2.7 follows easily from Lemma 2.3 and (2.15). □

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

$$ \frac{\alpha_{4}}{(a+n-1)^{6}}\leq \gamma (a)-\mu_{n}(a)< \frac{\beta _{4}}{(a+n-1)^{6}} $$
(2.16)

holds for all \(a>0\) and \(n\geq 1\).

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)

From (2.6) and (2.17) we clearly see that inequality (2.16) is equivalent to

$$ \alpha_{4}\leq f_{4}(n+a-1)< \beta_{4}. $$
(2.18)

Therefore, Theorem 2.8 follows easily from Lemma 2.4 and (2.18). □

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

$$\begin{aligned}& \frac{1}{2(a+n-1)}-\frac{1}{12(a+n-\beta_{1})^{2}}< y_{n}(a)-\gamma (a) \\& \hphantom{\frac{1}{2(a+n-1)}-\frac{1}{12(a+n-\beta_{1})^{2}}}\leq \frac{1}{2(a+n-1)}-\frac{1}{12(a+n-\alpha_{1})^{2}}, \end{aligned}$$
(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)

hold for all \(a>0\) and \(n\geq 3\).

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

$$\begin{aligned}& c_{1}=f_{1}(3)=1/\sqrt{12(\gamma +\log 3)-20}-3=0.015998 \ldots \,, \\& c_{2}=f_{2}(3)=1/ \sqrt{44-24(\gamma +\log 7-\log 2)}-3=0.5242567\ldots \,, \\& c_{3}=f_{3}(6)=-6+1/\sqrt[3]{48(\gamma -49/20+\log 937-\log 144)}=0.242347\ldots \end{aligned}$$

and

$$\begin{aligned}& c_{4}=f_{4}(1)=\gamma -23/40=0.00221566\ldots \,. \end{aligned}$$

Then

$$\begin{aligned}& \gamma (1)=\gamma , \qquad \alpha_{n}(1)=\gamma_{n}- \frac{1}{2n}=S_{n}, \qquad \beta_{n}(1)=R_{n}, \\& \lambda_{n}(1)=T_{n}, \qquad \mu_{n}(1)= \gamma_{n}-\frac{1}{2n}+\frac{1}{12n ^{2}}-\frac{1}{120n^{4}}. \end{aligned}$$

Therefore, Theorems 2.5–2.8 lead to Corollaries 2.1–2.5 immediately.

Corollary 2.1

The double inequality

$$ \frac{1}{2n}-\frac{1}{12n^{2}}< \gamma_{n}-\gamma \leq \frac{1}{2n}-\frac{1}{12(n+c _{1})^{2}} $$
(2.22)

holds for all \(n\geq 3\).

Corollary 2.2

The double inequality

$$ \frac{1}{12(n+c_{1})^{2}}\leq \gamma -S_{n}< \frac{1}{12n^{2}} $$
(2.23)

holds for all \(n\geq 3\).

Corollary 2.3

The double inequality

$$ \frac{1}{24(n+c_{2})^{2}}\leq R_{n}-\gamma < \frac{1}{24(n+1/2)^{2}} $$
(2.24)

holds for all \(n\geq 3\).

Corollary 2.4

The double inequality

$$ \frac{1}{48(n+c_{3})^{2}}\leq \gamma -T_{n}< \frac{1}{48(n+83/360)^{2}} $$
(2.25)

holds for all \(n\geq 6\).

Corollary 2.5

The double inequality

$$ \frac{1}{2n}-\frac{1}{12n^{2}}+\frac{1}{120n^{4}}-\frac{1}{252n^{6}} < \gamma_{n}-\gamma \leq \frac{1}{2n}-\frac{1}{12n^{2}}+ \frac{1}{120n ^{4}}-\frac{c_{4}}{n^{6}} $$
(2.26)

holds for all \(n\geq 1\).

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

$$ n> \biggl( \frac{1+\sqrt{1-4800[1-12(\gamma -1/2)]c_{4}}}{20[1-12( \gamma -1/2)]} \biggr) ^{1/2}=1.00000000006823\ldots $$

being the solution of the inequality

$$ \frac{1}{2n}-\frac{1}{12n^{2}}+\frac{1}{120n^{4}}-\frac{c_{4}}{n^{6}}< \frac{1}{2n}-\frac{\gamma -1/2}{n^{2}}. $$

3 Results and discussion

As the natural generalization of the Euler–Mascheroni constant

$$ \gamma =\lim_{n\rightarrow \infty } \biggl( 1+\frac{1}{2}+ \frac{1}{3}+ \cdots +\frac{1}{n}-\log n \biggr) =0.5772156649\ldots\, , $$

the generalized Euler–Mascheroni constant is defined by

$$ \gamma (a)=\lim_{n\rightarrow \infty } \biggl[ \frac{1}{a}+ \frac{1}{a+1}+ \cdots +\frac{1}{a+n-1}-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] $$

for \(a>0\).

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

$$ \gamma (a)=\lim_{n\rightarrow \infty } \biggl[ \frac{1}{a}+ \frac{1}{a+1}+ \cdots +\frac{1}{a+n-1}-\log \biggl( \frac{a+n-1}{a} \biggr) \biggr] $$

and improve some well-known bounds for the Euler–Mascheroni constant,

$$ \gamma =\lim_{n\rightarrow \infty } \biggl( 1+\frac{1}{2}+ \frac{1}{3}+ \cdots +\frac{1}{n}-\log n \biggr) =0.5772156649\ldots\,. $$