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
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)> z^{N-1}\sum _{k=2}^{2p+1}\frac{c_{k}(z)}{N ^{k}}, \quad 0< z< 1. \end{aligned}$$
(3.1)
If
\(N\geq -\frac{c_{2q+2}(z)}{c_{2q+1}(z)}\), then we have
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)< z^{N-1}\sum _{k=2}^{2q}\frac{c_{k}(z)}{N ^{k}},\quad 0< z< 1. \end{aligned}$$
(3.2)
In the above inequalities, the coefficients
\(c_{k}(z)\)
are explicitly expressed by
$$\begin{aligned} c_{k}(z)=\frac{(-1)^{k}}{k} \biggl[ \frac{1-z}{z}b_{k}(z)-kb_{k-1}(z) \biggr], \quad k\geq 2, \end{aligned}$$
(3.3)
where
\(b_{k}(z)\)
are the Eulerian fractions introduced in the second section.
Proof
Write
$$\begin{aligned} f(t)=\frac{1}{t}-\ln \frac{t+1}{t}. \end{aligned}$$
(3.4)
After simple calculations, we arrive at
$$\begin{aligned} f^{(i)}(t) &=(-1)^{i}(i-1)! \biggl[ \frac{i}{t^{i+1}}+ \frac{1}{(t+1)^{i}}-\frac{1}{t^{i}} \biggr] \end{aligned}$$
(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)
which implies that
$$\begin{aligned} (-1)^{i}f^{(i)}(t)>0,\quad t>0. \end{aligned}$$
(3.7)
By (3.4) and (3.6), for each \(i\geq 0\), we also have
$$\begin{aligned} \lim_{t\rightarrow +\infty } z^{t}f^{(i)}(t)=0,\quad 0< z< 1. \end{aligned}$$
(3.8)
Therefore, according to (2.12), for \(0< z<1\), there exists \(\theta \in (0,1)\) such that
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z) &=\frac{1}{z} \sum_{k=N}^{\infty } \biggl( \frac{1}{k}- \ln \frac{k+1}{k} \biggr)z^{k} \\ &=z^{N-1} \Biggl[\sum_{k=0}^{m-1} \frac{b_{k}(z)}{k!}f^{(k)}(N)+\theta \frac{b_{m}(z)}{m!}f^{(m)}(N) \Biggr], \end{aligned}$$
(3.9)
where m is a positive integer. It follows from (3.4) and (3.5) that
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)={}&z^{N-1} \Biggl\{ b_{0}(z) \biggl(\frac{1}{N}- \ln \frac{N+1}{N} \biggr) \\ &{}+\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{k} \biggl(\frac{k}{N^{k+1}}+ \frac{1}{(N+1)^{k}}-\frac{1}{N^{k}} \biggr) \\ &{}+\theta \frac{(-1)^{m}b_{m}(z)}{m} \biggl(\frac{m}{N^{m+1}}+ \frac{1}{(N+1)^{m}}-\frac{1}{N^{m}} \biggr) \Biggr\} . \end{aligned}$$
(3.10)
By the Taylor expansion we have
$$\begin{aligned} &\ln \frac{N+1}{N}=\sum_{j=1}^{m} \frac{(-1)^{j-1}}{jN^{j}}+\frac{(-1)^{m}}{(m+1)N ^{m+1}}\frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}}, \end{aligned}$$
(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)
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
