1 Introduction
For \(x>0\), the classical Euler gamma function Γ and psi (digamma) function ψ are defined by
respectively. Furthermore, the derivatives \(\psi^{\prime}, \psi ^{\prime\prime}, \psi^{\prime\prime\prime}, \ldots\) are known as polygamma functions.
$$ \Gamma ( x ) = \int_{0}^{\infty}t^{x-1}e^{-t}\,dt,\qquad \psi ( x ) =\frac{\Gamma^{\prime} ( x ) }{\Gamma (x ) }, $$
(1.1)
As an important role played in many branches, such as mathematical physics, probability, statistics, and engineering, the gamma and polygamma functions have attracted the attention of many scholars. Recently, many authors showed numerous interesting inequalities for the digamma (psi) function ψ and the Euler constant defined by
where \(\sum_{k=1}^{n}\frac{1}{k}:=H_{n}\) is called the harmonic number. In particular, there has many approximation formulas for psi function and harmonic number, which can be found in [1‐18], and closely related references therein.
$$ \gamma=\lim_{n\rightarrow\infty} \Biggl( \sum_{k=1}^{n} \frac{1}{k}-\ln n \Biggr) =0.577215664\cdots, $$
Anzeige
We would like to mention DeTemple and Wang’s paper [19] for half-integer approximation formulas
and
It was also proved in [13, 20], [21], Lemma 1.7, that
for \(x>0\), where \(\frac{1}{2}\) and \(e^{-\gamma}\) are the best possible constants, and \(\gamma=0.577215664\cdots\) is the Euler-Mascheroni constant. Thanks to formula (1.4) and the relation \(H_{n}=\gamma+\psi ( n+1 ) \), we have
for any \(n\in\mathbb{N}\). In 2011, Batir [22] further proved that
for all \(x>0\), where \(e^{-2\gamma}\) and \(1/3\) are the best possible. As a direct consequence, he showed that, for \(n\in\mathbb{N}\),
$$ \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac{1}{24 ( n+1 ) ^{2}}< H_{n}< \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac{1}{24n^{2}} $$
(1.2)
$$ \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac{1}{24n^{2}}+ \frac{7}{960} \frac{1}{ ( n+1 ) ^{4}}< H_{n}< \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac{1}{24n^{2}}+\frac{7}{960} \frac{1}{n^{4}}. $$
(1.3)
$$ \ln \biggl( x+\frac{1}{2} \biggr) < \psi ( x+1 ) \leq\ln \bigl( x+e^{-\gamma} \bigr) $$
(1.4)
$$ \gamma+\ln \biggl( n+\frac{1}{2} \biggr) < H_{n}\leq\gamma+\ln \bigl( n+e^{1-\gamma}-1 \bigr) $$
(1.5)
$$ \frac{1}{2}\ln \bigl( x^{2}+x+e^{-2\gamma} \bigr) \leq\psi ( x+1 ) < \frac{1}{2}\ln \biggl( x^{2}+x+\frac{1}{3} \biggr) $$
(1.6)
$$ \gamma+\frac{1}{2}\ln \bigl( n^{2}+n+e^{2-2\gamma}-2 \bigr) \leq H_{n}< \gamma+\frac{1}{2}\ln \biggl( n^{2}+n+ \frac{1}{3} \biggr) . $$
(1.7)
Batir [23, 24] provided another interesting bound for the psi function:
where \(x>0\), \(a\leq\ln2\), and \(b\geq0\). Consequently, the double inequality
for all \(n\in\mathbb{N}\) was attained in Corollary 2.2 in [23]. Later, this inequality was sharpened to
for all \(n\in\mathbb{N}\) by Alzer [25].
$$ a-\ln \bigl( e^{1/x}-1 \bigr) < \psi ( x ) < b-\ln \bigl( e^{1/x}-1 \bigr) , $$
(1.8)
$$ \ln\frac{\pi^{2}}{6}-\ln \bigl( e^{1/ ( n+1 ) }-1 \bigr) < H_{n}< \gamma-\ln \bigl( e^{1/ ( n+1 ) }-1 \bigr) $$
(1.9)
$$ 1+\ln ( \sqrt{e}-1 ) -\ln \bigl( e^{1/ ( n+1 ) }-1 \bigr) < H_{n}< \gamma-\ln \bigl( e^{1/ ( n+1 ) }-1 \bigr) $$
(1.10)
For reader’s convenience, here we name this class of bounds for the psi function and harmonic numbers the Batir-type bounds and call the corresponding inequality a Batir-type inequality.
Also, inequalities (1.10) are equivalent to
with the best constants \(a=e ( \sqrt{e}-1 ) \approx1.7634\) and \(b=e^{\gamma}\approx1.7810\) (see also [26]). For a more general result, see [3], Theorem 1.3. Similarly, in the context, we call inequalities (1.11) the Alzer-type ones.
$$ a\leq e^{H_{n+1}}-e^{H_{n}}< b $$
(1.11)
Anzeige
Batir [22] further proved some new Batir-type inequalities for the psi functions and harmonic number, in particular,
with the best constants \(a=2\) and \(b=e^{-2\gamma} ( e^{2}-1 ) \). This implies that, for \(n\in\mathbb{N}\), we have
where \(a=2\) and \(b=e^{2-2\gamma} ( e-1 ) -2\approx2.0024\) are the best possible. Obviously, it is equivalent to the double inequalities
Clearly, it is an Alzer-type inequality.
$$ \frac{1}{2}\ln\frac{2x+b}{e^{2/ ( x+1 ) }-1}\leq\psi ( x+1 ) \leq\frac{1}{2}\ln \frac{2x+b}{e^{2/ ( x+1 ) }-1} \quad\mbox{for } x\geq0 $$
$$ \frac{1}{2}\ln\frac{2n+b}{e^{2/ ( n+1 ) }-1}\leq H_{n}\leq \frac{1}{2}\ln\frac{2n+b}{e^{2/ ( n+1 ) }-1}, $$
$$ e^{2\gamma} ( 2n+2 ) < e^{2H_{n+1}}-e^{2H_{n}}\leq e^{2\gamma } ( 2n+2.0024\cdots) . $$
On the other hand, Villarino [27], Theorem 1.7, proved that the sequence
is increasing for \(n\in\mathbb{N}\), and meanwhile DeTemple and Wang [19] by an approximation argument for the harmonic number showed, for \(n\in\mathbb{N}\), the following inequality:
with the best constants \(\frac{21}{5}\) and \(1/ ( 1-\ln3+\ln2+\psi ( 1 ) ) -54\approx3.7393\). Yang et al. [28], Theorem 2, further showed that the function
is strictly completely monotonic on \(( 0,\infty ) \) if and only if \(a\geq7/40\).
$$ d_{n}=\frac{1}{\psi ( n+1 ) -\ln ( n+1/2 ) }-\frac {1}{24} \biggl( n+ \frac{1}{2} \biggr) ^{2} $$
$$ \frac{1}{24 ( n+1/2 ) ^{2}+21/5}< H_{n}-\ln \biggl( n+\frac {1}{2} \biggr) -\gamma< \frac{1}{24 ( n+1/2 ) ^{2}+1/ ( 1-\ln 3+\ln 2-\gamma ) -54} $$
$$ x\mapsto F_{a} ( x ) =24 \bigl( x^{2}+a \bigr) \bigl[ \psi ( x+1/2 ) -\ln x \bigr] -1 $$
Motivated by all these recent papers, the aim of this paper is to investigate the monotonicity and convexity of the function related to the psi function and present some new general, Batir-type, and Alzer-type inequalities for the psi function and harmonic number.
