1 Introduction
As it is known, defining some new approximations toward fundamental constants plays an important role in the field of mathematical constants. One of the most famous constants is Euler’s constant \(\gamma =0.577215\dots \), which is defined as the limit of the sequence
and has numerous applications in many areas of pure and applied mathematics, such as analysis, number theory, theory of probability, applied statistics, and special functions.
$$\begin{aligned}& \gamma_{n}=\sum_{k=1}^{n} \frac{1}{k} - \ln n \end{aligned}$$
(1.1)
Up until now, many authors have devoted great efforts and achieved much in the area of improving the convergence rate of the sequence \((\gamma_{n})_{n\geq 1}\). Among them, there are many inspiring achievements. For example, the estimate
was given in [1‐4].
$$\begin{aligned}& \frac{1}{2(n+1)}< \gamma_{n}-\gamma < \frac{1}{2n} \quad \mathrm{(Young)} \end{aligned}$$
(1.2)
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In [5, 6], a new sequence \((D_{n})_{n\geq 1}\) converging faster to γ was introduced, which is defined as
DeTemple also concluded that the speed of the new sequence to γ is of order \(n^{-2}\) since
Another modification was provided by Vernescu [7] as
who proved that
It is easy to conclude that though (1.3) and (1.5) only make slight modifications on the Euler’s sequence (1.1), but the convergent rates are significantly improved from \(n^{-1}\) to \(n^{-2}\).
$$\begin{aligned}& D_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n} - \ln \biggl(n+ \frac{1}{2} \biggr). \end{aligned}$$
(1.3)
$$\begin{aligned}& \frac{1}{24(n+1)^{2}}< D_{n}-\gamma < \frac{1}{24n^{2}} \quad \mathrm{(DeTemple)}. \end{aligned}$$
(1.4)
$$\begin{aligned}& V_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{2n}- \ln n, \end{aligned}$$
(1.5)
$$\begin{aligned}& \frac{1}{12(n+1)^{2}}< \gamma -V_{n}< \frac{1}{12n^{2}}. \end{aligned}$$
(1.6)
Moreover, Mortici obtained some sequences converging even faster than (1.1), (1.3), and (1.5). More specifically, Mortici [8] constructed the following two sequences:
Both (1.7) and (1.8) had been proved to converge to γ as \(n^{-3}\).
$$\begin{aligned}& u_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{(6-2 \sqrt{6})n}- \ln \biggl(n+\frac{1}{\sqrt{6}} \biggr), \end{aligned}$$
(1.7)
$$\begin{aligned}& v_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{(6+2 \sqrt{6})n}- \ln \biggl(n-\frac{1}{\sqrt{6}} \biggr). \end{aligned}$$
(1.8)
Moreover, Mortici [9] introduced the following class of sequences:
where \(a,b\in \mathbb{R}\), \(a>0\). They proved that, among the sequences \((\mu_{n}(a,b))_{n\geq 1}\), in the case of \(a=\sqrt{2}/2\) and \(b=(2+\sqrt{2})/4\) the privileged sequence offers the best approximations of γ since
Recently, Lu, Song, and Yu [10] provided some approximations of Euler’s constant. A new important sequence was defined as follows:
where
Two particular sequences were provided as
These two sequences converge faster than all other sequences mentioned since for all \(n\in \mathbb{N}\),
$$\begin{aligned}& \mu_{n}(a,b)=\sum_{k=1}^{n} \frac{1}{k} +\ln \bigl(e^{a/(n+b)}-1\bigr)- \ln a, \end{aligned}$$
(1.9)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3} \biggl( \mu_{n} \biggl(\frac{ \sqrt{2}}{2},\frac{2+\sqrt{2}}{4} \biggr)-\gamma \biggr)=\frac{ \sqrt{2}}{96}. \end{aligned}$$
(1.10)
$$\begin{aligned}& \gamma_{n,k}^{(s)}=1+\frac{1}{2}+ \frac{1}{3}+\cdots +\frac{1}{n}- \ln n - \frac{1}{k}\ln \biggl(1+\frac{a_{1}}{n+\frac{a_{2}n}{n+\frac{a _{3}n}{n+\frac{a_{4}n}{n+\ddots +a_{s}}}}} \biggr), \end{aligned}$$