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)={}&z^{N-1} \Biggl\{ \sum_{k=2}^{m} \frac{(-1)^{k}b _{0}(z)}{kN^{k}} +\frac{(-1)^{m+1}b_{0}(z)}{(m+1)N^{m+1}}\frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}} \\ &{}+\sum_{k=2}^{m}\frac{(-1)^{k-1}b_{k-1}(z)}{N^{k}}+ \sum_{k=1}^{m-1} \sum _{j=0}^{m-k}\frac{(-1)^{k+j}b_{k}(z)}{kN^{k+j}} \binom{k+j-1}{j} \\ &{}+\frac{(-1)^{m+1}}{N^{m+1}}\sum_{k=1}^{m-1} \frac{b_{k}(z)}{k}\frac{ \binom{m}{k-1}}{ (1+\frac{\phi _{k}}{N} )^{m+1}}-\sum_{k=1} ^{m-1}\frac{(-1)^{k}b_{k}(z)}{k}\frac{1}{N^{k}} +\epsilon \Biggr\} , \end{aligned}$$
(3.13)
where
$$\begin{aligned} \epsilon =\theta \frac{(-1)^{m}b_{m}(z)}{m} \biggl( \frac{m}{N^{m+1}}+ \frac{1}{(N+1)^{m}}-\frac{1}{N^{m}} \biggr). \end{aligned}$$
(3.14)
Taking \(u=k+j\), we obtain
$$\begin{aligned} &\sum_{k=1}^{m-1}\sum _{j=0}^{m-k}\frac{(-1)^{k+j}b_{k}(z)}{kN^{k+j}} \binom{k+j-1}{j} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{k=1}^{m-1} \sum _{j=1}^{m-k}\frac{(-1)^{k+j}b_{k}(z)}{kN^{k+j}} \binom{k+j-1}{j} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m}\frac{(-1)^{u}}{N ^{u}}\sum _{k=1}^{u-1}\frac{b_{k}(z)}{k} \binom{u-1}{k-1} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m-1}\frac{(-1)^{u}}{N ^{u}}\sum _{k=1}^{u-1}\frac{b_{k}(z)}{k} \binom{u-1}{k-1}+\frac{(-1)^{m}}{N ^{m}}\sum_{k=1}^{m-1} \binom{m-1}{k-1}\frac{b_{k}(z)}{k}. \end{aligned}$$
Since
$$\begin{aligned} \frac{1}{k}\binom{u-1}{k-1}=\frac{1}{u} \binom{u}{k}, \end{aligned}$$
we have
$$\begin{aligned} &\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m-1}\frac{(-1)^{u}}{N ^{u}}\sum _{k=1}^{u-1}\frac{b_{k}(z)}{k} \binom{u-1}{k-1} \\ &\quad =\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{kN^{k}}+ \sum_{u=2}^{m-1}\frac{(-1)^{u}}{uN ^{u}}\sum _{k=1}^{u-1}\binom{u}{k}b_{k}(z) \\ &\quad =-\frac{b_{1}(z)}{N}+\sum_{k=2}^{m-1} \frac{(-1)^{k}}{kN^{k}} \Biggl(b _{k}(z)+\sum _{j=1}^{k-1}\binom{k}{j}b_{j}(z) \Biggr) \\ &\quad=\sum_{k=1}^{m-1}\frac{(-1)^{k}}{kN^{k}} \sum_{j=1}^{k}\binom{k}{j}b _{j}(z). \end{aligned}$$
Therefore,
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)={}&z^{N-1} \Biggl\{ \sum _{k=2}^{m}\frac{(-1)^{k}b _{0}(z)}{kN^{k}} + \frac{(-1)^{m+1}b_{0}(z)}{(m+1)N^{m+1}}\frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}} \\ &{}+\sum_{k=2}^{m}\frac{(-1)^{k-1}b_{k-1}(z)}{N^{k}}+ \sum_{k=1}^{m-1}\frac{(-1)^{k}}{kN ^{k}}\sum _{j=1}^{k}\binom{k}{j}b_{j}(z) \\ &{}+\frac{(-1)^{m}}{N^{m}}\sum_{k=1}^{m-1} \binom{m-1}{k-1} \frac{b_{k}(z)}{k}+\frac{(-1)^{m+1}}{N^{m+1}}\sum _{k=1}^{m-1}\frac{b _{k}(z)}{k} \frac{\binom{m}{k-1}}{ (1+\frac{\phi _{k}}{N} ) ^{m+1}} \\ &{}-\sum_{k=1}^{m-1}\frac{(-1)^{k}b_{k}(z)}{k} \frac{1}{N^{k}} +\epsilon \Biggr\} . \end{aligned}$$
Because of (2.8) and (3.3) it follows that
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z) &=z^{N-1} \Biggl\{ \sum_{k=2}^{m-1} \frac{c _{k}(z)}{N^{k}} +T_{1}+T_{2}+\epsilon \Biggr\} , \end{aligned}$$
