1 Introduction
Somos’ quadratic recurrence constant σ arises in the study of the asymptotic behavior of the sequence (see [8, p. 446] and [21, 27]):
where \(g_{n}\) are recursively defined by
with the first few terms
The constant σ is usually defined by
or
See [9, 20‐22]. The constant σ appears in many important problems in pure and applied analysis and it has been investigated by a large number of researchers [3, 5, 9, 10, 12, 13, 15, 17‐19, 25, 30].
$$\begin{aligned} g_{n}\sim{}& \frac{\sigma ^{2^{n}}}{n} \biggl(1+ \frac{2}{n}-\frac{1}{n^{2}}+\frac{4}{n ^{3}}- \frac{21}{n^{4}}+\frac{138}{n^{5}}-\frac{1091}{n^{6}}+ \frac{10\text{,}088}{n ^{7}}-\frac{106\text{,}918}{n^{8}} \\ &{}+\frac{1\text{,}279\text{,}220}{n^{9}}-\frac{17\text{,}070\text{,}418}{n^{10}}+\frac{251\text{,}560\text{,}472}{n ^{11}}- \frac{4\text{,}059\text{,}954\text{,}946}{n^{12}}+\cdots \biggr)^{-1}, \end{aligned}$$
(1.1)
$$\begin{aligned} g_{0}=1,\qquad g_{n}=ng_{n-1}^{2},\quad n=1,2,\ldots, \end{aligned}$$
(1.2)
$$\begin{aligned} g_{0}=1,\qquad g_{1}=1,\qquad g_{2}=2,\qquad g_{3}=12, \qquad g_{4}=576,\qquad g_{5}=1 \text{,}658\text{,}880,\ldots . \end{aligned}$$
$$\begin{aligned} \sigma =\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots }}}}=\prod _{k=1} ^{\infty }k^{\frac{1}{2^{k}}}=1.66168794\cdots \end{aligned}$$
(1.3)
$$\begin{aligned} \sigma =\exp \biggl\{ - \int _{0}^{1}\frac{1-x}{(2-x)\ln x}\,dx \biggr\} = \exp \biggl\{ - \int _{0}^{1} \int _{0}^{1}\frac{x}{(2-xy)\ln (xy)}\,dx\,dy \biggr\} . \end{aligned}$$
(1.4)
Recently, Sondow and Hadjicostas [25] introduced and studied the generalized-Euler-constant function \(\gamma (z)\) defined by the power series
when \(|z|\leq 1\). There exist integral representations for the function
Its values include Euler’s constant \(\gamma =\gamma (1)\) and the “alternating Euler constant” \(\log \frac{4}{\pi }=\gamma (-1)\), see for example [23, 24]. In particular, at \(z=1/2\), the function takes the value
which is equivalent to
Mortici [15] proved that, for \(n\geq 1\), it follows that
where the partial sum of \(\gamma (z)\)
Lu and Song [13] improved Mortici’s estimate and proved that, for \(n\geq 1\),
You and Chen [30] improved these inequalities by using continued fraction. Very recently, Chen and Han [5] obtained new lower bounds for \(\gamma (1/2)-\gamma _{n}(1/2)\):
In their paper, Chen and Han pointed out that the lower bound in (1.11) is sharper than the one in (1.10) for \(n\geq 24\), and the upper bound in (1.11) is sharper than the one in (1.10) for \(n\geq 18\). Moreover, they gave the following asymptotic expansion:
with a recursive formula for successively determining the coefficients
$$\begin{aligned} \gamma (z)=\sum_{k=1}^{\infty }z^{k-1} \biggl(\frac{1}{k}-\ln \frac{k+1}{k} \biggr) \end{aligned}$$
(1.5)
$$\begin{aligned} \gamma (z)= \int _{0}^{1}\frac{1-x+\ln x}{(1-xz)\ln x}\,dx=- \int _{0}^{1} \int _{0}^{1}\frac{1-x}{(1-xyz)\ln (xy)}\,dx\,dy. \end{aligned}$$
(1.6)
$$\begin{aligned} \gamma \biggl(\frac{1}{2} \biggr)=2\ln \frac{2}{\sigma }, \end{aligned}$$
(1.7)
$$\begin{aligned} \sigma =2\exp \biggl\{ -\frac{1}{2}\gamma \biggl( \frac{1}{2} \biggr) \biggr\} . \end{aligned}$$
(1.8)
$$\begin{aligned} \frac{270(n+1)}{2^{n}(270n^{3}+1530n^{2}+1065n+6293)}< \gamma \biggl(\frac{1}{2} \biggr)- \gamma _{n} \biggl(\frac{1}{2} \biggr)< \frac{18}{2^{n}(18n^{2}+84n-13)}, \end{aligned}$$
(1.9)
$$\begin{aligned} \gamma _{n}(z)=\sum_{k=1}^{n} z^{k-1} \biggl(\frac{1}{k}-\ln \frac{k+1}{k} \biggr),\quad \vert z \vert \leq 1. \end{aligned}$$
$$\begin{aligned} \frac{690n^{2}+3524n+145}{6(2^{n})(n+1)^{2}(115n^{2}+894n+779)}&< \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{n} \biggl(\frac{1}{2} \biggr) \\ &< \frac{48n+127}{3(2^{n})(16n+85)(n+1)^{2}}. \end{aligned}$$
(1.10)
$$\begin{aligned} &\frac{1}{2^{n}} \biggl(\frac{1}{(n+1)^{2}}- \frac{8}{3(n+1)^{3}}+ \frac{23}{2(n+1)^{4}}-\frac{332}{5(n+1)^{5}}+ \frac{479}{(n+1)^{6}}- \frac{29\text{,}024}{7(n+1)^{7}} \biggr) \\ & \quad< \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{n} \biggl( \frac{1}{2} \biggr) \\ &\quad < \frac{1}{2^{n}} \biggl(\frac{1}{(n+1)^{2}}- \frac{8}{3(n+1)^{3}}+ \frac{23}{2(n+1)^{4}}-\frac{332}{5(n+1)^{5}}+ \frac{479}{(n+1)^{6}} \biggr). \end{aligned}$$
(1.11)
$$\begin{aligned} &\gamma \biggl(\frac{1}{2} \biggr)-\gamma _{n} \biggl(\frac{1}{2} \biggr) \\ &\quad \sim \frac{1}{2^{n}} \biggl\{ \frac{a_{2}}{(n+1)^{2}}+\frac{a _{3}}{(n+1)^{3}}+ \frac{a_{4}}{(n+1)^{4}}+\frac{a_{5}}{(n+1)^{5}}+\frac{a _{6}}{(n+1)^{6}}+\cdots \biggr\} ,\quad n\rightarrow \infty \end{aligned}$$
(1.12)
$$\begin{aligned} \begin{aligned} &a_{2}=1, \\ &a_{k}=\frac{2(-1)^{k}}{k}+\sum_{j=2}^{k-1}(-1)^{k-j} \binom{k-1}{k-j}a_{j},\quad k\geq 3. \end{aligned} \end{aligned}$$
(1.13)
Anzeige
Recently, there have been several interesting works related to approximations of Euler’s constant \(\gamma =\gamma (1)\), see for example [4, 16, 28] and the references therein. Also, some works related to approximations of \(\gamma (z)\) at special values \(1/2\), \(1/3\), and \(1/4\) appear in [3, 5, 13‐15, 29, 30]. Motivated by these, the first aim of this paper is to give an approximation for \(\gamma (z)\) when \(0< z<1\). Specifically, we give a general inequality for the error bound of \(\gamma (z)-\gamma _{N-1}(z)\). The second aim is to establish an asymptotic expansion for \(\gamma (z)- \gamma _{N-1}(z)\) whose coefficients can be computed explicitly and recursively. In particular, we generalize inequality (1.11) and asymptotic expansion (1.12) due to Chen and Han. Using the relation between the generalized Somos’ quadratic recurrence constant and the generalized-Euler-constant function established by Sondow and Hadjicostas [25], we find approximate estimates for the generalized Somos’ quadratic recurrence constant and its natural logarithm, respectively. Moreover, two asymptotic expansions for the natural logarithm of the generalized Somos’ quadratic recurrence constant are presented.