(1.11)
$$\begin{aligned}& a_{1}=\frac{k}{2}, \qquad a_{2}=\frac{2-3k}{12}, \qquad a_{3}= \frac{3k^{2}+4}{12(3k-2)}, \\& a_{4}=-\frac{15k^{4}-30k^{3}+60k^{2}-104k+96}{20(3k-2)(3k ^{2}+4)},\qquad \dots. \end{aligned}$$
$$\begin{aligned}& \gamma_{n,1}^{(2)}=1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}- \ln n - \ln \biggl(1+\frac{a_{1}}{n+a_{2}} \biggr), \end{aligned}$$
(1.12)
$$\begin{aligned}& \gamma_{n,2}^{(3)}=1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}- \ln n - \frac{1}{2}\ln \biggl(1+ \frac{a_{1}}{n+\frac{a_{2}n}{n+a_{3}}} \biggr). \end{aligned}$$
(1.13)
$$\frac{7}{288(n+1)^{3}}< \gamma -\gamma_{n,1}^{(2)}< \frac{7}{288n ^{3}} \quad \mbox{and} \quad \frac{1}{180(n+1)^{4}}< \gamma - \gamma_{n,2}^{(3)}< \frac{1}{180(n-1)^{4}}. $$
On the other hand, Lu [11] introduced the following class of sequences:
where \(k,s\in \mathbb{N}\). They also proved that, among the sequences \((K_{n,k}^{(s)})_{n\geq 1}\), in the case of
the privileged sequence offers the best approximations of γ since when \(s=1\),
when \(s=2\),
when \(s=3\),
$$\begin{aligned} K_{n,k}^{(s)} = &1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}- \ln n \\ &{}- \frac{1}{k}\ln \biggl(1+\frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+ \frac{a _{3}}{n^{3}}+\cdots +\frac{a_{s}}{n^{s}} \biggr), \end{aligned}$$
(1.14)
$$\begin{aligned}& a_{1}=\frac{k}{2}, \qquad a_{2}=\frac{k(3k-2)}{24}, \qquad a_{3}= \frac{k^{2}(k-2)}{48}, \\& a_{4}=\frac{k(15k^{3}-60k^{2}+20k+48)}{5760},\qquad \dots, \end{aligned}$$
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{2}\bigl(K_{n,k}^{(1)}- \gamma\bigr)= \frac{3k-2}{24}; \end{aligned}$$
(1.15)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}\bigl(K_{n,k}^{(2)}- \gamma\bigr)=\frac{k ^{2}-2k}{48}; \end{aligned}$$
(1.16)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{4}\bigl(K_{n,k}^{(3)}- \gamma\bigr)=\frac{15k ^{3}-60k^{2}+20k+48}{5760}. \end{aligned}$$
(1.17)
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These works motivated our study. In this paper, our main goal is to modify the sequence based on the early works of DeTemple, Moritici, and Lu and provide a new convergent sequence of relatively simple form with higher speed.
2 The main results
Lemma 2.1
For any fixed
\(a,b\in \mathbb{R}\), we have the following convergent sequence for Euler’s constant:
Moreover, for
\(a=1\)
and
\(b=0\), we have
for
\(a=1\)
and
\(b=1/2\), we have
for
\(a=2\)
and
\(b=0\), we have
for
\(a=6-2\sqrt{6}\)
and
\(b={1}/{\sqrt{6}}\), we have
and for
\(a=6+2\sqrt{6}\)
and
\(b=-{1}/{\sqrt{6}}\), we have
$$\begin{aligned}& N_{n,a,b}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}+ \frac{1}{an}- \ln (n+b). \end{aligned}$$
(2.1)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n(N_{(n,1,0)}-\gamma)= \frac{1}{2}; \end{aligned}$$
(2.2)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{2}(N_{(n,1,\frac{1}{2})}- \gamma)=\frac{1}{24}; \end{aligned}$$
(2.3)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{2}(N_{(n,2,0)}- \gamma)= \frac{1}{24}; \end{aligned}$$
(2.4)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}(N_{(n,6-2\sqrt{6},\frac{1}{ \sqrt{6}})}- \gamma)=-\frac{1}{18\sqrt{6}}; \end{aligned}$$
(2.5)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}(N_{(n,6+2\sqrt{6},-\frac{1}{ \sqrt{6}})}- \gamma)=\frac{1}{18\sqrt{6}}. \end{aligned}$$
(2.6)
Using Lemma 2.1, we have the following conclusion.