(3.15)
where
$$\begin{aligned} &T_{1}=\frac{(-1)^{m}b_{0}(z)}{mN^{m}}+\frac{(-1)^{m+1}b_{0}(z)}{(m+1)N ^{m+1}} \frac{1}{ (1+\frac{\phi _{0}}{N} )^{m+1}}, \end{aligned}$$
(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)
Observe that
$$\begin{aligned} T_{1}+T_{2}=\frac{(-1)^{m}}{N^{m}} \Biggl\{ \frac{1}{m}\sum_{k=0}^{m-2} \binom{m}{k}b_{k}(z)-\frac{1}{(m+1)N}\sum _{k=0}^{m-1}b_{k}(z) \frac{ \binom{m+1}{k}}{ (1+\frac{\phi _{k}}{N} )^{m+1}} \Biggr\} , \end{aligned}$$
(3.18)
which implies that
$$\begin{aligned} (-1)^{m}(T_{1}+T_{2})>0, \end{aligned}$$
(3.19)
if
$$\begin{aligned} N\geq \frac{\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}b_{k}(z)}{ \frac{1}{m}\sum_{k=0}^{m-2}\binom{m}{k}b_{k}(z)}. \end{aligned}$$
(3.20)
According to (2.8) and (3.3), (3.20) is equivalent to
$$\begin{aligned} N\geq -\frac{c_{m+1}(z)}{c_{m}(z)}. \end{aligned}$$
(3.21)
Thus, if \(m=2q+1\) and \(N\geq -\frac{c_{m+1}(z)}{c_{m}(z)}\), then
$$\begin{aligned} T_{1}+T_{2}< 0, \quad \epsilon < 0. \end{aligned}$$
As a consequence, by (3.15) we have
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)< z^{N-1}\sum _{k=2}^{2q}\frac{c_{k}(z)}{N ^{k}}. \end{aligned}$$
Similarly, if \(m=2p+2\) and \(N\geq -\frac{c_{m+1}(z)}{c_{m}(z)}\), then
$$\begin{aligned} T_{1}+T_{2}>0,\quad \epsilon > 0, \end{aligned}$$
which implies
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)> z^{N-1}\sum _{k=2}^{2p+1}\frac{c_{k}(z)}{N ^{k}}. \end{aligned}$$
This completes the proof. □
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
$$\begin{aligned} c_{2}(z)=\frac{1}{2} \biggl[\frac{1-z}{z}b_{2}(z)-2b_{1}(z) \biggr]= \frac{1}{2(1-z)}. \end{aligned}$$
For \(k\geq 3\), by (2.7), (2.8), and (3.3), we calculate the right-hand side of (3.23):
$$\begin{aligned} &\frac{z}{1-z}\sum_{j=1}^{k-1}(-1)^{k-j} \binom{k-1}{j-1}c_{j}(z)+ \frac{(-1)^{k}}{k(1-z)} \\ &\quad=(-1)^{k} \Biggl\{ \frac{z}{1-z}\sum _{j=1}^{k-1}\frac{1}{j} \binom{k-1}{j-1} \biggl[\frac{1-z}{z}b_{j}(z)-jb_{j-1}(z) \biggr]+ \frac{1}{k(1-z)} \Biggr\} \\ &\quad=(-1)^{k} \Biggl\{ \frac{1}{k}\sum _{j=0}^{k-1}\binom{k}{j}b_{j}(z)- \frac{1-z}{z}\sum_{j=1}^{k-1} \binom{k-1}{j-1}b_{j-1}(z) \Biggr\} \\ &\quad=\frac{(-1)^{k}}{k} \biggl\{ \frac{1-z}{z}b_{k}(z)-kb_{k-1}(z) \biggr\} \\ &\quad =c_{k}(z). \end{aligned}$$
Thus, the proof is complete. □
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
$$\begin{aligned} (-1)^{k}c_{k}(z)>0, \quad k\geq 2. \end{aligned}$$
(3.24)
In particular, if
\(\frac{1}{7}< z<1\), then
$$\begin{aligned} \bigl\vert c_{k}(z) \bigr\vert < \bigl\vert c_{k+1}(z) \bigr\vert ,\quad k\geq 2. \end{aligned}$$