2 Preliminaries
Let us recall the following classical result [2, 7, 11]:
where \(A_{n}(z)\) is the Eulerian polynomial of degree n, and it can be defined by the exponential generating function
By (2.2), \(A_{n}(x)\) may be explicitly written as
where \(A(n,k)\) are known as the Eulerian numbers, the numbers of permutations of the numbers 1 to n in which exactly k elements are greater than the previous element, and they can be expressed by
Therefore, one may easily obtain
$$\begin{aligned} \sum_{k=0}^{\infty }k^{n}z^{k}= \frac{A_{n}(z)}{(1-z)^{n+1}},\quad \vert z \vert < 1, \end{aligned}$$
(2.1)
$$\begin{aligned} \sum_{n=0}^{\infty }A_{n}(z) \frac{t^{n}}{n!}= \frac{1-z}{1-ze^{t(1-z)}}. \end{aligned}$$
(2.2)
$$\begin{aligned} \begin{aligned} &A_{0}(z)=1, \\ &A_{n}(z)=\sum_{k=1}^{n}A(n,k)z^{k},\quad n\geq 1, \end{aligned} \end{aligned}$$
(2.3)
$$\begin{aligned} A(n,k)=\sum_{i=1}^{k}(-1)^{i} \binom{n+1}{i}(k-i)^{n}. \end{aligned}$$
(2.4)
$$\begin{aligned} A_{0}(z)=1, \qquad A_{1}(z)=z, \qquad A_{2}(z)=z^{2}+z,\qquad A_{3}(z)=z^{3}+4z^{2}+z, \ldots \end{aligned}$$
For \(z\neq 1\), denote the Eulerian fraction
From (2.2) it is clear that
which implies that \(b_{k}(z)\) can be computed by the following recurrence relation:
Also, we can obtain an explicit expression [11] as follows:
where \(S(k,j)\) are the Stirling numbers of the second kind.
$$\begin{aligned} b_{n}(z)=\frac{A_{n}(z)}{(1-z)^{n+1}}. \end{aligned}$$
(2.5)
$$\begin{aligned} \sum_{k=0}^{\infty }b_{k}(z) \frac{t^{k}}{k!}=\frac{1}{1-ze^{t}}, \end{aligned}$$
(2.6)
$$\begin{aligned} &b_{0}(z)=\frac{1}{1-z}, \end{aligned}$$
(2.7)
$$\begin{aligned} &b_{k}(z)=\frac{z}{1-z}\sum_{j=0}^{k-1} \binom{k}{j}b_{j}(z), \quad k \geq 1. \end{aligned}$$
(2.8)
$$\begin{aligned} b_{k}(z)=\sum_{j=0}^{k} j!S(k,j)\frac{z^{j}}{(1-z)^{j+1}}, \end{aligned}$$
(2.9)
It is worth noting that a general summation formula for power series using the Eulerian fractions was established in the recent paper [26].
Anzeige
Lemma 2.1
(cf. Theorem 4.2 [26])
Let
m
be a positive integer, and both
P
and
Q
be nonnegative integers. Suppose that
\(f(x)\)
has the
\((m+1)\)th continuous derivative on
\([P,Q]\). For
\(0< z<1\), \(f^{(m)}(x)\)
and
\(f^{(m+1)}(x)\)
keep opposite signs in
\((P,Q)\). Then there exists a number
\(\theta \in (0,1)\)
such that
$$\begin{aligned} \sum_{k=P}^{Q-1}f(k)z^{k}= \sum_{k=0}^{m-1}\frac{b_{k}(z)}{k!} \bigl[z ^{P}f^{(k)}(P)-z^{Q}f^{(k)}(Q) \bigr]+\theta \frac{b_{m}(z)}{m!} \bigl[z^{P}f^{(m)}(P)-z^{Q}f^{(m)}(Q) \bigr]. \end{aligned}$$
(2.10)
Remark 2.1
It is well known that the Euler–Maclaurin formula [1] is usually applied to treat the summation \(S=\sum_{k=a}^{b}f(k)\), and it was also used to approximate the generalized-Euler-constant function \(\gamma (z)\), see [12]. It seems that the above lemma is more suitable to treat a general summation formula for power series with the form \(S=\sum_{k=a}^{b}f(k)z^{k}\ (0< z<1)\). The lemma provides us a new approach to estimate \(\gamma (z)\), and it will play a key role in the next section.
Remark 2.2
In particular, if
as \(Q\rightarrow +\infty \), then (2.10) can be rewritten as
Obviously, taking \(f(x)=x^{n}\), \(P=0\), and \(m=n+1\), (2.12) reduces to (2.1).
$$\begin{aligned} \lim_{Q\rightarrow +\infty }z^{Q}f^{(k)}(Q)=0,\quad 0\leq k\leq m, \end{aligned}$$
(2.11)
$$\begin{aligned} \sum_{k=P}^{\infty }f(k)z^{k}=z^{P} \Biggl\{ \sum_{k=0}^{m-1} \frac{b _{k}(z)}{k!}f^{(k)}(P)+\theta \frac{b_{m}(z)}{m!}f^{(m)}(P) \Biggr\} . \end{aligned}$$
(2.12)
3 Estimate of the generalized-Euler-constant function
Let N be a positive integer. In this section, by using (2.12) we first give a general inequality for the error bound of \(\gamma (z)-\gamma _{N-1}(z)\) when \(0< z<1\).