Corollary 2.2
The fastest possible sequence
\((N_{n,a,b})_{n\geq 1}\)
is obtained only for
and
$$\textstyle\begin{cases} \frac{1}{a}-b-\frac{1}{2}=0,\\ -\frac{1}{a}+b^{2}+b+ \frac{1}{3}=0,\end{cases} $$
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}(N_{n,a,b}- \gamma)= \frac{1}{3} \biggl(\frac{1}{a}-b^{3}- \frac{3b^{3}}{2}-b- \frac{1}{4} \biggr). \end{aligned}$$
(2.7)
Theorem 2.3
For any fixed
\(s\in \mathbb{N}\), there exist
\(k\in \mathbb{N}\)
and
\(a, b \in \mathbb{R}\)
such that the following sequence converges to Euler’s constant:
where
Furthermore, let
Then we also have, for
\(s=1\),
for
\(s=2\),
and for
\(s=3\),
$$\begin{aligned} \gamma \approx \gamma_{n,k,a,b}^{(s)} =& 1+ \frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n-1}+ \frac{1}{an}- \ln (n+b) \\ &{}- \frac{1}{k}\ln \biggl(1+\frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+ \frac{a_{3}}{n^{3}}+ \cdots +\frac{a_{s}}{n^{s}} \biggr), \end{aligned}$$
(2.8)
$$\begin{aligned}& a_{1}=k \biggl(\frac{1}{a}-b-\frac{1}{2} \biggr), \\& a_{2}=-\frac{1}{2a}+\frac{b^{2}}{2}+ \frac{b}{2}+\frac{1}{6}+\frac{(2k-2abk-ak)^{2}}{8a^{2}k}+\frac{2k-2abk-ak}{4ak}, \\& a_{3}=\frac{k}{3a}-\frac{b^{3}k}{3}- \frac{b^{2}k}{2}- \frac{bk}{3}-\frac{k}{12}-\frac{a_{1}}{3}- \frac{a_{1}^{2}}{2}+a_{2}+a _{1}a_{2}- \frac{a_{1}^{3}}{3},\qquad \dots. \end{aligned}$$
$$\begin{aligned}& \gamma_{n,k,a,b}^{(1)}=1+\frac{1}{2}+ \frac{1}{3} +\cdots + \frac{1}{n-1}+\frac{1}{an}- \ln (n+b)- \frac{1}{k}\ln \biggl(1+\frac{a _{1}}{n} \biggr), \end{aligned}$$
(2.9)
$$\begin{aligned}& \gamma_{n,k,a,b}^{(2)}=1+ \frac{1}{2}+ \frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{an}- \ln (n+b)- \frac{1}{k}\ln \biggl(1+\frac{a _{1}}{n}+\frac{a_{2}}{n^{2}} \biggr), \end{aligned}$$
(2.10)
$$\begin{aligned}& \gamma_{n,k,a,b}^{(3)}=1+ \frac{1}{2}+ \frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{an}- \ln (n+b)- \frac{1}{k}\ln \biggl(1+\frac{a _{1}}{n}+\frac{a_{2}}{n^{2}}+ \frac{a_{3}}{n^{3}} \biggr). \end{aligned}$$
(2.11)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3} \bigl( \gamma_{n,k,a,b}^{(1)}- \gamma \bigr) \\& \quad =\frac{4k-4ab^{3}k-6ab^{2}k-4abk-ak-4aa_{1}-6aa _{1}^{2}-4aa_{1}^{3}}{12ak}; \end{aligned}$$
(2.12)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{4} \bigl( \gamma_{n,k,a,b}^{(2)}- \gamma \bigr)=-\frac{1}{4a}+ \frac{b^{4}}{4}+\frac{b^{2}}{2}+ \frac{b}{4}+\frac{1}{20}; \end{aligned}$$