2 Preliminaries
In this section, let us recall a few of involving lemmas and some basic facts.
Lemma 1
([29], pp.258-260)
Let
\(x>0\)
and
\(n\in\mathbb{N}\). Then
$$ \psi^{ ( n ) }(x+1)-\psi^{ ( n ) }(x)=\frac{ ( -1 ) ^{n}n!}{x^{n+1}}. $$
(2.1)
Lemma 2
([29], pp.258-260)
As
\(x\rightarrow\infty\), we have
$$\begin{aligned}& \psi ( x ) \thicksim\ln x-\frac{1}{2x}-\frac {1}{12x^{2}}+ \frac{1}{120x^{4}}-\frac{1}{252x^{6}}, \end{aligned}$$
(2.2)
$$\begin{aligned}& \psi^{\prime} ( x ) \thicksim\frac{1}{x}+\frac{1}{2x^{2}}+ \frac{1}{6x^{3}}-\frac{1}{30x^{5}}+\frac{1}{42x^{7}}, \end{aligned}$$
(2.3)
$$\begin{aligned}& \psi^{\prime\prime} ( x ) \thicksim-\frac{1}{x^{2}}-\frac {1}{x^{3}}- \frac{1}{2x^{4}}+\frac{1}{6x^{6}}-\frac{1}{6x^{8}}. \end{aligned}$$
(2.4)
Lemma 3
([30])
Let
f
be a function on an interval
I
such that
\(\lim_{x\rightarrow\infty}f(x)=0\). If
\(f(x+1)-f(x)>0\)
for all
\(x\in ( a,\infty ) \), then
\(f(x)<0\). Conversely, if
\(f(x+1)-f(x)<0\)
for all
\(x\in ( a,\infty ) \), then
\(f(x)>0\).
Lemma 4
([31], Lemma 7)
For
\(n\in\mathbb{N}\)
and
\(m\in\mathbb{N}\cup\{0\}\)
with
\(n>m\), let
\(P_{n} ( t ) \)
be a polynomial with
n
degrees defined by
where
\(a_{n},a_{m}>0\)
and
\(a_{i}\geq0\)
for
\(0\leq i\leq n-1\)
with
\(i\neq m\). Then there exists an unique number
\(t_{m+1}\in ( 0,\infty ) \)
satisfying
\(P_{n} ( t ) =0\)
such that
\(P_{n} ( t ) <0\)
for
\(t\in ( 0,t_{m+1} ) \)
and
\(P_{n} ( t ) >0\)
for
\(t\in ( t_{m+1},\infty ) \).
$$ P_{n} ( t ) =\sum_{i=m+1}^{n}a_{i}t^{i}- \sum_{i=0}^{m}a_{i}t^{i}, $$
(2.5)
Lemma 5
Let
u
be the function on
\(( -\infty,\infty ) \)
defined by
Then, for
\(x\neq-1/2\), we have
$$ u(x)=\frac{5}{120x^{2}+21}. $$
(2.6)
$$ p ( x ) =-\frac{1}{(x+1/2)^{2}}-u^{\prime}(x+1)+u^{\prime}(x)< 0. $$
(2.7)
Proof
Differentiation leads to
Factoring gives
Replacing x by \(( t-1/2 ) \) and factoring, we get
where \(t=x+1/2\).
$$ u^{\prime}(x)=-\frac{400}{3}\frac{x}{ ( 40x^{2}+7 ) ^{2}}. $$
(2.8)
$$\begin{aligned} p ( x ) =&-\frac{1}{(x+1/2)^{2}}+\frac{400}{3}\frac{x+1}{ ( 40(x+1)^{2}+7 ) ^{2}}- \frac{400}{3}\frac{x}{ ( 40x^{2}+7 ) ^{2}}\\ =&-\frac{1}{(x+1/2)^{2}}-\frac{400}{3}\frac{4{,}800x^{4}+9{,}600x^{3}+6{,}960x^{2}+2{,}160x-49}{ ( 40(x+1)^{2}+7 ) ^{2} ( 40x^{2}+7 ) ^{2}}. \end{aligned}$$
$$ p ( x ) =-\frac{7{,}680{,}000t^{8}-384{,}000t^{6}+2{,}851{,}200t^{4}-531{,}760t^{2}+250{,}563}{3t^{2} ( 40 ( t-1/2 ) ^{2}+7 ) ^{2} ( 40 ( t+1/2 ) ^{2}+7 ) ^{2}}, $$
Note that the numerator of this fraction can be written as
and the desired result easily follows. □
$$\begin{aligned} p_{1} ( t ) =&7{,}680{,}000t^{8}-384{,}000t^{6}+2{,}851{,}200t^{4}-531{,}760t^{2}+250{,}563 \\ =&4{,}800t^{4} \bigl( 40t^{2}-1 \bigr) ^{2}+ \frac{7{,}692{,}800}{3}t^{4}+\frac {1}{3} \bigl( 920t^{2}-867 \bigr) ^{2}>0, \end{aligned}$$
(2.9)
Lemma 6
Let
u
and
p
be defined by (2.6) and (2.7). Suppose that
q
and
r
are defined on
\(( -1/2,\infty ) \)
by
Then, for
\(x>-1/2\), we have
$$\begin{aligned}& q ( x ) =\frac{2}{(x+1/2)^{3}}-u^{\prime\prime }(x+1)+u^{\prime \prime}(x),\\& r ( x ) =-\frac{1}{(x+1/2)^{2}}-u^{\prime}(x+1)-u^{\prime}(x). \end{aligned}$$
$$ S ( x ) :=-\frac{2}{(x+1/2)^{2}}+r ( x+1 ) +\frac {q ( x+1 ) }{p ( x+1 ) }-r ( x ) - \frac{q ( x ) }{p ( x ) }< 0. $$
(2.10)
Proof
An immediate computation yields
Then we get
Substituting \(p ( x )\), \(q ( x )\), \(r ( x ) \) into \(S ( x ) \) and factoring it give
where
$$ u^{\prime\prime}(x)=\frac{400}{3}\frac{120x^{2}-7}{ ( 40x^{2}+7 ) ^{3}}. $$
(2.11)
$$\begin{aligned}& q ( x ) =\frac{2}{(x+1/2)^{3}}-\frac{400}{3}\frac {120(x+1)^{2}-7}{ ( 40(x+1)^{2}+7 ) ^{3}}+ \frac{400}{3}\frac{120x^{2}-7}{ ( 40x^{2}+7 ) ^{3}}, \end{aligned}$$
(2.12)
$$\begin{aligned}& r ( x ) =-\frac{1}{(x+1/2)^{2}}+\frac{400}{3}\frac{x+1}{ ( 40(x+1)^{2}+7 ) ^{2}}+ \frac{400}{3}\frac{x}{ ( 40x^{2}+7 ) ^{2}}. \end{aligned}$$
(2.13)
$$ S ( x ) =-\frac{16\times10^{11}}{3}\frac{S_{2} ( x ) }{S_{1} ( x ) }, $$