Theorem 3.1
Let
p
and
q
be any positive integers. If
\(N\geq -\frac{c_{2p+3}(z)}{c _{2p+2}(z)}\), then we have
If
\(N\geq -\frac{c_{2q+2}(z)}{c_{2q+1}(z)}\), then we have
In the above inequalities, the coefficients
\(c_{k}(z)\)
are explicitly expressed by
where
\(b_{k}(z)\)
are the Eulerian fractions introduced in the second section.
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)> z^{N-1}\sum _{k=2}^{2p+1}\frac{c_{k}(z)}{N ^{k}}, \quad 0< z< 1. \end{aligned}$$
(3.1)
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)< z^{N-1}\sum _{k=2}^{2q}\frac{c_{k}(z)}{N ^{k}},\quad 0< z< 1. \end{aligned}$$
(3.2)
$$\begin{aligned} c_{k}(z)=\frac{(-1)^{k}}{k} \biggl[ \frac{1-z}{z}b_{k}(z)-kb_{k-1}(z) \biggr], \quad k\geq 2, \end{aligned}$$
(3.3)
Proof
Write
After simple calculations, we arrive at
which implies that
By (3.4) and (3.6), for each \(i\geq 0\), we also have
Therefore, according to (2.12), for \(0< z<1\), there exists \(\theta \in (0,1)\) such that
where m is a positive integer. It follows from (3.4) and (3.5) that
By the Taylor expansion we have
where \(0<\phi _{k}<1\) for \(k=0,1,\ldots,m-1\). When \(m\geq 3\), substituting (3.11) and (3.12) into (3.10) yields
where
Taking \(u=k+j\), we obtain
Since
we have
Therefore,
Because of (2.8) and (3.3) it follows that
where
Observe that
which implies that
if
According to (2.8) and (3.3), (3.20) is equivalent to
Thus, if \(m=2q+1\) and \(N\geq -\frac{c_{m+1}(z)}{c_{m}(z)}\), then
As a consequence, by (3.15) we have
Similarly, if \(m=2p+2\) and \(N\geq -\frac{c_{m+1}(z)}{c_{m}(z)}\), then
which implies
This completes the proof. □
$$\begin{aligned} f(t)=\frac{1}{t}-\ln \frac{t+1}{t}. \end{aligned}$$
(3.4)
$$\begin{aligned} f^{(i)}(t) &=(-1)^{i}(i-1)! \biggl[ \frac{i}{t^{i+1}}+ \frac{1}{(t+1)^{i}}-\frac{1}{t^{i}} \biggr] \end{aligned}$$
(3.5)
$$\begin{aligned} &=\frac{(-1)^{i}(i-1)!}{t^{i+1}(t+1)^{i}}\sum_{k=0}^{i} \binom{i}{k} \biggl(i-\frac{k}{i-k+1} \biggr)t^{k}, \quad i \geq 1, \end{aligned}$$
(3.6)
$$\begin{aligned} (-1)^{i}f^{(i)}(t)>0,\quad t>0. \end{aligned}$$
(3.7)
$$\begin{aligned} \lim_{t\rightarrow +\infty } z^{t}f^{(i)}(t)=0,\quad 0< z< 1. \end{aligned}$$
(3.8)
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z) &=\frac{1}{z} \sum_{k=N}^{\infty } \biggl( \frac{1}{k}- \ln \frac{k+1}{k} \biggr)z^{k} \\ &=z^{N-1} \Biggl[\sum_{k=0}^{m-1} \frac{b_{k}(z)}{k!}f^{(k)}(N)+\theta \frac{b_{m}(z)}{m!}f^{(m)}(N) \Biggr], \end{aligned}$$
(3.9)
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)={}&z^{N-1} \Biggl\{ b_{0}(z) \biggl(\frac{1}{N}- \ln \frac{N+1}{N} \biggr) \\ &{}+\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{k} \biggl(\frac{k}{N^{k+1}}+ \frac{1}{(N+1)^{k}}-\frac{1}{N^{k}} \biggr) \\ &{}+\theta \frac{(-1)^{m}b_{m}(z)}{m} \biggl(\frac{m}{N^{m+1}}+ \frac{1}{(N+1)^{m}}-\frac{1}{N^{m}} \biggr) \Biggr\} . \end{aligned}$$
(3.10)
$$\begin{aligned} &\ln \frac{N+1}{N}=\sum_{j=1}^{m} \frac{(-1)^{j-1}}{jN^{j}}+\frac{(-1)^{m}}{(m+1)N ^{m+1}}\frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}}, \end{aligned}$$
(3.11)
$$\begin{aligned} &\frac{1}{(N+1)^{k}}=\frac{1}{N^{k}} \Biggl\{ \sum _{j=0}^{m-k} \binom{k+j-1}{j} \frac{(-1)^{j}}{N^{j}}+\binom{m}{k-1}\frac{(-1)^{m+1-k}}{N ^{m+1-k} (1+\frac{\phi _{k}}{N} )^{m+1}} \Biggr\} , \end{aligned}$$
(3.12)
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)={}&z^{N-1} \Biggl\{ \sum_{k=2}^{m} \frac{(-1)^{k}b _{0}(z)}{kN^{k}} +\frac{(-1)^{m+1}b_{0}(z)}{(m+1)N^{m+1}}\frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}} \\ &{}+\sum_{k=2}^{m}\frac{(-1)^{k-1}b_{k-1}(z)}{N^{k}}+ \sum_{k=1}^{m-1} \sum _{j=0}^{m-k}\frac{(-1)^{k+j}b_{k}(z)}{kN^{k+j}} \binom{k+j-1}{j} \\ &{}+\frac{(-1)^{m+1}}{N^{m+1}}\sum_{k=1}^{m-1} \frac{b_{k}(z)}{k}\frac{ \binom{m}{k-1}}{ (1+\frac{\phi _{k}}{N} )^{m+1}}-\sum_{k=1} ^{m-1}\frac{(-1)^{k}b_{k}(z)}{k}\frac{1}{N^{k}} +\epsilon \Biggr\} , \end{aligned}$$