(2.13)
$$\begin{aligned} \lim_{n\rightarrow \infty }n^{5}\bigl( \gamma_{n,a,b,k}^{(3)}- \gamma\bigr) = &\frac{1}{5a}- \frac{b^{5}}{5}-\frac{b^{4}}{2}-\frac{2b ^{3}}{3}-\frac{b^{2}}{2}- \frac{b}{5}-\frac{1}{30}- \frac{a_{1}}{5k} \\ &{}+\frac{a_{2}}{k}-\frac{3a_{3}}{5k}+\frac{2a_{1}a _{2}}{k}+ \frac{a_{2}a_{3}}{k}-\frac{2a_{1}a_{3}}{k} \\ &{}-\frac{a_{1} ^{2}a_{3}}{k}-\frac{a_{1}a_{2}^{2}}{k}+\frac{2a_{1}^{2}a_{2}}{k}+ \frac{a _{1}^{3}a_{2}}{k}-\frac{a_{1}^{4}}{2k} \\ &{}-\frac{2a_{1}^{3}}{3k}-\frac{4a _{1}^{5}}{25}. \end{aligned}$$
(2.14)
Lemma 2.4
If
\((x_{n})_{n\geq 1}\)
converges to zero and there exists the limit
with
\(s>1\), then
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{s}(x_{n}-x_{n+1})=l \in [- \infty,+\infty ] \end{aligned}$$
(2.15)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{s-1}x_{n}= \frac{l}{s-1}. \end{aligned}$$
(2.16)
3 The proof of Theorem 2.3
Based on the argument of Theorem 2.1 in [13] or Theorem 5 in [14], we need to find the value of \(a_{1}\in \mathbb{R} \) that produces the most accurate approximation of the form
To measure the accuracy of this approximation, a method is to say that an approximation (3.1) is better as \(N_{n,a,b}-\gamma \) faster converges to zero. Using (3.1), we have
Developing in power series in \(1/n\), we have
From Lemma 2.4 we know that the speed of convergence of the sequence \((N_{n,a,b})_{n\geq 1}\) is even higher than the value s satisfying (2.15). Thus, using Lemma 2.4, we have: If \(-\frac{1}{a}+b^{2}+b+\frac{1}{3}\neq 0 \), then the rate of convergence of the sequence \((N_{n,a,b}-\gamma)_{n \geq 1}\) is \(n^{-2}\) since
If \(-\frac{1}{a}+b^{2}+b+\frac{1}{3}=0 \), then from (3.3) we have
and the rate of convergence of the sequence \((N_{n,a,b}-\gamma)_{n \geq 1}\) is \(n^{-3}\) since
Moreover, for \(a=1\) and \(b=0\), we have
for \(a=1\) and \(b=1/2\), we have
for \(a=2\) and \(b=0\), we have
for \(a=6-2\sqrt{6}\) and \(b={1}/{\sqrt{6}}\), we have
and for \(a=6+2\sqrt{6}\) and \(b=-{1}/{\sqrt{6}}\), we have
$$\begin{aligned}& N_{n,a,b}=1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n-1}+ \frac{1}{an}- \ln (n+b). \end{aligned}$$
(3.1)
$$\begin{aligned} N_{n,a,b}-N_{n+1,a,b} = &-\frac{1}{n}+ \frac{1}{an}-\frac{1}{a(n+1)} \\ &{}- \ln (n+b)+\ln (n+1+b). \end{aligned}$$
(3.2)
