## 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 [25]

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

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 [610].

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 [1724] 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 2.52.8 lead to Corollaries 2.12.5 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,

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