(3.13)
$$\begin{aligned} \epsilon =\theta \frac{(-1)^{m}b_{m}(z)}{m} \biggl( \frac{m}{N^{m+1}}+ \frac{1}{(N+1)^{m}}-\frac{1}{N^{m}} \biggr). \end{aligned}$$
(3.14)
$$\begin{aligned} &\sum_{k=1}^{m-1}\sum _{j=0}^{m-k}\frac{(-1)^{k+j}b_{k}(z)}{kN^{k+j}} \binom{k+j-1}{j} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{k=1}^{m-1} \sum _{j=1}^{m-k}\frac{(-1)^{k+j}b_{k}(z)}{kN^{k+j}} \binom{k+j-1}{j} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m}\frac{(-1)^{u}}{N ^{u}}\sum _{k=1}^{u-1}\frac{b_{k}(z)}{k} \binom{u-1}{k-1} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m-1}\frac{(-1)^{u}}{N ^{u}}\sum _{k=1}^{u-1}\frac{b_{k}(z)}{k} \binom{u-1}{k-1}+\frac{(-1)^{m}}{N ^{m}}\sum_{k=1}^{m-1} \binom{m-1}{k-1}\frac{b_{k}(z)}{k}. \end{aligned}$$
$$\begin{aligned} \frac{1}{k}\binom{u-1}{k-1}=\frac{1}{u} \binom{u}{k}, \end{aligned}$$
$$\begin{aligned} &\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m-1}\frac{(-1)^{u}}{N ^{u}}\sum _{k=1}^{u-1}\frac{b_{k}(z)}{k} \binom{u-1}{k-1} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m-1}\frac{(-1)^{u}}{uN ^{u}}\sum _{k=1}^{u-1}\binom{u}{k}b_{k}(z) \\ &\quad =-\frac{b_{1}(z)}{N}+\sum_{k=2}^{m-1} \frac{(-1)^{k}}{kN^{k}} \Biggl(b _{k}(z)+\sum _{j=1}^{k-1}\binom{k}{j}b_{j}(z) \Biggr) \\ &\quad=\sum_{k=1}^{m-1}\frac{(-1)^{k}}{kN^{k}} \sum_{j=1}^{k}\binom{k}{j}b _{j}(z). \end{aligned}$$
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)={}&z^{N-1} \Biggl\{ \sum _{k=2}^{m}\frac{(-1)^{k}b _{0}(z)}{kN^{k}} + \frac{(-1)^{m+1}b_{0}(z)}{(m+1)N^{m+1}}\frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}} \\ &{}+\sum_{k=2}^{m}\frac{(-1)^{k-1}b_{k-1}(z)}{N^{k}}+ \sum_{k=1}^{m-1}\frac{(-1)^{k}}{kN ^{k}}\sum _{j=1}^{k}\binom{k}{j}b_{j}(z) \\ &{}+\frac{(-1)^{m}}{N^{m}}\sum_{k=1}^{m-1} \binom{m-1}{k-1} \frac{b_{k}(z)}{k}+\frac{(-1)^{m+1}}{N^{m+1}}\sum _{k=1}^{m-1}\frac{b _{k}(z)}{k} \frac{\binom{m}{k-1}}{ (1+\frac{\phi _{k}}{N} ) ^{m+1}} \\ &{}-\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{k} \frac{1}{N^{k}} +\epsilon \Biggr\} . \end{aligned}$$
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z) &=z^{N-1} \Biggl\{ \sum_{k=2}^{m-1} \frac{c _{k}(z)}{N^{k}} +T_{1}+T_{2}+\epsilon \Biggr\} , \end{aligned}$$
(3.15)
$$\begin{aligned} &T_{1}=\frac{(-1)^{m}b_{0}(z)}{mN^{m}}+\frac{(-1)^{m+1}b_{0}(z)}{(m+1)N ^{m+1}} \frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}}, \end{aligned}$$
(3.16)
$$\begin{aligned} &T_{2}=\frac{(-1)^{m}}{mN^{m}}\sum_{k=1}^{m-2} \binom{m}{k}b_{k}(z)+\frac{(-1)^{m+1}}{(m+1)N ^{m+1}}\sum _{k=1}^{m-1}b_{k}(z) \frac{\binom{m+1}{k}}{ (1+\frac{ \phi _{k}}{N} )^{m+1}}. \end{aligned}$$
(3.17)
$$\begin{aligned} T_{1}+T_{2}=\frac{(-1)^{m}}{N^{m}} \Biggl\{ \frac{1}{m}\sum_{k=0}^{m-2} \binom{m}{k}b_{k}(z)-\frac{1}{(m+1)N}\sum _{k=0}^{m-1}b_{k}(z) \frac{ \binom{m+1}{k}}{ (1+\frac{\phi _{k}}{N} )^{m+1}} \Biggr\} , \end{aligned}$$
(3.18)
$$\begin{aligned} (-1)^{m}(T_{1}+T_{2})>0, \end{aligned}$$
(3.19)
$$\begin{aligned} N\geq \frac{\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}b_{k}(z)}{ \frac{1}{m}\sum_{k=0}^{m-2}\binom{m}{k}b_{k}(z)}. \end{aligned}$$
(3.20)
$$\begin{aligned} N\geq -\frac{c_{m+1}(z)}{c_{m}(z)}. \end{aligned}$$
(3.21)
$$\begin{aligned} T_{1}+T_{2}< 0, \quad \epsilon < 0. \end{aligned}$$
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)< z^{N-1}\sum _{k=2}^{2q}\frac{c_{k}(z)}{N ^{k}}. \end{aligned}$$
$$\begin{aligned} T_{1}+T_{2}>0,\quad \epsilon > 0, \end{aligned}$$
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)> z^{N-1}\sum _{k=2}^{2p+1}\frac{c_{k}(z)}{N ^{k}}. \end{aligned}$$
The coefficients \(c_{k}(z)\) in Theorem 3.1 are computed explicitly. In the following theorem, we provide an alternative approach to compute the coefficients recursively.