$$\begin{aligned} N_{n,a,b}-N_{n+1,a,b} = & \biggl(\frac{1}{a}-b- \frac{1}{2} \biggr)\frac{1}{n ^{2}}+ \biggl(-\frac{1}{a}+b^{2}+b+ \frac{1}{3} \biggr)\frac{1}{n^{3}} \\ &{}+ \biggl(\frac{1}{a}-b^{3}- \frac{3b^{2}}{2}-b-\frac{1}{4} \biggr)\frac{1}{n^{4}} \\ &{}+ \biggl(-\frac{1}{a}+b^{4}+2b^{3}+2b ^{2}+b+1 \biggr)\frac{1}{n^{5}}+O \biggl( \frac{1}{n^{6}} \biggr). \end{aligned}$$
(3.3)
(i)
If \(\frac{1}{a}-b-\frac{1}{2}\neq 0 \), then the convergence rate of the sequence \((N_{n,a,b}-\gamma)_{n \geq 1}\) is \(1/n\) since
$$\lim_{n\rightarrow \infty }n(N_{n,a,b}-\gamma)= \frac{1}{a}-b- \frac{1}{2}\neq 0. $$
(ii)
If \(\frac{1}{a}-b-\frac{1}{2}=0 \), then from (3.3) we have
$$\begin{aligned} N_{n,a,b}-N_{n+1,a,b} = & \biggl(-\frac{1}{a}+b^{2}+b+ \frac{1}{3} \biggr)\frac{1}{n^{3}}+ \biggl(\frac{1}{a}-b^{3}- \frac{3b^{2}}{2}-b- \frac{1}{4} \biggr)\frac{1}{n^{4}} \\ &{}+ \biggl(-\frac{1}{a}+b^{4}+2b ^{3}+2b^{2}+b+1 \biggr) \frac{1}{n^{5}}+O \biggl(\frac{1}{n^{6}} \biggr). \end{aligned}$$
$$\lim_{n\rightarrow \infty }n^{2}(N_{n,a,b}- \gamma)=- \frac{1}{2a}+\frac{b^{2}}{2}+\frac{b}{2}+ \frac{1}{6}. $$
$$\begin{aligned} N_{n,a,b}-N_{n+1,a,b} = & \biggl(\frac{1}{a}-b^{3}- \frac{3b^{2}}{2}-b- \frac{1}{4} \biggr)\frac{1}{n^{4}} \\ &{}+ \biggl(-\frac{1}{a}+b^{4}+2b ^{3}+2b^{2}+b+1 \biggr) \frac{1}{n^{5}} +O \biggl(\frac{1}{n^{6}} \biggr), \end{aligned}$$
$$\lim_{n\rightarrow \infty }n^{3}(N_{n,a,b}- \gamma)= \frac{1}{3a}-\frac{b^{3}}{3}-\frac{b^{2}}{2}- \frac{b}{3}- \frac{1}{12}. $$
$$\begin{aligned}& \lim_{n\rightarrow \infty }n(N_{(n,1,0)}-\gamma)= \frac{1}{2}; \end{aligned}$$
(3.4)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{2}(N_{(n,1,\frac{1}{2})}- \gamma)=\frac{1}{24}; \end{aligned}$$
(3.5)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{2}(N_{(n,2,0)}- \gamma)= \frac{1}{24}; \end{aligned}$$
(3.6)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}(N_{(n,6-2\sqrt{6},\frac{1}{ \sqrt{6}})}- \gamma)=-\frac{1}{18\sqrt{6}}; \end{aligned}$$
(3.7)
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}(N_{(n,6+2\sqrt{6},-\frac{1}{ \sqrt{6}})}- \gamma)=\frac{1}{18\sqrt{6}}. \end{aligned}$$
(3.8)
Proof of Theorem 2.3
We define the sequence \((\gamma_{n,a,b,k}^{(s)})_{n\geq 1}\) by the relations
and
Using a similar method as in (3.1)–(3.3), we have
The fastest possible sequence \((\gamma_{n,a,b,k}^{(1)})_{n\geq 1}\) is obtained when
Then we have
and the rate of convergence is \(n^{-3}\).