(3.25)
Proof
By (2.8), we have
$$\begin{aligned} c_{k}(z)=\frac{(-1)^{k}}{k}\sum _{j=0}^{k-2}\binom{k}{j}b_{j}(z), \end{aligned}$$
(3.26)
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
$$\begin{aligned} \bigl\vert c_{k+1}(z) \bigr\vert - \bigl\vert c_{k}(z) \bigr\vert =\sum_{j=0}^{k-2} \biggl[\frac{1}{k+1} \binom{k+1}{j}-\frac{1}{k} \binom{k}{j} \biggr]b_{j}(z)+\frac{k}{2}b _{k-1}(z). \end{aligned}$$
(3.27)
For \(1\leq j\leq k-2\),
$$\begin{aligned} \frac{1}{k+1}\binom{k+1}{j}-\frac{1}{k} \binom{k}{j}= \biggl[\frac{1}{k+1-j}- \frac{1}{k} \biggr] \binom{k}{j}\geq 0. \end{aligned}$$
(3.28)
Thus,
$$\begin{aligned} \bigl\vert c_{k+1}(z) \bigr\vert - \bigl\vert c_{k}(z) \bigr\vert \geq \biggl[\frac{1}{k+1}- \frac{1}{k} \biggr]b _{0}(z)+\frac{k}{2}b_{k-1}(z). \end{aligned}$$
(3.29)
From (2.8) it follows that
$$\begin{aligned} b_{k-1}(z)>\frac{z}{1-z}b_{0}(z). \end{aligned}$$
(3.30)
By (3.29) and (3.30), we obtain
$$\begin{aligned} \bigl\vert c_{k+1}(z) \bigr\vert - \bigl\vert c_{k}(z) \bigr\vert > \biggl[\frac{1}{k+1}- \frac{1}{k}+ \frac{kz}{2(1-z)} \biggr]b_{0}(z). \end{aligned}$$
(3.31)
It is easy to verify that when \(1/7< z<1\) there holds
$$\begin{aligned} \frac{1}{k+1}-\frac{1}{k}+ \frac{kz}{2(1-z)}>0, \end{aligned}$$
(3.32)
which implies that (3.25) is true. □
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
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)> z^{N-1}\sum _{k=2}^{2p+1}\frac{c_{k}(z)}{N ^{k}},\quad 0< z< 1. \end{aligned}$$
(3.33)
If
\(N\geq (2q+1) ((q+\frac{1}{2})\frac{z}{1-z}+\frac{1}{3} )\), then
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)< z^{N-1}\sum _{k=2}^{2q}\frac{c_{k}(z)}{N ^{k}}, \quad 0< z< 1. \end{aligned}$$
(3.34)
Proof
By (2.8) we have
$$\begin{aligned} &\sum_{k=0}^{m-2}\frac{b_{k}(z)}{m} \binom{m}{k}-\frac{1}{N}\sum_{k=0} ^{m-1}\frac{b_{k}(z)}{m+1}\binom{m+1}{k} \\ &\quad =\sum_{k=0}^{m-2}\frac{b_{k}(z)}{m} \binom{m}{k}-\frac{1}{N}\sum_{k=0} ^{m-2}\frac{b_{k}(z)}{m+1}\binom{m+1}{k}-\frac{m}{2N} \frac{z}{1-z} \sum_{k=0}^{m-2} \binom{m-1}{k}b_{k}(z) \\ &\quad =\frac{1}{N}\sum_{k=0}^{m-2}b_{k}(z) \binom{m}{k} \biggl[\frac{N}{m}- \frac{1}{m+1-k}- \frac{m-k}{2}\frac{z}{1-z} \biggr]. \end{aligned}$$
If \(N\geq m (\frac{1}{3}+\frac{m}{2}\frac{z}{1-z} )\), we have
$$\begin{aligned} \frac{N}{m}-\frac{1}{m+1-k}-\frac{m-k}{2} \frac{z}{1-z}\geq 0 \end{aligned}$$
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. □
Remark 3.2
In Theorems 3.1 and 3.4, the conditions for N are sufficient but not necessary. It is possible to relax the conditions for N. Even we conjecture that, for some fixed \(z\in (0,1)\), Theorem 3.1 is true for any positive integer N.