$$\begin{aligned} S_{1} ( x ) =& ( 2x+1 ) ^{2} ( 2x+3 ) ^{2} \bigl( 40x^{2}+7 \bigr) ^{2} \bigl( 40x^{2}+160x+167 \bigr) ^{2} \\ &{}\times \bigl(7{,}680{,}000x^{8}+92{,}160{,}000x^{7}+483{,}456{,}000x^{6}+1{,}448{,}064{,}000x^{5} \\ &{}+2{,}711{,}491{,}200x^{4} +3{,}257{,}107{,}200x^{3}+2{,}458{,}239{,}440x^{2} \\ &{} +1{,}069{,}159{,}920x+205{,}944{,}303 \bigr) \\ &{}\times \bigl(7{,}680{,}000x^{8}+30{,}720{,}000x^{7} +53{,}376{,}000x^{6}+52{,}608{,}000x^{5} \\ &{}+35{,}011{,}200x^{4} +18{,}182{,}400x^{3}+6{,}745{,}040x^{2} \\ &{}+1{,}301{,}840x+319{,}823\bigr),\\ S_{2} ( x ) =&19{,}544{,}408{,}064x^{20}+390{,}888{,}161{,}280x^{19}+\frac{18{,}277{,}115{,}756{,}544}{5}x^{18} \\ &{}+ \frac{106{,}181{,}831{,}688{,}192}{5}x^{17}+\frac{2{,}147{,}345{,}768{,}669{,}184}{25}x^{16} \\ &{}+\frac{6{,}422{,}868{,}775{,}010{,}304}{25}x^{15}+ \frac{368{,}287{,}175{,}087{,}671{,}296}{625}x^{14} \\ &{}+\frac{662{,}967{,}302{,}010{,}630{,}144}{625}x^{13}+\frac{4{,}756{,}138{,}453{,}310{,}742{,}528}{3{,}125}x^{12} \\ &{}+ \frac{5{,}496{,}272{,}101{,}145{,}296{,}896}{3{,}125}x^{11}+\frac{25{,}764{,}390{,}625{,}415{,}987{,}616}{15{,}625}x^{10} \\ &{}+\frac{3{,}939{,}903{,}200{,}496{,}190{,}272}{3{,}125}x^{9}+ \frac{12{,}333{,}289{,}847{,}706{,}921{,}772}{15{,}625}x^{8} \\ &{}+\frac{6{,}339{,}553{,}647{,}515{,}390{,}816}{15{,}625}x^{7}+\frac{26{,}760{,}964{,}338{,}980{,}254{,}467}{156{,}250}x^{6} \\ &{}+ \frac{4{,}616{,}788{,}558{,}072{,}176{,}841}{78{,}125}x^{5}+\frac{513{,}733{,}037{,}814{,}725{,}250{,}509}{31{,}250{,}000}x^{4} \\ &{}+\frac{27{,}954{,}545{,}230{,}825{,}852{,}509}{7{,}812{,}500}x^{3}+ \frac{44{,}860{,}757{,}315{,}321{,}422{,}071}{78{,}125{,}000}x^{2} \\ &{}+\frac{2{,}404{,}936{,}823{,}928{,}444{,}981}{39{,}062{,}500}x+\frac{389{,}355{,}305{,}888{,}516{,}211{,}027}{100{,}000{,}000{,}000}. \end{aligned}$$
(2.14)
We further prove that \(S_{1} ( x ) ,S_{2} ( x ) >0\) for \(x>-1/2\). In fact, replacing x by \(( t-1/2 ) \) in (2.14) and arranging, we obtain
where \(p_{1} ( t ) \) is defined by (2.9), and \(t=x+1/2>0\). Clearly, \(S_{1} ( x ) >0\) for \(x>-1/2\).
$$\begin{aligned} S_{1} ( x ) =&16t^{2} ( t+1 ) ^{2} \bigl( 40t^{2}-40t+17 \bigr) ^{2} \bigl( 120t+40t^{2}+97 \bigr) ^{2}\\ &{}\times \bigl(7{,}680{,}000t^{8}+61{,}440{,}000t^{7}+214{,}656{,}000t^{6} +427{,}776{,}000t^{5}\\ &{}+534{,}691{,}200t^{4}+433{,}804{,}800t^{3}+225{,}855{,}440t^{2}+69{,}477{,}280t\\ &{}+9{,}866{,}003\bigr) \times p_{1} ( t ) , \end{aligned}$$
Similarly, we have
where
It is clear that \(S_{3} ( t ) >0\) for \(t=x+1/2>0\). To prove that \(S_{2} ( x ) >0\) for \(x>-1/2\), it suffices to prove that \(S_{4} ( t ) >0\) for \(t>0\). In fact, it is easy to check that
has a positive zero point \(t_{0}\in ( 1/3,1/2 ) \) such that \(S_{4}^{\prime} ( t ) <0\) for \(t\in ( 0,t_{0} ) \) and \(S_{4}^{\prime} ( t ) <0\) for \(t\in ( t_{0},\infty ) \). Since \(S_{4} ( 0 ) ,S_{4} ( \infty ) >0\) and
so we get \(S_{4} ( t ) >0\) for \(t>0\), which proves that \(S_{2} ( x ) >0\) for \(x>-1/2\) and completes the proof. □
$$ S_{2} ( x ) =S_{3} ( t ) +t^{2}S_{4} ( t ) , $$
$$\begin{aligned} S_{3} ( t ) =&19{,}544{,}408{,}064t^{20}+195{,}444{,}080{,}640t^{19}+\frac{4{,}351{,}725{,}010{,}944}{5}t^{18}\\ &{}+\frac{11{,}314{,}743{,}607{,}296}{5}t^{17}+\frac{94{,}415{,}373{,}656{,}064}{25}t^{16}+ \frac{104{,}269{,}760{,}888{,}832}{25}t^{15}\\ &{}+\frac{1{,}887{,}440{,}973{,}367{,}296}{625}t^{14}+\frac{888{,}059{,}638{,}659{,}072}{625}t^{13}\\ &{}+ \frac{1{,}921{,}294{,}271{,}712{,}768}{3{,}125}t^{12}+\frac{1{,}795{,}217{,}217{,}788{,}928}{3{,}125}t^{11}\\ &{}+\frac{8{,}617{,}726{,}188{,}296{,}736}{15{,}625}t^{10}+ \frac{928{,}558{,}757{,}330{,}976}{3{,}125}t^{9}\\ &{}+\frac{1{,}217{,}964{,}858{,}530{,}932}{15{,}625}t^{8}+\frac{353{,}703{,}953{,}859{,}088}{15{,}625}t^{7}\\ &{}+ \frac{3{,}731{,}661{,}240{,}517{,}347}{156{,}250}t^{6}+\frac{151{,}574{,}123{,}955{,}363{,}957}{156{,}250{,}000}t\\ &{}+\frac{20{,}166{,}045{,}810{,}484{,}915{,}467}{100{,}000{,}000{,}000},\\ S_{4} ( t ) =&\frac{1{,}496{,}705{,}911{,}425{,}721}{156{,}250}t^{3}-\frac{78{,}452{,}580{,}264{,}405{,}241}{31{,}250{,}000}t^{2}\\ &{}- \frac{25{,}060{,}239{,}724 662{,}741}{15{,}625{,}000}t+\frac{11{,}608{,}440{,}209{,}633{,}547}{9{,}765{,}625}. \end{aligned}$$