$$\begin{aligned} \gamma \approx \gamma_{n,k,a,b}^{(s)} = &1+ \frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n-1}+ \frac{1}{an}- \ln (n+b) \\ &{}- \frac{1}{k}\ln \biggl(1+\frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+ \frac{a_{3}}{n^{3}}+ \cdots +\frac{a_{s}}{n^{s}} \biggr) \end{aligned}$$
(3.9)
$$\begin{aligned} \gamma \approx \gamma_{n,k,a,b}^{(1)} = &1+ \frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n-1}+ \frac{1}{an}- \ln (n+b) \\ &{}- \frac{1}{k}\ln \biggl(1+\frac{a_{1}}{n} \biggr). \end{aligned}$$
(3.10)
$$\begin{aligned}& \gamma_{n,k,a,b}^{(1)}-\gamma_{n+1,k,a,b}^{(1)} \\& \quad =\frac{-3k+3ab ^{2}k+3abk+ak+3aa_{1}+3aa_{1}^{2}}{3akn^{3}} \\& \qquad {}+\frac{2k-2abk-ak-2aa _{1}}{2akn^{2}} \\& \qquad {}+\frac{4k-4ab^{3}k-6ab^{2}k-4abk-ak-4aa _{1}-6aa_{1}^{2}-4aa_{1}^{3}}{4akn^{4}} \\& \qquad {}+\frac{-5k+5ab^{4}k+10ab ^{3}k+10ab^{2}k+5abk+ak-5aa_{1}+10aa_{1}^{2}}{5akn^{5}} \\& \qquad {}+\frac{10aa _{1}^{2}+10aa_{1}^{3}+5aa_{1}^{4}}{5akn^{5}}+O \biggl(\frac{1}{n^{6}} \biggr). \end{aligned}$$
(3.11)
$$\textstyle\begin{cases} \frac{2k-2abk-ak-2aa_{1}}{2ak}=0, \\ \frac{-3k+3ab^{2}k+3abk+ak+3aa _{1}+3aa_{1}^{2}}{3ak}=0.\end{cases} $$
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}\bigl( \gamma_{n,a,b,k}^{(1)}- \gamma\bigr) \\& \quad =\frac{4k-4ab^{3}k-6ab^{2}k-4abk-ak-4aa_{1}-6aa_{1} ^{2}-4aa_{1}^{3}}{12ak}, \end{aligned}$$
For example, for \(a=2\) and \(b=1/(2\sqrt{3})\),
and the rate of convergence is \(n^{-3}\).
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{3}\bigl( \gamma_{n,2,\frac{1}{2\sqrt{3}},k}^{(1)}-\gamma\bigr)=\frac{-k+3 \sqrt{3}k+3\sqrt{3}k^{2}-k^{3}}{72\sqrt{3}k}, \end{aligned}$$
Next, we define the second sequence with the previous conclusions:
where \(a_{1}=\frac{2k-2abk-ak}{2a} \).
$$\begin{aligned} \gamma_{n,k,a,b}^{(2)} = &1+\frac{1}{2}+ \frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{an}- \ln (n+b) \\ &{}- \frac{1}{k}\ln \biggl(1+\frac{a _{1}}{n}+\frac{a_{2}}{n^{2}} \biggr), \end{aligned}$$
(3.12)
Then we get the equation
Taking
we obtain the fastest sequence \((\gamma_{n,a,b,k}^{(2)})_{n\geq 1}\) with convergent rate \(n^{-4}\) since
Moreover, for
we define the third sequence with the previous conclusions:
Then we have the equality
Taking
we obtain the fastest sequence \((\gamma_{n,a,b,k}^{(3)})_{n\geq 1}\) with convergent rate \(n^{-5}\) since