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
$$\begin{aligned} & z^{N-1}\sum_{k=2}^{2p+1} \frac{(-1)^{k}}{kN^{k}} \biggl[\frac{1-z}{z}b _{k}(z)-kb_{k-1}(z) \biggr]< \gamma (z)-\gamma _{N-1}(z) \\ &\quad < z^{N-1}\sum_{k=2}^{2q} \frac{(-1)^{k}}{kN^{k}} \biggl[\frac{1-z}{z}b _{k}(z)-kb_{k-1}(z) \biggr]. \end{aligned}$$
(3.35)
In particular, when \(z=1/2\), there holds
$$\begin{aligned} & \biggl(\frac{1}{2} \biggr)^{N-1}\sum _{k=2}^{2p+1}\frac{(-1)^{k}}{kN ^{k}} \biggl[b_{k} \biggl(\frac{1}{2} \biggr)-kb_{k-1} \biggl(\frac{1}{2} \biggr) \biggr] \\ &\quad < \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl(\frac{1}{2} \biggr) \\ &\quad < \biggl(\frac{1}{2} \biggr)^{N-1}\sum _{k=2}^{2q}\frac{(-1)^{k}}{kN ^{k}} \biggl[b_{k} \biggl(\frac{1}{2} \biggr)-kb_{k-1} \biggl(\frac{1}{2} \biggr) \biggr], \end{aligned}$$
(3.36)
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.
$$\begin{aligned} &\frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}} \biggr)< \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl(\frac{1}{2} \biggr)< \frac{1}{2^{N-1}}\frac{1}{N^{2}},\quad (p=q=1), \\ &\frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}}-\frac{332}{5N^{5}} \biggr)\\ &\quad < \gamma \biggl( \frac{1}{2} \biggr)- \gamma _{N-1} \biggl( \frac{1}{2} \biggr) \\ &\quad < \frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}} \biggr),\quad (p=q=2), \\ &\frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}}-\frac{332}{5N^{5}}+\frac{479}{N^{6}}- \frac{29\text{,}024}{7N^{7}} \biggr)\\ &\quad < \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl(\frac{1}{2} \biggr) \\ &\quad < \frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}-\frac{8}{3N^{3}}+ \frac{23}{2N ^{4}}-\frac{332}{5N^{5}}+\frac{479}{N^{6}} \biggr),\quad (p=q=3). \end{aligned}$$
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:
$$\begin{aligned} \gamma (z)-\gamma _{N-1}(z)=z^{N-1}\sum _{k=2}^{m-1}\frac{c_{k}(z)}{N ^{k}}+O \biggl(\frac{z^{N-1}}{N^{m}} \biggr),\quad N\rightarrow \infty , \end{aligned}$$
(3.37)
where
\(c_{k}(z)\)
are described as in (3.3).
Proof
By (3.15) we immediately obtain (3.37). □
Remark 3.5
When \(z=1/2\), we recover the result of Chen and Han [5] about asymptotic expansion of \(\gamma (\frac{1}{2} )- \gamma _{N-1} (\frac{1}{2} )\):
$$\begin{aligned} \gamma \biggl(\frac{1}{2} \biggr)-\gamma _{N-1} \biggl( \frac{1}{2} \biggr)= \frac{1}{2^{N-1}} \biggl(\frac{1}{N^{2}}- \frac{8}{3N^{3}}+\frac{23}{2N ^{4}}-\frac{332}{5N^{5}}+ \frac{479}{N^{6}}-\frac{29\text{,}024}{7N^{7}}+ \cdots \biggr), \end{aligned}$$
as \(N\rightarrow \infty \).