$$\begin{aligned} S_{4}^{\prime} ( t ) =&\frac{4{,}490{,}117{,}734{,}277{,}163}{156{,}250}t^{2}- \frac{78{,}452{,}580{,}264{,}405{,}241}{15{,}625{,}000}t\\ &{}-\frac{25{,}060{,}239{,}724{,}662{,}741}{15{,}625{,}000} \end{aligned}$$
$$\begin{aligned} S_{4} ( t_{0} ) =&\frac{1{,}496{,}705{,}911{,}425{,}721}{156{,}250}t_{0}^{3}+ \frac{11{,}608{,}440{,}209{,}633{,}547}{9{,}765{,}625} \\ &{}- \biggl( \frac{78{,}452{,}580{,}264{,}405{,}241}{31{,}250{,}000}t_{0}^{2}+ \frac{25{,}060{,}239{,}724{,}662{,}741}{15{,}625{,}000}t_{0} \biggr) \\ >&\frac{1{,}496{,}705{,}911{,}425{,}721}{156{,}250} \biggl( \frac{1}{3} \biggr) ^{3}+\frac{11{,}608{,}440{,}209{,}633{,}547}{9{,}765{,}625} \\ &{}- \biggl( \frac{78{,}452{,}580{,}264{,}405{,}241}{31{,}250{,}000} \biggl( \frac {1}{2} \biggr) ^{2}+\frac{25{,}060{,}239{,}724{,}662{,}741}{15{,}625{,}000}\frac {1}{2} \biggr)\\ =&\frac{1{,}922{,}580{,}540{,}937{,}065{,}541}{16{,}875{,}000{,}000}>0, \end{aligned}$$
3 Monotonicity and convexity
In this section, we state and prove Theorems 1-3 on the monotonicity and convexity of three important functions \(f_{1}(x)\), \(f_{2}(x)\), and \(f_{3}(x)\) concerning the psi function, respectively.
Theorem 1
The function
is decreasing from
\(( -1/2,\infty ) \)
onto
\(( 0,1/2 ) \)
and convex on
\(( -1/2,\infty ) \).
$$ x\mapsto f_{1} ( x ) =\exp \biggl( \psi \biggl( x+\frac {1}{2} \biggr) -\frac{1}{24}\frac{1}{x^{2}+7/40} \biggr) -x $$
Proof
With (2.6) in hand, \(f_{1} ( x ) \) can be written as
Differentiation of this formula yields
By (2.1) this yields
where \(p ( x )\), \(q ( x )\), \(r ( x ) \) are defined by (2.7), (2.12), (2.13), respectively.
$$ f_{1} ( x ) =e^{\psi(x+1/2)-u(x)}-x. $$
$$\begin{aligned}& f_{1}^{\prime} ( x )= \bigl( \psi^{\prime }(x+1/2)-u^{\prime }(x) \bigr) e^{\psi(x+1/2)-u(x)}-1, \end{aligned}$$
(3.1)
$$\begin{aligned}& \begin{aligned}[b] f_{1}^{\prime\prime} ( x ) &=\frac{\psi^{\prime\prime }(x+1/2)-u^{\prime\prime}(x)+ ( \psi^{\prime}(x+1/2)-u^{\prime }(x) ) ^{2}}{e^{-\psi(x+1/2)+u(x)}} \\ &:=\frac{g ( x ) }{e^{-\psi(x+1/2)+u(x)}}. \end{aligned} \end{aligned}$$
(3.2)
$$\begin{aligned} & g ( x+1 ) -g ( x ) \\ &\quad= \bigl[ \psi(x+3/2,2)-u^{\prime \prime}(x+1) \bigr] + \bigl[ \psi(x+3/2,1)-u^{\prime}(x+1) \bigr] ^{2} \\ &\qquad{}- \bigl\{ \psi(x+1/2,2)-u^{\prime\prime}(x)+ \bigl[ \psi (x+1/2,1)-u^{\prime}(x) \bigr] ^{2} \bigr\} \\ &\quad= p ( x ) \biggl[ 2\psi(x+1/2,1)+r ( x ) +\frac {q ( x ) }{p ( x ) } \biggr] \\ &\quad:=p ( x ) h ( x ) , \end{aligned}$$
(3.3)
Similarly, we get
By Lemma 6 we have \(h ( x+1 ) -h ( x ) <0\). It follows from \(\lim_{x\rightarrow\infty}h ( x ) =0\) and Lemma 3 that \(h ( x ) >\lim_{x\rightarrow\infty}h ( x ) =0\). Thanks to inequality (3.3), \(p ( x ) <0\), and Lemma 5, it follows that
which implies by Lemma 3 that \(g ( x ) >\lim_{x\rightarrow \infty}g ( x ) =0\). Thus, in combination with (3.2), this leads to \(f_{1}^{\prime\prime} ( x ) >0\), that is, \(f_{1}\) is convex on \(( -1/2,\infty ) \), and \(f_{1}^{\prime}\) is increasing on \(( -1/2,\infty ) \). Utilizing the asymptotic formulas (2.2)-(2.3), this yields
Therefore, we get that \(f_{1}^{\prime} ( x ) <\lim_{x\rightarrow\infty }f_{1}^{\prime} ( x ) =0\), which implies that \(f_{1} ( x ) \) is decreasing on \(( -1/2,\infty ) \). Moreover, we conclude that
which completes the proof. □
$$ h ( x+1 ) -h ( x ) =-\frac{2}{(x+1/2)^{2}}+r ( x+1 ) +\frac{q ( x+1 ) }{p ( x+1 ) }-r ( x ) - \frac{q ( x ) }{p ( x ) }=S ( x ) . $$
$$ g ( x+1 ) -g ( x ) =p ( x ) h ( x ) < 0, $$
$$ \lim_{x\rightarrow\infty}f_{1}^{\prime} ( x ) =\lim _{x\rightarrow\infty}f_{1} ( x ) =0. $$
$$ 0=\lim_{x\rightarrow\infty}f_{1} ( x ) < f_{1} ( x ) < \lim_{x\rightarrow-1/2^{+}}f_{1} ( x ) =\frac{1}{2}, $$
Theorem 2
The function
is decreasing from
\(( 0,\infty ) \)
onto
\(( 0,e^{-\gamma }/4 ) \)
and convex on
\(( 1/2,\infty ) \).