□
$$\begin{aligned}& \gamma_{n,k,a,b}^{(2)}-\gamma_{n+1,k,a,b}^{(2)} \\& \quad = \biggl(- \frac{1}{a}+b^{2}+b+ \frac{1}{3}+\frac{a_{1}^{2}}{k}+ \frac{a_{1}}{k}-\frac{2a_{2}}{k} \biggr)\frac{1}{n^{3}} \\& \qquad {}+ \biggl( \frac{1}{a}-b^{3}- \frac{3b^{2}}{2}-b-\frac{1}{4}-\frac{3a_{1} ^{2}}{2k}+\frac{3a_{1}a_{2}}{k} \biggr)\frac{1}{n^{4}} \\& \qquad {}+ \biggl(-\frac{a _{1}^{3}}{k}-\frac{a_{1}}{k}+\frac{3a_{2}}{k} \biggr)\frac{1}{n^{4}}+ \biggl(-\frac{1}{a}+b^{4}+2b^{3} \\& \qquad {}+2b^{2}+b+\frac{1}{5}+ \frac{a _{1}}{k}-\frac{4a_{2}}{k}+\frac{2a_{1}^{2}}{k}+\frac{2a_{2}^{2}}{k} \\& \qquad {}-\frac{6a_{1}a_{2}}{k}+\frac{2a_{1}^{3}}{k}- \frac{4a_{1}^{2}a_{2}}{k} + \frac{a_{1}^{4}}{k} \biggr)\frac{1}{n^{5}} +O \biggl(\frac{1}{n^{6}} \biggr). \end{aligned}$$
(3.13)
$$\textstyle\begin{cases} a_{1}=\frac{k}{a}-bk-\frac{k}{2}, \\ a_{2}=-\frac{k}{2a}+\frac{b ^{2}k}{2}+\frac{bk}{2}+\frac{k}{6}+\frac{a_{1}^{2}}{2}+ \frac{a_{1}}{2},\\ \frac{1}{a}-b^{3}-\frac{3b^{2}}{2}-b- \frac{1}{4}-\frac{3a_{1}^{2}}{2k}+\frac{3a_{1}a_{2}}{k}-\frac{a_{1} ^{3}}{k}-\frac{a_{1}}{k}+\frac{3a_{2}}{k}=0, \end{cases} $$
$$\begin{aligned} \lim_{n\rightarrow \infty }n^{4}\bigl( \gamma_{n,a,b,k}^{(2)}- \gamma\bigr) = &-\frac{1}{4a}+ \frac{b^{4}}{4}+\frac{b^{3}}{2}+\frac{2b ^{2}}{4}+\frac{b}{4}+ \frac{1}{20}+\frac{a_{1}}{4k}-\frac{a_{2}}{k}+\frac{a _{1}^{2}}{2k} \\ &{}+\frac{a_{2}^{2}}{2k}-\frac{3a_{1}a_{2}}{2k}+\frac{a _{1}^{3}}{2k} - \frac{a_{1}^{2}a_{2}}{k} +\frac{a_{1}^{4}}{4k}. \end{aligned}$$
$$\textstyle\begin{cases} a_{1}=\frac{k}{a}-bk-\frac{k}{2},\\ a_{2}=-\frac{k}{2a}+\frac{b^{2}k}{2}+\frac{bk}{2}+\frac{k}{6}+\frac{a_{1}^{2}}{2}+ \frac{a_{1}}{2}, \end{cases} $$
$$\begin{aligned} \gamma_{n,k,a,b}^{(3)} = &1+\frac{1}{2}+ \frac{1}{3}+\cdots + \frac{1}{n-1}+\frac{1}{an}- \ln (n+b) - \frac{1}{k}\ln \biggl(1+\frac{a _{1}}{n}+\frac{a_{2}}{n^{2}}+ \frac{a_{3}}{n^{3}} \biggr). \end{aligned}$$
(3.14)
$$\begin{aligned}& \gamma_{n,k,a,b}^{(3)}-\gamma_{n+1,k,a,b}^{(3)} \\& \quad = \biggl(\frac{1}{a}-b ^{3}- \frac{3b^{2}}{2}-b-\frac{1}{4}-\frac{a_{1}}{k}-\frac{3a _{1}^{2}}{2k}+ \frac{3a_{2}}{k}-\frac{3a_{3}}{k}+\frac{3a_{1}a_{2}}{k}-\frac{a_{1}^{3}}{k} \biggr) \\& \qquad {}\frac{1}{n^{4}}+ \biggl(- \frac{1}{a}+b ^{4}+2b^{3}+2b^{2}+b+ \frac{1}{5}+\frac{a_{1}}{k}-\frac{4a _{2}}{k} \\& \qquad {}+\frac{2a_{2}^{2}}{k}+\frac{6a_{3}}{k}+ \frac{2a_{1}^{2}}{k}+ \frac{2a_{1}^{3}}{k}-\frac{6a_{1}a_{2}}{k}+\frac{4a _{1}a_{3}}{k}-\frac{4a_{1}^{2}a_{2}}{k}+ \frac{a_{1}^{4}}{k} \biggr)\frac{1}{n ^{5}} \\& \qquad {}+ \biggl(\frac{1}{a}-b^{5}- \frac{5b^{4}}{2}-\frac{10b ^{3}}{3}-\frac{5b^{2}}{2}-b-\frac{1}{6}- \frac{a_{1}}{k}+\frac{5a _{2}}{k}-\frac{3a_{3}}{k} \\& \qquad {} +\frac{10a_{1}a_{2}}{k}+\frac{5a_{2}a _{3}}{k}-\frac{10a_{1}a_{3}}{k}- \frac{5a_{1}^{2}a_{3}}{k}-\frac{5a _{1}a_{2}^{2}}{k}+\frac{10a_{1}^{2}a_{2}}{k} \\& \qquad {}+\frac{5a_{1}^{3}a _{2}}{k}-\frac{5a_{1}^{4}}{2k}-\frac{10a_{1}^{3}}{3k}- \frac{4a_{1} ^{5}}{5} \biggr)\frac{1}{n^{6}}+O \biggl(\frac{1}{n^{7}} \biggr). \end{aligned}$$
(3.15)
$$\textstyle\begin{cases} a_{3}=\frac{k}{3a}-\frac{b^{3}k}{3}-\frac{b^{2}k}{2}- \frac{bk}{3}-\frac{k}{12}-\frac{a_{1}}{3}-\frac{a_{1}^{2}}{2}+a_{2}+a _{1}a_{2}-\frac{a_{1}^{3}}{3}, \\ -\frac{1}{a}+b^{4}+2b^{3}+2b ^{2}+b+\frac{1}{5}+\frac{a_{1}}{k}-\frac{4a_{2}}{k}+\frac{2a_{2} ^{2}}{k}+\frac{6a_{3}}{k}+\frac{2a_{1}^{2}}{k}+\frac{2a_{1}^{3}}{k}, \\ -\frac{6a_{1}a_{2}}{k}+\frac{4a_{1}a_{3}}{k}-\frac{4a_{1}^{2}a _{2}}{k}+\frac{a_{1}^{4}}{k}=0, \end{cases} $$
$$\begin{aligned}& \lim_{n\rightarrow \infty }n^{5}\bigl( \gamma_{n,a,b,k}^{(3)}- \gamma\bigr) \\& \quad =\frac{1}{5a}-\frac{b^{5}}{5}-\frac{b^{4}}{2}- \frac{2b ^{3}}{3}-\frac{b^{2}}{2}-\frac{b}{5}-\frac{1}{30}- \frac{a_{1}}{5k}+\frac{a_{2}}{k}-\frac{3a_{3}}{5k}+ \frac{2a_{1}a_{2}}{k}+ \frac{a_{2}a_{3}}{k} \\& \qquad {}-\frac{2a_{1}a_{3}}{k}-\frac{a _{1}^{2}a_{3}}{k}-\frac{a_{1}a_{2}^{2}}{k}+ \frac{2a_{1}^{2}a_{2}}{k}+\frac{a _{1}^{3}a_{2}}{k}-\frac{a_{1}^{4}}{2k}-\frac{2a_{1}^{3}}{3k}- \frac{4a _{1}^{5}}{25}. \end{aligned}$$
Acknowledgements
This work was supported by the National Natural Science Foundation of China (Nos. U1706227, 11201095), the Youth Scholar Backbone Supporting Plan Project of Harbin Engineering University, the Fundamental Research Funds for the Central Universities (No. HEUCFM181102), the Postdoctoral research startup foundation of Heilongjiang (No. LBH-Q14044), and the Science Research Funds for Overseas Returned Chinese Scholars of Heilongjiang Province (No. LC201502).
Competing interests
The authors declare that they have no competing interests.
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