Theorem 3.2
The coefficients
\(c_{k}(z)\)
in Theorem 3.1
can be determined by the following recurrence relation:
$$\begin{aligned} &c_{2}(z)=\frac{1}{2(1-z)}, \end{aligned}$$
(3.22)
$$\begin{aligned} &c_{k}(z)=\frac{z}{1-z}\sum_{j=2}^{k-1}(-1)^{k-j} \binom{k-1}{j-1}c _{j}(z)+\frac{(-1)^{k}}{k(1-z)},\quad k\geq 3. \end{aligned}$$
(3.23)
Proof
From (3.3) and (2.9) it is easy to verify that
For \(k\geq 3\), by (2.7), (2.8), and (3.3), we calculate the right-hand side of (3.23):
Thus, the proof is complete. □
$$\begin{aligned} c_{2}(z)=\frac{1}{2} \biggl[\frac{1-z}{z}b_{2}(z)-2b_{1}(z) \biggr]= \frac{1}{2(1-z)}. \end{aligned}$$
$$\begin{aligned} &\frac{z}{1-z}\sum_{j=1}^{k-1}(-1)^{k-j} \binom{k-1}{j-1}c_{j}(z)+ \frac{(-1)^{k}}{k(1-z)} \\ &\quad=(-1)^{k} \Biggl\{ \frac{z}{1-z}\sum _{j=1}^{k-1}\frac{1}{j} \binom{k-1}{j-1} \biggl[\frac{1-z}{z}b_{j}(z)-jb_{j-1}(z) \biggr]+ \frac{1}{k(1-z)} \Biggr\} \\ &\quad=(-1)^{k} \Biggl\{ \frac{1}{k}\sum _{j=0}^{k-1}\binom{k}{j}b_{j}(z)- \frac{1-z}{z}\sum_{j=1}^{k-1} \binom{k-1}{j-1}b_{j-1}(z) \Biggr\} \\ &\quad=\frac{(-1)^{k}}{k} \biggl\{ \frac{1-z}{z}b_{k}(z)-kb_{k-1}(z) \biggr\} \\ &\quad =c_{k}(z). \end{aligned}$$
Remark 3.1
By the above recurrence relation, we obtain the first few cases of \(c_{k}(z)\):
$$\begin{aligned} &c_{2}(z)=\frac{1}{2(1-z)},\qquad c_{3}(z)=- \frac{2z+1}{3(1-z)^{2}}, \\ &c_{4}(z)=\frac{3z^{2}+8z+1}{4(1-z)^{3}}, \qquad c_{5}(z)=- \frac{4z ^{3}+33z^{2}+22z+1}{5(1-z)^{4}}, \\ &c_{6}(z)=\frac{5z^{4}+104z^{3}+198z^{2}+52z+1}{6(1-z)^{5}}, \\ &c_{7}(z)=- \frac{6z^{5}+285z^{4}+1208z^{3}+906z^{2}+114z+1}{7(1-z)^{6}}. \end{aligned}$$
Observing the above, we find the following properties for \(c_{k}(z)\).
Theorem 3.3
For
\(0< z<1\), the coefficients
\(c_{k}(z)\)
in Theorem 3.1
satisfy
In particular, if
\(\frac{1}{7}< z<1\), then
$$\begin{aligned} (-1)^{k}c_{k}(z)>0, \quad k\geq 2. \end{aligned}$$
(3.24)
$$\begin{aligned} \bigl\vert c_{k}(z) \bigr\vert < \bigl\vert c_{k+1}(z) \bigr\vert ,\quad k\geq 2. \end{aligned}$$
(3.25)
Proof
By (2.8), we have
which implies that (3.24) is true because \(b_{j}(z)>0\) for all \(j\geq 0\) and \(0< z<1\). By (3.26), we further have
For \(1\leq j\leq k-2\),
Thus,
From (2.8) it follows that
By (3.29) and (3.30), we obtain
It is easy to verify that when \(1/7< z<1\) there holds
which implies that (3.25) is true. □
$$\begin{aligned} c_{k}(z)=\frac{(-1)^{k}}{k}\sum _{j=0}^{k-2}\binom{k}{j}b_{j}(z), \end{aligned}$$
(3.26)
$$\begin{aligned} \bigl\vert c_{k+1}(z) \bigr\vert - \bigl\vert c_{k}(z) \bigr\vert =\sum_{j=0}^{k-2} \biggl[\frac{1}{k+1} \binom{k+1}{j}-\frac{1}{k} \binom{k}{j} \biggr]b_{j}(z)+\frac{k}{2}b _{k-1}(z). \end{aligned}$$
(3.27)
$$\begin{aligned} \frac{1}{k+1}\binom{k+1}{j}-\frac{1}{k} \binom{k}{j}= \biggl[\frac{1}{k+1-j}- \frac{1}{k} \biggr] \binom{k}{j}\geq 0. \end{aligned}$$
(3.28)
$$\begin{aligned} \bigl\vert c_{k+1}(z) \bigr\vert - \bigl\vert c_{k}(z) \bigr\vert \geq \biggl[\frac{1}{k+1}- \frac{1}{k} \biggr]b _{0}(z)+\frac{k}{2}b_{k-1}(z). \end{aligned}$$
(3.29)
$$\begin{aligned} b_{k-1}(z)>\frac{z}{1-z}b_{0}(z). \end{aligned}$$
(3.30)
$$\begin{aligned} \bigl\vert c_{k+1}(z) \bigr\vert - \bigl\vert c_{k}(z) \bigr\vert > \biggl[\frac{1}{k+1}- \frac{1}{k}+ \frac{kz}{2(1-z)} \biggr]b_{0}(z). \end{aligned}$$
(3.31)
$$\begin{aligned} \frac{1}{k+1}-\frac{1}{k}+ \frac{kz}{2(1-z)}>0, \end{aligned}$$
(3.32)
The following theorem gives the conditions for N in a simpler form than those in Theorem 3.1.
Theorem 3.4
Let
p
and
q
be any positive integers. If
\(N\geq (2p+2) (\frac{(p+1)z}{1-z}+ \frac{1}{3} )\), then
If
\(N\geq (2q+1) ((q+\frac{1}{2})\frac{z}{1-z}+\frac{1}{3} )\), then
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)> z^{N-1}\sum _{k=2}^{2p+1}\frac{c_{k}(z)}{N ^{k}},\quad 0< z< 1. \end{aligned}$$
(3.33)
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)< z^{N-1}\sum _{k=2}^{2q}\frac{c_{k}(z)}{N ^{k}}, \quad 0< z< 1. \end{aligned}$$
(3.34)
Proof
By (2.8) we have
If \(N\geq m (\frac{1}{3}+\frac{m}{2}\frac{z}{1-z} )\), we have
for all \(k=0,1,\ldots,m-2\). This implies (3.19) is true. The proof left is similar to that of Theorem 3.1. □
$$\begin{aligned} &\sum_{k=0}^{m-2}\frac{b_{k}(z)}{m} \binom{m}{k}-\frac{1}{N}\sum_{k=0} ^{m-1}\frac{b_{k}(z)}{m+1}\binom{m+1}{k} \\ &\quad =\sum_{k=0}^{m-2}\frac{b_{k}(z)}{m} \binom{m}{k}-\frac{1}{N}\sum_{k=0} ^{m-2}\frac{b_{k}(z)}{m+1}\binom{m+1}{k}-\frac{m}{2N} \frac{z}{1-z} \sum_{k=0}^{m-2} \binom{m-1}{k}b_{k}(z) \\ &\quad =\frac{1}{N}\sum_{k=0}^{m-2}b_{k}(z) \binom{m}{k} \biggl[\frac{N}{m}- \frac{1}{m+1-k}- \frac{m-k}{2}\frac{z}{1-z} \biggr]. \end{aligned}$$
$$\begin{aligned} \frac{N}{m}-\frac{1}{m+1-k}-\frac{m-k}{2} \frac{z}{1-z}\geq 0 \end{aligned}$$
Remark 3.2
Remark 3.3
In a nutshell, according to Theorems 3.1 and 3.4, for sufficiently large N and any positive integers p and q, we have
In particular, when \(z=1/2\), there holds
which generalizes the result (1.11) due to Chen and Han [5] for large N. In view of (3.36), a variety of simple inequalities for \(\gamma (\frac{1}{2} )-\gamma _{N-1} (\frac{1}{2} )\) are derived for sufficiently large N by choosing different parameters p and q.