$$ x\mapsto f_{2} ( x ) =\exp\psi \biggl( x+\frac{1}{2} \biggr) -x \exp \biggl( \frac{1}{24}\frac{1}{x^{2}+7/40} \biggr) $$
Proof
We have
Noting that
we have
A simple calculation leads to
which completes the proof. □
$$\begin{aligned} f_{2} ( x ) =&\exp \biggl( \frac{1}{24}\frac {1}{x^{2}+7/40} \biggr) \biggl\{ \exp \biggl[ \psi \biggl( x+\frac{1}{2} \biggr) - \frac {1}{24}\frac{1}{x^{2}+7/40} \biggr] -x \biggr\} \\ =&\exp \biggl( \frac{1}{24}\frac{1}{x^{2}+7/40} \biggr) f_{1} ( x ) :=f_{0} ( x ) f_{1} ( x ) . \end{aligned}$$
$$\begin{aligned}& f_{0}^{\prime} ( x ) =-\frac{1}{12}\frac{x}{ ( x^{2}+7/40 ) ^{2}} \exp \biggl( \frac{1}{24}\frac {1}{x^{2}+7/40} \biggr) \leq0\quad\mbox{for }x \geq0,\\& f_{0}^{\prime\prime} ( x ) =\frac{1}{57{,}600}\frac{14{,}400x^{4}+2{,}080x^{2}-147}{ ( x^{2}+7/40 ) ^{4}} \exp \biggl( \frac{1}{24}\frac{1}{x^{2}+7/40} \biggr) >0\quad\mbox{for }x\geq \frac{1}{2}, \end{aligned}$$
$$\begin{aligned}& f_{2}^{\prime} ( x ) =f_{0}^{\prime} ( x ) f_{1} ( x ) +f_{0} ( x ) f_{1}^{\prime} ( x ) < 0\quad\mbox{for }x\geq0,\\& f_{2}^{\prime\prime} ( x ) =f_{0}^{\prime\prime} ( x ) f_{1} ( x ) +2f_{0}^{\prime} ( x ) f_{1}^{\prime } ( x ) +f_{0} ( x ) f_{1}^{\prime\prime} ( x ) >0\quad\mbox{for }x\geq\frac{1}{2}. \end{aligned}$$
$$ \lim_{x\rightarrow\infty}f_{2} ( x ) =\lim_{x\rightarrow \infty } \exp \biggl( \frac{1}{24}\frac{1}{x^{2}+7/40} \biggr) \lim _{x\rightarrow \infty}f_{1} ( x ) =0\quad\mbox{and}\quad \lim _{x\rightarrow 0+}f_{2} ( x ) =\frac{1}{4}e^{-\gamma}, $$
Theorem 3
Let
\(a\geq0\). Then the function
is decreasing and convex on
\(( 0,\infty ) \)
if and only if
\(a\geq 7/40\).
$$ x\mapsto f_{3} ( x ) =\psi \biggl( x+\frac{1}{2} \biggr) -\ln x- \frac{1}{24}\frac{1}{x^{2}+a} $$
Proof
The necessity is obvious; it follows from the inequality \(\lim_{x\rightarrow\infty }x^{5}f_{3}^{\prime} ( x ) \leq0\). Indeed, using the asymptotic formulas (2.2), we have
which yields \(a\geq7/40\).
$$ \lim_{x\rightarrow\infty}x^{5}f_{3}^{\prime} ( x ) =\frac {7}{240}-\frac{1}{6}a\leq0, $$
We now are a position to prove the sufficiency. By differentiation we have
Using (2.1), we have
where
$$\begin{aligned}& f_{3}^{\prime} ( x ) =\psi^{\prime} \biggl( x+ \frac{1}{2} \biggr) -\frac{1}{x}+\frac{1}{12} \frac{x}{ ( x^{2}+a ) ^{2}}, \\& f_{3}^{\prime\prime} ( x ) =\psi^{\prime\prime} \biggl( x+ \frac{1}{2} \biggr) +\frac{1}{x^{2}}+\frac{1}{12} \frac{1}{ ( x^{2}+a ) ^{2}}-\frac{1}{3}\frac{x^{2}}{ ( x^{2}+a ) ^{3}}. \end{aligned}$$
$$\begin{aligned} f_{3}^{\prime\prime} ( x+1 ) -f_{3}^{\prime\prime} ( x ) =& \psi^{\prime\prime} \biggl( x+\frac{3}{2} \biggr) +\frac {1}{ ( x+1 ) ^{2}}+ \frac{1}{12}\frac{1}{ ( ( x+1 ) ^{2}+a ) ^{2}}-\frac{1}{3}\frac{x^{2}}{ ( ( x+1 ) ^{2}+a ) ^{3}} \\ &{}-\psi^{\prime\prime} \biggl( x+\frac{1}{2} \biggr) -\frac {1}{x^{2}}- \frac{1}{12}\frac{1}{ ( x^{2}+a ) ^{2}}+\frac{1}{3}\frac{x^{2}}{ ( x^{2}+a ) ^{3}} \\ =&\frac{2}{ ( x+1/2 ) ^{3}}+\frac{1}{ ( x+1 ) ^{2}}+\frac{1}{12} \frac{1}{ ( ( x+1 ) ^{2}+a ) ^{2}} \\ &{}-\frac{1}{3}\frac{x^{2}}{ ( ( x+1 ) ^{2}+a ) ^{3}}- \frac{1}{x^{2}}- \frac{1}{12}\frac{1}{ ( x^{2}+a ) ^{2}}+\frac {1}{3}\frac{x^{2}}{ ( x^{2}+a ) ^{3}} \\ =&-\frac{P ( x ) }{12x^{2} ( 2x+1 ) ^{3} ( x^{2}+a ) ^{3} ( x+1 ) ^{2} ( x^{2}+2x+a+1 ) ^{3}}, \end{aligned}$$
$$\begin{aligned} P ( x ) = {}& 12 ( 40a-7 ) x^{12}+72 ( 40a-7 ) x^{11}+ ( 1{,}536a^{2}+7{,}576a-1{,}290 ) x^{10}\\ &{}+ ( 7{,}680a^{2}+11{,}480a-1{,}830 ) x^{9}+ ( 2{,}016a^{3}+16{,}788a^{2}+11{,}038a-1{,}563 ) x^{8}\\ &{}+ ( 8{,}064a^{3}+21{,}072a^{2}+6{,}952a-816 ) x^{7}\\ &{}+ ( 1{,}440a^{4}+14{,}112a^{3}+16{,}710a^{2}+2{,}840a-252 ) x^{6}\\ &{}+ ( 4{,}320a^{4}+14{,}112a^{3}+8{,}634a^{2}+724a-42 ) x^{5}\\ &{}+ ( 576a^{5}+5{,}652a^{4}+8{,}862a^{3}+2{,}853a^{2}+115a-3 ) x^{4}\\ &{}+ ( 1{,}152a^{5}+4{,}104a^{4}+3{,}612a^{3}+540a^{2}+14a ) x^{3}\\ &{}+ ( 96a^{6}+936a^{5}+1{,}764a^{4}+966a^{3}+39a^{2}+a ) x^{2}\\ &{}+ ( 96a^{6}+360a^{5}+432a^{4}+168a^{3} ) x+ ( 12a^{6}+36a^{5}+36a^{4}+12a^{3} ) . \end{aligned}$$
It is easy to check that the coefficients of the polynomial \(P ( x ) \) are nonnegative for \(a\geq7/40\). Then we have \(f_{3}^{\prime \prime} ( x+1 ) -f_{3}^{\prime\prime} ( x ) <0\) for \(x>0 \). By Lemma 3 we get that \(f_{3}^{\prime\prime} ( x ) >\lim_{x\rightarrow\infty}f_{3}^{\prime\prime} ( x ) =0 \), which implies that \(f_{3}^{\prime} ( x ) \) is increasing on \(( 0,\infty ) \). Therefore, \(f_{3}^{\prime} ( x ) <\lim_{x\rightarrow\infty}f_{3}^{\prime} ( x ) =0\), which completes our proof. □
4 Inequalities for the psi function and harmonic number
Denote \(F_{i} ( x ) =f_{i} ( x+1/2 ) \) (\(i=1,2,3\)) in Theorems 1 and 2. Then, since \(F_{1}\) is decreasing, \(F_{1} ( 0 ) =e^{-\gamma-5/51}\), and \(\lim_{x\rightarrow \infty}F ( x ) =1/2\), we get the following conclusion.