$$\begin{aligned} & z^{N-1}\sum_{k=2}^{2p+1} \frac{(-1)^{k}}{kN^{k}} \biggl[\frac{1-z}{z}b _{k}(z)-kb_{k-1}(z) \biggr]< \gamma (z)-\gamma _{N-1}(z) \\ &\quad < z^{N-1}\sum_{k=2}^{2q} \frac{(-1)^{k}}{kN^{k}} \biggl[\frac{1-z}{z}b _{k}(z)-kb_{k-1}(z) \biggr]. \end{aligned}$$
(3.35)
$$\begin{aligned} & \biggl(\frac{1}{2} \biggr)^{N-1}\sum _{k=2}^{2p+1}\frac{(-1)^{k}}{kN ^{k}} \biggl[b_{k} \biggl(\frac{1}{2} \biggr)-kb_{k-1} \biggl(\frac{1}{2} \biggr) \biggr] \\ &\quad < \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl(\frac{1}{2} \biggr) \\ &\quad < \biggl(\frac{1}{2} \biggr)^{N-1}\sum _{k=2}^{2q}\frac{(-1)^{k}}{kN ^{k}} \biggl[b_{k} \biggl(\frac{1}{2} \biggr)-kb_{k-1} \biggl(\frac{1}{2} \biggr) \biggr], \end{aligned}$$
(3.36)
$$\begin{aligned} &\frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}} \biggr)< \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl(\frac{1}{2} \biggr)< \frac{1}{2^{N-1}}\frac{1}{N^{2}},\quad (p=q=1), \\ &\frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}}-\frac{332}{5N^{5}} \biggr)\\ &\quad < \gamma \biggl( \frac{1}{2} \biggr)- \gamma _{N-1} \biggl( \frac{1}{2} \biggr) \\ &\quad < \frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}} \biggr),\quad (p=q=2), \\ &\frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}}-\frac{332}{5N^{5}}+\frac{479}{N^{6}}- \frac{29\text{,}024}{7N^{7}} \biggr)\\ &\quad < \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl(\frac{1}{2} \biggr) \\ &\quad < \frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}}-\frac{332}{5N^{5}}+\frac{479}{N^{6}} \biggr),\quad (p=q=3). \end{aligned}$$
Remark 3.4
The inequalities for \(\gamma (\frac{1}{t} )-\gamma _{N-1} (\frac{1}{t} )\) were investigated by Ma and Chen [14]. In particular, some inequalities for the cases \(t=3\) and \(t=4\) were presented, see [14, 15, 29]. As examples, here we apply Theorem 3.1 to obtain several simple and new inequalities related to \(\gamma (\frac{1}{3} )- \gamma _{N-1} (\frac{1}{3} )\) and \(\gamma (\frac{1}{4} )- \gamma _{N-1} (\frac{1}{4} )\) for sufficiently large N.
Case
\(t=3\):
$$\begin{aligned} &\frac{1}{3^{N-1}} \biggl(\frac{3}{4N^{2}}-\frac{5}{4N^{3}} \biggr)< \gamma \biggl(\frac{1}{3} \biggr)-\gamma _{N-1} \biggl(\frac{1}{3} \biggr)< \frac{1}{3^{N-1}}\frac{3}{4N^{2}}, \\ &\frac{1}{3^{N-1}} \biggl(\frac{3}{4N^{2}}-\frac{5}{4N^{3}}+ \frac{27}{8N ^{4}}-\frac{123}{10N^{5}} \biggr)\\ &\quad < \gamma \biggl( \frac{1}{3} \biggr)- \gamma _{N-1} \biggl( \frac{1}{3} \biggr) \\ &\quad < \frac{1}{3^{N-1}} \biggl(\frac{3}{4N^{2}}-\frac{5}{4N^{3}}+ \frac{27}{8N ^{4}} \biggr), \\ &\frac{1}{3^{N-1}} \biggl(\frac{3}{4N^{2}}-\frac{5}{4N^{3}}+ \frac{27}{8N ^{4}}-\frac{123}{10N^{5}}+\frac{56}{N^{6}}- \frac{17\text{,}127}{56N^{7}} \biggr)\\ &\quad < \gamma \biggl(\frac{1}{3} \biggr)-\gamma _{N-1} \biggl(\frac{1}{3} \biggr) \\ &\quad < \frac{1}{3^{N-1}} \biggl(\frac{3}{4N^{2}}-\frac{5}{4N^{3}}+ \frac{27}{8N ^{4}}-\frac{123}{10N^{5}}+\frac{56}{N^{6}} \biggr). \end{aligned}$$
Case
\(t=4\):
$$\begin{aligned} &\frac{1}{4^{N-1}} \biggl(\frac{2}{3N^{2}}-\frac{8}{9N^{3}} \biggr)< \gamma \biggl(\frac{1}{4} \biggr)-\gamma _{N-1} \biggl(\frac{1}{4} \biggr)< \frac{1}{4^{N-1}}\frac{2}{3N^{2}}, \\ &\frac{1}{4^{N-1}} \biggl(\frac{2}{3N^{2}}-\frac{8}{9N^{3}}+ \frac{17}{9N ^{4}}-\frac{736}{135N^{5}} \biggr)\\ &\quad< \gamma \biggl( \frac{1}{4} \biggr)- \gamma _{N-1} \biggl( \frac{1}{4} \biggr) \\ &\quad < \frac{1}{4^{N-1}} \biggl(\frac{2}{3N^{2}}-\frac{8}{9N^{3}}+ \frac{17}{9N ^{4}} \biggr), \\ &\frac{1}{4^{N-1}} \biggl(\frac{2}{3N^{2}}-\frac{8}{9N^{3}}+ \frac{17}{9N ^{4}}-\frac{736}{135N^{5}}+\frac{1594}{81N^{6}}- \frac{48\text{,}296}{567N^{7}} \biggr)\\ &\quad< \gamma \biggl(\frac{1}{4} \biggr)- \gamma _{N-1} \biggl(\frac{1}{4} \biggr) \\ &\quad < \frac{1}{4^{N-1}} \biggl(\frac{2}{3N^{2}}-\frac{8}{9N^{3}}+ \frac{17}{9N ^{4}}-\frac{736}{135N^{5}}+\frac{1594}{81N^{6}} \biggr). \end{aligned}$$
Theorem 3.5
Let
\(m\geq 3\)
be a positive integer. For
\(0< z<1\), we have the following asymptotic expansion:
where
\(c_{k}(z)\)
are described as in (3.3).
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)=z^{N-1}\sum _{k=2}^{m-1}\frac{c_{k}(z)}{N ^{k}}+O \biggl(\frac{z^{N-1}}{N^{m}} \biggr),\quad N\rightarrow \infty , \end{aligned}$$
(3.37)
Remark 3.5
When \(z=1/2\), we recover the result of Chen and Han [5] about asymptotic expansion of \(\gamma (\frac{1}{2} )- \gamma _{N-1} (\frac{1}{2} )\):
as \(N\rightarrow \infty \).
$$\begin{aligned} \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl( \frac{1}{2} \biggr)= \frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}- \frac{8}{3N^{3}}+\frac{23}{2N ^{4}}-\frac{332}{5N^{5}}+ \frac{479}{N^{6}}-\frac{29\text{,}024}{7N^{7}}+ \cdots \biggr), \end{aligned}$$
4 Estimate of the generalized Somos’ quadratic recurrence constant
Sondow and Hadjicostas [25] generalized Somos’ quadratic recurrence constant (1.3) as
Then they established the relation between the generalized Somos’ quadratic recurrence constant \(\sigma _{t}\) and the function \(\gamma (\frac{1}{t} )\):
Recently, Coffey [6] obtained the following integral and series representations for \(\ln \sigma _{t}\):
and
in terms of the polylogarithm function. Using (4.2) and Theorem 3.1, it is natural that we give an estimate for the generalized Somos’ quadratic recurrence constant \(\sigma _{t}\):
where the bounds
Equivalently, (4.5) leads to a general estimate for \(\ln \sigma _{t}\).
$$\begin{aligned} \sigma _{t}=\sqrt[t]{1\sqrt[t]{2\sqrt[t]{3 \sqrt[t]{4\cdots }}}}= \prod_{k=1}^{\infty }k^{\frac{1}{t^{k}}}. \end{aligned}$$
(4.1)
$$\begin{aligned} \gamma \biggl(\frac{1}{t} \biggr)=t\ln \frac{t}{(t-1)\sigma _{t}^{t-1}},\quad t>1. \end{aligned}$$
(4.2)
$$\begin{aligned} \ln \sigma _{t}= \int _{0}^{\infty } \biggl(\frac{e^{-x}}{t-1}+ \frac{1}{1-te ^{x}} \biggr)\frac{dx}{x} \end{aligned}$$
(4.3)
$$\begin{aligned} \ln \sigma _{t}=\frac{1}{t-1}\sum _{k=1}^{\infty }\frac{(-1)^{k-1}}{k}Li _{k} \biggl(\frac{1}{t} \biggr)=\frac{1}{t-1}\sum _{k=1}^{\infty } \frac{1}{k} \biggl[Li_{k} \biggl(\frac{1}{t} \biggr)-1 \biggr] \end{aligned}$$
(4.4)
$$\begin{aligned} L_{t}< \sigma _{t} < U_{t}, \end{aligned}$$
(4.5)
$$\begin{aligned} &L_{t}= \Biggl\{ \frac{t}{t-1}\exp \Biggl\{ - \frac{\gamma _{N-1} (\frac{1}{t} )}{t}-\frac{1}{t^{N}}\sum_{k=2}^{2q} \frac{c_{k} (\frac{1}{t} )}{N^{k}} \Biggr\} \Biggr\} ^{\frac{1}{t-1}}, \end{aligned}$$
(4.6)
$$\begin{aligned} &U_{t}= \Biggl\{ \frac{t}{t-1}\exp \Biggl\{ - \frac{\gamma _{N-1} (\frac{1}{t} )}{t}-\frac{1}{t^{N}}\sum_{k=2}^{2p+1} \frac{c_{k} (\frac{1}{t} )}{N^{k}} \Biggr\} \Biggr\} ^{\frac{1}{t-1}}. \end{aligned}$$
(4.7)
Theorem 4.1
Let
p
and
q
be any positive integers. For sufficiently large
N, we have
where
\(c_{k}(z)\)
are described as in (3.3).
$$\begin{aligned} &\frac{1}{t-1} \Biggl\{ - \frac{\gamma _{N-1} (\frac{1}{t} )}{t}-\ln \biggl(1- \frac{1}{t} \biggr)-\frac{1}{t^{N}}\sum _{k=2}^{2q}\frac{c_{k} (\frac{1}{t} )}{N ^{k}} \Biggr\} \\ &\quad< \ln \sigma _{t} < \frac{1}{t-1} \Biggl\{ - \frac{\gamma _{N-1} (\frac{1}{t} )}{t}-\ln \biggl(1- \frac{1}{t} \biggr)-\frac{1}{t^{N}}\sum _{k=2}^{2p+1}\frac{c_{k} (\frac{1}{t} )}{N ^{k}} \Biggr\} , \end{aligned}$$
(4.8)
As a consequence, we easily obtain an asymptotic expansion for \(\ln \sigma _{t}\).
Theorem 4.2
As
\(N\rightarrow \infty \), the following asymptotic expansion holds:
where
\(m\geq 3\).
$$\begin{aligned} \ln \sigma _{t}=\frac{1}{t-1} \Biggl\{ - \frac{\gamma _{N-1} (\frac{1}{t} )}{t}-\ln \biggl(1- \frac{1}{t} \biggr)- \frac{1}{t^{N}}\sum_{k=2}^{m-1} \frac{c_{k} (\frac{1}{t} )}{N ^{k}} \Biggr\} +O \biggl(\frac{1}{N^{m}t^{N}} \biggr), \end{aligned}$$
(4.9)
Remark 4.1
Theorems 4.1 and 4.2 show that the function
is a good approximation of \(\ln \sigma _{t}\) with the error term \(O (1/(N^{m}t^{N}) )\). In particular, for large t, the approximation effect is great.
$$\begin{aligned} \frac{1}{t-1} \Biggl\{ - \frac{\gamma _{N-1} (\frac{1}{t} )}{t}-\ln \biggl(1- \frac{1}{t} \biggr)-\frac{1}{t^{N}}\sum _{k=2}^{m-1}\frac{c_{k} (\frac{1}{t} )}{N ^{k}} \Biggr\} \end{aligned}$$
When \(t=2\), we have the following estimate and asymptotic expansion for lnσ.