Corollary 1
(i)
For
\(x>-1/2\), we have
$$ \psi(x+1)>\ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24} \frac{1}{x^{2}+x+17/40}. $$
(ii)
For
\(x\geq0\), we have
where the constants
\(1/2\)
and
\(\alpha_{0}=e^{-\gamma-5/51}\approx0.50903\)
are the best possible.
$$ \ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24}\frac {1}{x^{2}+x+17/40}< \psi (x+1)< \ln ( x+\alpha_{0} ) +\frac{1}{24} \frac{1}{x^{2}+x+17/40}, $$
(4.1)
Remark 1
Comparing (4.1) with (1.4), we find that, for \(x>0\),
In fact, it suffices to show that the last inequality is valid for \(x>0 \). Let
By differentiation we have
where
It is easy to check that \(a_{4},a_{3}>0\) and \(a_{2},a_{1},a_{0}<0\). Then by Lemma 4 we see that there is \(x_{0}>0\) such that \(D_{2} ( x ) <0\) for \(x\in ( 0,x_{0} ) \) and \(D_{2} ( x ) >0\) for \(x\in ( x_{0},\infty ) \). This indicates that \(D_{1}\) is decreasing on \(( 0,x_{0} ) \) and increasing on \(( x_{0},\infty ) \). Therefore, we conclude that, for \(x>0\),
$$\begin{aligned} \ln \biggl( x+\frac{1}{2} \biggr) < &\ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24}\frac{1}{x^{2}+x+17/40}< \psi(x+1) \\ < &\ln ( x+\alpha_{0} ) +\frac{1}{24}\frac {1}{x^{2}+x+17/40}< \ln \bigl( x+e^{-\gamma} \bigr). \end{aligned}$$
$$ D_{1} ( x ) =\ln ( x+\alpha_{0} ) +\frac {1}{24} \frac{1}{x^{2}+x+17/40}< \ln \bigl( x+e^{-\gamma} \bigr) . $$
$$\begin{aligned} D_{1}^{\prime} ( x ) =&\frac{1}{x+\alpha_{0}}-\frac {1}{24} \frac{2x+1}{ ( x^{2}+x+17/40 ) ^{2}}-\frac{1}{x+\beta} \\ =&\frac{D_{2} ( x ) }{3 ( x+\alpha_{0} ) ( x+\beta ) ( 40x^{2}+40x+17 ) ^{2}}, \end{aligned}$$
$$\begin{aligned} D_{2} ( x ) ={}&4{,}800 ( e^{-\gamma}-\alpha_{0} ) x^{4}+400 ( 24e^{-\gamma}-24\alpha_{0}-1 ) x^{3}-40 ( 232\alpha_{0}-212e^{-\gamma}+5 ) x^{2} \\ &{}-40 ( 107\alpha_{0}-97e^{-\gamma}+10\alpha_{0}e^{-\gamma} ) x- ( 867\alpha_{0}-867e^{-\gamma}+200\alpha_{0}e^{-\gamma} )\\ :={}&a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}. \end{aligned}$$
$$ D_{1} ( x ) < \max \bigl( D_{1} ( 0 ) ,D_{1} ( \infty ) \bigr) =0. $$
Remark 2
Similarly, we get the following inequalities:
for \(x>0\). A direct computation shows that
which implies that the approximation formula of the psi function given in (4.1) is superior to (1.6).
$$\begin{aligned} \frac{1}{2}\ln \bigl( x^{2}+x+e^{-2\gamma} \bigr) < &\ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24}\frac{1}{x^{2}+x+17/40}< \psi(x+1)\\ < &\frac{1}{2}\ln \biggl( x^{2}+x+\frac{1}{3} \biggr) < \ln ( x+\alpha _{0} ) +\frac{1}{24}\frac{1}{x^{2}+x+17/40} \end{aligned}$$
$$\begin{aligned}& \lim_{x\rightarrow\infty}\frac{\psi(x+1)-\frac{1}{2}\ln ( x^{2}+x+\frac{1}{3} ) }{x^{-4}} =-\frac{1}{180}, \end{aligned}$$
(4.2)
$$\begin{aligned}& \lim_{x\rightarrow\infty}\frac{\psi(x+1)-\ln ( x+\frac{1}{2} ) -\frac{1}{24}\frac{1}{x^{2}+x+17/40}}{x^{-6}} =\frac{2{,}071}{806{,}400}, \end{aligned}$$
(4.3)
Since \(F_{1} ( 1 ) =e^{286/291-\gamma}\), by the relation \(\psi(n+1)=H_{n}-\gamma\) we deduce the following:
Corollary 2
For
\(n\in\mathbb{N}\), we have
where
\(1/2\)
and
\(\alpha_{1}=e^{286/291-\gamma}\approx0.50021\)
are the best possible.
$$ \gamma+\frac{1}{24}\frac{1}{n^{2}+n+17/40}+\ln ( n+1/2 ) < H_{n}< \gamma+\frac{1}{24}\frac{1}{n^{2}+n+17/40}+\ln ( n+\alpha _{1} ) , $$
Since \(F_{2}\) is decreasing on \(( -1/2,\infty ) \), \(\lim_{x\rightarrow\infty}F_{2} ( x ) =0\), and
we get the following:
$$ F_{2} ( 0 ) =e^{-\gamma}-\frac{1}{2}e^{5/51}=\frac{1}{2}\beta _{0}, \qquad F_{2} ( 1 ) =e^{1-\gamma}-\frac {3}{2}e^{5/291}=\frac{1}{2}\beta_{1},$$
Corollary 3
For
\(x\geq0\), we have
where
\(\beta_{0}=2e^{-\gamma}-e^{5/51}\approx0.019913\)
is the best constant.
$$\begin{aligned} \ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24}\frac{1}{x^{2}+x+17/40} < &\psi ( x+1 ) < \ln \biggl( x+\frac{1}{2} \biggr) +\frac {1}{24}\frac{1}{x^{2}+x+17/40}\\ &{}+\ln \biggl[ 1+\frac{\beta_{0}}{2x+1}\exp \biggl( -\frac{1}{24} \frac {1}{x^{2}+x+17/40} \biggr) \biggr] , \end{aligned}$$
Corollary 4
For
\(n\in\mathbb{N}\), we have
where
\(\beta_{1}=2e^{1-\gamma}-3e^{5/291}\approx0.00041845\)
is the best constant.
$$\begin{aligned} \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac{1}{24} \frac{1}{n^{2}+n+17/40} < &H_{n}< \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac {1}{24}\frac{1}{n^{2}+n+17/40} \\ &{}+\ln \biggl[ 1+\frac{\beta_{1}}{2n+1}\exp \biggl( -\frac{1}{24} \frac {1}{n^{2}+n+17/40} \biggr) \biggr] , \end{aligned}$$
For \(a=7/40\), since \(F_{3}\) is decreasing on \(( -1/2,\infty ) \), \(\lim_{x\rightarrow\infty}F_{3} ( x ) =0\), and
we deduce the following:
$$ F_{3} ( 0 ) =\ln2-\frac{5}{51}-\gamma=\delta_{0},\qquad F_{3} ( 1 ) =\frac{286}{291}-\ln\frac{3}{2}-\gamma=\delta _{1}, $$
Corollary 5
For
\(x\geq0\), we have
where
\(\delta_{0}^{\ast}=0\)
and
\(\delta_{0}=\ln2-5/51-\gamma\approx 0.017892\)
are the best constants.
$$ \ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24} \frac{1}{x^{2}+x+17/40}+\delta_{0}^{\ast}< \psi ( x+1 ) < \ln \biggl( x+\frac{1}{2} \biggr) +\frac{1}{24} \frac{1}{x^{2}+x+17/40}+\delta_{0}, $$
Corollary 6
For
\(n\in\mathbb{N}\), we have
where
\(\delta_{1}^{\ast}=0\)
and
\(\delta_{1}=286/291-\ln ( 3/2 ) -\gamma\approx0.00013710\)
are the best constants.
$$ \gamma+\ln \biggl( n+\frac{1}{2} \biggr) +\frac{1}{24} \frac{1}{n^{2}+n+17/40}+\delta_{1}^{\ast}< H_{n}< \gamma+ \ln \biggl( n+\frac {1}{2} \biggr) +\frac{1}{24} \frac{1}{n^{2}+n+17/40}+\delta_{1}, $$
From Theorem 1 we can obtain the following Batir-type inequalities for the psi function and harmonic number.
Corollary 7
For
\(x\geq0\), we have
with the best constants
\(c_{0}=\ln ( e^{5{,}347/4{,}947}-1 ) -5/51-\gamma\approx-0.0088601\)
and
\(c_{0}^{\ast}=0\), where
$$ \frac{1}{24 ( x^{2}+x+17/40 ) }-\ln \bigl( e^{Q ( x ) }-1 \bigr) +c_{0}< \psi ( x+1 ) < \frac{1}{24 ( x^{2}+x+17/40 ) }-\ln \bigl( e^{Q ( x ) }-1 \bigr) +c_{0}^{\ast} $$
$$ Q ( x ) =\frac{4{,}800x^{4}+19{,}200x^{3}+28{,}480x^{2}+18 560x+5{,}347}{3 ( x+1 ) ( 40x^{2}+40x+17 ) ( 40x^{2}+120x+97 ) }. $$
Proof
Let \(G_{1} ( x ) =F_{1} ( x+1 ) -F_{1} ( x ) =f_{1} ( x+3/2 ) -f_{1} ( x+1/2 ) \). Since \(f_{1}\) is convex on \(( -1/2,\infty ) \), we have
which means that \(G_{1}\) is increasing on \(( 0,\infty ) \). Considering
we have
which attains the desired inequality. □
$$ G_{1}^{\prime} ( x ) =f_{1}^{\prime} ( x+3/2 ) -f_{1}^{\prime} ( x+1/2 ) =f_{1}^{\prime\prime} ( x+1/2+\theta ) >0, $$
$$\begin{aligned}& \begin{aligned}[b] G_{1} ( x ) ={}&\exp \biggl( \psi ( x+2 ) -\frac {1}{24}\frac{1}{ ( x+3/2 ) ^{2}+7/40} \biggr) \\ &{}-\exp \biggl( \psi ( x+1 ) -\frac{1}{24} \frac{1}{ ( x+1/2 ) ^{2}+7/40} \biggr) -1 \\ ={}&\exp \biggl( \psi ( x+1 ) -\frac{1}{24}\frac{1}{ ( x+1/2 ) ^{2}+7/40} \biggr) \bigl( e^{Q ( x ) }-1 \bigr) -1, \end{aligned}\\& G_{1} ( 0 ) =\exp \biggl( \frac{286}{291}-\gamma \biggr) -e \biggl( -\frac{5}{51}-\gamma \biggr) -1\quad\mbox{and}\quad\lim _{x\rightarrow\infty }G_{1} ( x ) =0, \end{aligned}$$
$$ G_{1} ( 0 ) < \exp \biggl( \psi ( x+1 ) -\frac{1}{24}\frac{1}{ ( x+1/2 ) ^{2}+7/40} \biggr) \bigl( e^{Q ( x ) }-1 \bigr) -1< 0, $$
The increasing property of \(G_{1}\) and
yield a sharp bound of harmonic number.
$$ G_{1} ( 1 ) =\exp \biggl( \frac{2{,}303}{1{,}542}-\gamma \biggr) -\exp \biggl( \frac{286}{291}-\gamma \biggr) -1 $$
Corollary 8
For
\(n\in\mathbb{N}\), we have
where
\(c_{1}=286/291-\gamma+\ln ( e^{76{,}387/149{,}574}-1 ) \approx-0.00018438\)
and
\(c_{1}^{\ast}=0\)
are the best constants.
$$\begin{aligned} &\gamma+\frac{1}{24 ( n^{2}+n+17/40 ) }-\ln \bigl( e^{Q ( n ) }-1 \bigr) +c_{1}\\ &\quad< H_{n}< \gamma+\frac{1}{24 ( n^{2}+n+17/40 ) }-\ln \bigl( e^{Q ( n ) }-1 \bigr) +c_{1}^{\ast}, \end{aligned}$$
Remark 3
Note that since \(\psi ( n+1 ) =H_{n}-\gamma\), \(G_{1} ( n ) \) is written as
Then by \(G_{1} ( 1 ) \leq G_{1} ( n ) < G_{1} ( \infty ) \) we have the following Alzer-type inequalities:
where
$$\begin{aligned} G_{1} ( n ) ={}&\exp \biggl( H_{n+1}-\gamma-\frac{1}{24} \frac {1}{ ( n+3/2 ) ^{2}+7/40} \biggr) \\ &{}-\exp \biggl( H_{n}-\gamma- \frac {1}{24}\frac{1}{ ( n+1/2 ) ^{2}+7/40} \biggr) -1. \end{aligned}$$
$$ 1.7807\approx\exp \biggl( \frac{2{,}303}{1{,}542} \biggr) -\exp \biggl( \frac {286}{291} \biggr) < e^{H_{n+1}-u_{n+1}}-e^{H_{n}-u_{n}}< e^{\gamma} \approx1.7811, $$
$$ u_{n}=\frac{1}{24}\frac{1}{ ( n+1/2 ) ^{2}+7/40}. $$
(4.4)
Remark 4
Similarly, it is easy to check that
is increasing on \(( 0,\infty ) \). Then, from
we derive other Alzer-type inequalities:
where \(d_{0}=0\) and \(d_{1}=3e^{5/291}/2-5e^{5/771}/2+e^{1-\gamma} ( e^{1/2}-1 ) \approx-0.00018779\) are the best constants with \(u_{n}\) as in (4.4).