Corollary 4.1
Let
p
and
q
be any positive integers. For sufficiently large
N, we have
$$\begin{aligned} -\frac{\gamma _{N-1} (\frac{1}{2} )}{2}+\ln 2- \frac{1}{2^{N}}\sum _{k=2}^{2q}\frac{c_{k} (\frac{1}{2} )}{N ^{k}}< \ln \sigma < - \frac{\gamma _{N-1} (\frac{1}{2} )}{2}+ \ln 2-\frac{1}{2^{N}}\sum _{k=2}^{2p+1}\frac{c_{k} (\frac{1}{2} )}{N ^{k}}. \end{aligned}$$
(4.10)
Corollary 4.2
As
\(N\rightarrow \infty \), the following asymptotic expansion holds:
where
\(m\geq 3\).
$$\begin{aligned} \ln \sigma =-\frac{\gamma _{N-1} (\frac{1}{2} )}{2}+\ln 2- \frac{1}{2^{N}} \sum_{k=2}^{m-1}\frac{c_{k} (\frac{1}{2} )}{N ^{k}}+O \biggl(\frac{1}{N^{m}2^{N}} \biggr), \end{aligned}$$
(4.11)
In fact, according to Definition (4.1) and Lemma 2.1 or (2.12), we find an alternative approach to get the estimate for \(\ln \sigma _{t}\).
Theorem 4.3
Let
N
be any positive integer. Then, for
\(t>1\)
and
\(m\geq 1\), we have
where
\(0<\kappa <1\)
and
\(\lambda _{n}(z)=\sum_{k=1}^{n}(\ln k)z^{k}\).
$$\begin{aligned} \ln \sigma _{t}=\lambda _{N-1} \biggl( \frac{1}{t} \biggr)-\frac{\ln N}{t ^{N-1}(t-1)}-\frac{1}{t^{N}} \Biggl\{ \sum_{k=1}^{m-1}\frac{(-1)^{k}b _{k} (\frac{1}{t} )}{kN^{k}}+ \kappa \frac{(-1)^{m}b_{m} (\frac{1}{t} )}{mN^{m}} \Biggr\} , \end{aligned}$$
(4.12)
It should be noted that the number N in Theorem 4.3 is not restricted except N is a positive integer. By Theorem 4.3, it is clear that the following corollary is true.
Corollary 4.3
For
\(t>1\)
and
\(m\geq 1\), we have
as
\(N\rightarrow \infty \).
$$\begin{aligned} \ln \sigma _{t}=\lambda _{N-1} \biggl( \frac{1}{t} \biggr)-\frac{\ln N}{t ^{N-1}(t-1)}-\frac{1}{t^{N}}\sum _{k=1}^{m-1}\frac{(-1)^{k}b_{k}( \frac{1}{t})}{kN^{k}}+O \biggl(\frac{1}{N^{m}t^{N}} \biggr), \end{aligned}$$
(4.13)
Also, we can obtain a new type of approximation for \(\ln \sigma _{t}\) similar to Theorem 3.1.
Corollary 4.4
For any positive integers
p, q, and
N, there holds
$$\begin{aligned} &\lambda _{N-1} \biggl(\frac{1}{t} \biggr)- \frac{\ln N}{t^{N-1}(t-1)}-\frac{1}{t ^{N}}\sum_{k=1}^{2p} \frac{(-1)^{k}b_{k}(\frac{1}{t})}{kN^{k}} \\ &\quad< \ln \sigma _{t} < \lambda _{N-1} \biggl(\frac{1}{t} \biggr)- \frac{\ln N}{t^{N-1}(t-1)}-\frac{1}{t ^{N}}\sum_{k=1}^{2q-1} \frac{(-1)^{k}b_{k}(\frac{1}{t})}{kN^{k}}. \end{aligned}$$
(4.14)
In particular, when \(t=2\) we have
and
as \(N\rightarrow \infty \).
$$\begin{aligned} & \lambda _{N-1} \biggl(\frac{1}{2} \biggr)- \frac{\ln N}{2^{N-1}}- \frac{1}{2^{N}}\sum_{k=1}^{2p} \frac{(-1)^{k}b_{k}(\frac{1}{2})}{kN ^{k}} \\ &\quad< \ln \sigma < \lambda _{N-1} \biggl(\frac{1}{2} \biggr)- \frac{\ln N}{2^{N-1}}- \frac{1}{2^{N}}\sum_{k=1}^{2q-1} \frac{(-1)^{k}b_{k}(\frac{1}{2})}{kN ^{k}}, \end{aligned}$$
(4.15)
$$\begin{aligned} \ln \sigma =\lambda _{N-1} \biggl( \frac{1}{2} \biggr)- \frac{\ln N}{2^{N-1}}-\frac{1}{2^{N}}\sum _{k=1}^{m-1}\frac{(-1)^{k}b _{k}(\frac{1}{2})}{kN^{k}}+O \biggl(\frac{1}{N^{m}2^{N}} \biggr), \end{aligned}$$
(4.16)
Finally, we get a new estimate for \(\gamma (z)\) when \(0< z<1\).
Corollary 4.5
For any positive integers
p, q, and
N, there holds
$$\begin{aligned} (1-z)z^{N-2}\sum_{k=1}^{2q-1} \frac{(-1)^{k}b_{k}(z)}{kN^{k}} &< \gamma (z)- \biggl(z^{N-2}\ln N- \frac{1-z}{z^{2}}\lambda _{N-1}(z)-\frac{ \ln (1-z)}{z} \biggr) \\ &< (1-z)z^{N-2}\sum_{k=1}^{2p} \frac{(-1)^{k}b_{k}(z)}{kN^{k}}. \end{aligned}$$
(4.17)
Proof
In a similar manner, we obtain the following asymptotic expansion for \(\gamma (z)\).
Corollary 4.6
Let
\(m\geq 1\)
be a positive integer. For
\(0< z<1\), we have
as
\(N\rightarrow \infty \).
$$\begin{aligned} \gamma (z)={}&z^{N-2}\ln N-\frac{1-z}{z^{2}} \lambda _{N-1}(z)-\frac{ \ln (1-z)}{z} \\ & {}+(1-z)z^{N-2}\sum_{k=1}^{m-1} \frac{(-1)^{k}b_{k}(z)}{kN^{k}}+O \biggl(\frac{z ^{N-2}}{N^{m}} \biggr), \end{aligned}$$
(4.18)
Remark 4.2
Remark 4.3
If we approximate \(\gamma (z)\) by the function \(z^{N-2}\ln N-\frac{1-z}{z ^{2}}\lambda _{N-1}(z)-\frac{\ln (1-z)}{z}\) as \(N\rightarrow \infty \), then we arrive at the error bound \(O(z^{N-2}/N)\). By Theorem 3.5, we conclude that it is better that we approximate \(\gamma (z)\) by its partial sum \(\gamma _{N-1}(z)\) because the error bound is \(O(z^{N-1}/N ^{2})\).
Acknowledgements
The author sincerely appreciates the anonymous referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.
Competing interests
The author declares that they have no competing interests.
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