$$ G_{2} ( x ) =F_{2} ( x+1 ) -F_{2} ( x ) =f_{2} ( x+3/2 ) -f_{2} ( x+1/2 ) $$
$$ -0.00018779\approx\frac{3}{2}e^{5/291}-\frac{5}{2}e^{5/771}+e^{1-\gamma } \bigl( e^{1/2}-1 \bigr) =G_{2} ( 1 ) \leq G_{2} ( n ) < G_{2} ( \infty ) =0 $$
$$ \biggl( n+\frac{3}{2} \biggr) u_{n+1}- \biggl( n+ \frac{1}{2} \biggr) u_{n}+d_{1}< e^{H_{n+1}}-e^{H_{n}}< \biggl( n+\frac{3}{2} \biggr) u_{n+1}- \biggl( n+ \frac{1}{2} \biggr) u_{n}+d_{0}, $$
Using the increasing property of \(F_{1}^{\prime}\), and noting that \(F_{1}^{\prime} ( -1^{+} ) =-1\), \(F_{1}^{\prime} ( 0 ) = ( \pi^{2}/6+200/867 ) e^{-\gamma-5/51}-1\), and \(F_{1} ( \infty ) =0\), we get
which implies the following:
$$ \biggl( \frac{1}{6}\pi^{2}+\frac{200}{867} \biggr) e^{-\gamma -5/51}-1< \bigl( \psi^{\prime}(x+1)-u^{\prime}(x) \bigr) e^{\psi(x+1)-u(x)}-1< 0, $$
Corollary 9
(i)
For
\(x>-1\), we have
$$ \psi^{\prime}(x+1)< -\frac{1}{12}\frac{x+1/2}{ ( x^{2}+x+17/40 ) ^{2}}+\exp \biggl( - \psi(x+1)+\frac{1}{24 ( x^{2}+x+17/40 ) } \biggr) . $$
(ii)
For
\(x\geq0\), we have the double inequalities
where
\(\lambda_{1}= ( \pi^{2}/6+200/867 ) e^{-\gamma -5/51}\approx0.95474\)
and
\(\lambda_{2}=1\)
are the best constants.
$$\begin{aligned} &{-}\frac{1}{12}\frac{x+1/2}{ ( x^{2}+x+17/40 ) ^{2}}+\lambda _{1}\exp \biggl( - \psi(x+1)+\frac{1}{24 ( x^{2}+x+17/40 ) } \biggr)\\ &\quad< \psi^{\prime}(x+1)< -\frac{1}{12}\frac{x+1/2}{ ( x^{2}+x+17/40 ) ^{2}}\\ &\qquad{}+ \lambda_{2}\exp \biggl( -\psi(x+1)+\frac {1}{24 ( x^{2}+x+17/40 ) } \biggr) , \end{aligned}$$
Remark 5
Elezovic et al. [32] proved the inequality
for \(x>0\). It has been improved by Batir [22] as
for \(x>0\) with the best constants \(a^{\ast}=1/2\) and \(b^{\ast}=\pi ^{2}e^{-2\gamma}/6\). The last corollary gives another improvement of (4.5).
$$ \psi^{\prime} ( x ) < e^{-\psi ( x ) } $$
(4.5)
$$ \bigl( x+a^{\ast} \bigr) e^{-2\psi ( x ) }< \psi^{\prime } ( x+1 ) < \bigl( x+b^{\ast} \bigr) e^{-2\psi ( x ) } $$
From the proof of Theorem 1 we see that \(g ( x ) >0\) for \(x>-1/2\), which can be written as the following corollary.
Corollary 10
For
\(x>0\), we have
where
\(u ( x ) \)
is defined by (2.6).
$$ \psi^{\prime\prime}(x)-u^{\prime\prime}(x-1/2)+ \bigl( \psi^{\prime }(x)-u^{\prime}(x-1/2) \bigr) ^{2}>0, $$
(4.6)
Remark 6
Batir [23] showed that, for \(x>0\),
Therefore, inequality (4.6) can be written as
where
Indeed, this result is optimal due to
where
A numeric computation shows that \(\Delta ( x ) >0\) for \(0< x<2/5\) and \(x>3/2\), and so inequality (4.6) is better than (4.7).
$$ \psi^{\prime}(x)^{2}+\psi^{\prime\prime}(x)>0. $$
(4.7)
$$ \psi^{\prime\prime}(x)+\psi^{\prime}(x)^{2}>\Delta ( x ) , $$
$$\begin{aligned} \Delta ( x ) ={}&{-}\frac{400}{3}\frac{2x-1}{ ( 40x^{2}-40x+17 ) ^{2}} \\ &{}\times\biggl( \psi^{\prime}(x)-\frac{14{,}400x^{4}-28{,}800x^{3}+22{,}880x^{2}-8{,}480x+1{,}073}{3 ( 2x-1 ) ( 40x^{2}-40x+17 ) ^{2}} \biggr) . \end{aligned}$$
$$\begin{aligned} &\psi^{\prime\prime}(x)-\frac{400}{3}\frac{120 ( x-1/2 ) ^{2}-7}{ ( 40 ( x-1/2 ) ^{2}+7 ) ^{3}}+ \biggl( \psi^{\prime }(x)+\frac{400}{3}\frac{ ( x-1/2 ) }{ ( 40 ( x-1/2 ) ^{2}+7 ) ^{2}} \biggr) ^{2} \\ &\quad=\psi^{\prime\prime}(x)+\psi^{\prime}(x)^{2}-\frac{400}{3} \frac {1}{ ( 40x^{2}-40x+17 ) ^{2}}\Delta ( x ) , \end{aligned}$$
$$ \Delta ( x ) =- ( 2x-1 ) \psi^{\prime}(x)+\frac{14{,}400x^{4}-28{,}800x^{3}+22{,}880x^{2}-8{,}480x+1{,}073}{3 ( 40x^{2}-40x+17 ) ^{2}}. $$
From the inequalities \(f_{3}^{\prime} ( x ) <0\) and \(f_{3}^{\prime \prime} ( x ) >0\) on \(( 0,\infty ) \) for \(a=7/40\), which are given in the proof of Theorem 3, we have the following:
Corollary 11
For
\(x>0\), we have the following inequalities:
$$\begin{aligned}& \psi^{\prime} \biggl( x+\frac{1}{2} \biggr) < \frac{1}{3} \frac{4{,}800x^{4}+1{,}280x^{2}+147}{x ( 40x^{2}+7 ) ^{2}},\\& \psi^{\prime\prime} \biggl( x+\frac{1}{2} \biggr) >-\frac{1}{3} \frac{ 192{,}000x^{6}+52{,}800x^{4}+20{,}440x^{2}+1{,}029}{x^{2} ( 40x^{2}+7 ) ^{3}},\\& \psi^{\prime\prime\prime} \biggl( x+\frac{1}{2} \biggr) < 2\frac{2{,}560{,}000x^{8}+512{,}000x^{6}+694{,}400x^{4}+54{,}880x^{2}+2{,}401}{x^{3} ( 40x^{2}+7 ) ^{4}}. \end{aligned}$$
Acknowledgements
This paper was supported by the Natural Science Foundation of China under Grant 11371050. The authors would like to thank the referees for their valuable comments and suggestions which essentially improved the quality of this paper.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.