Abstract
We present the necessary and sufficient conditions such that the functions involving \(R ( x ) =\psi ( x+1/2 ) -\ln x\) with a parameter are completely monotonic on \(( 0,\infty )\), find three new sequences which are fast convergence toward the Euler-Mascheroni constant, and give a positive answer to the conjecture proposed by Chen (J. Math. Inequal. 3(1):79-91, 2009), where ψ is the digamma function.
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1 Introduction
A real-valued function f is said to be completely monotonic on the interval I if f has derivatives of all orders on I and satisfies
for all \(x\in I\) and \(n=0,1,2,\ldots \) . f is said to be strictly completely monotonic on I if inequality (1.1) is strict.
It is well known that f is completely monotonic on \((0,\infty)\) if and only if
where μ is a nonnegative measure on \([0,\infty)\) such that the integral is convergent for all \(x>0\) (see [1], p.161).
Let \(x>0\), then the classical Euler gamma function Γ and psi (digamma) function ψ are, respectively, defined by
The derivatives \(\psi^{\prime}, \psi^{\prime\prime}, \psi ^{\prime \prime\prime},\ldots\) are known as polygamma functions. Recently, the gamma and polygamma functions have attracted the attention of many researchers since they play important roles in many branches, such as mathematical physics, probability, statistics, and engineering.
Let \(H_{n}=\sum_{k=1}^{n}\frac{1}{k}\) be the harmonic number and \(D_{n}=H_{n}-\ln n\). Then the well-known Euler-Mascheroni constant \(\gamma=0.577215664\ldots\) can be expressed as \(\gamma=H_{n}-\psi(n+1)\) or \(\gamma=\lim_{n\rightarrow\infty}D_{n}\), and the double inequality
holds for all \(n\in\mathbb{N}\) (see [2, 3]). Therefore, the convergence rate of \(D_{n}\) is very slowly. Recently, many results involving the quicker convergence toward the Euler-Mascheroni constant can be found in the literature [4–27].
In 1993, DeTemple [7] introduced the DeTemple sequence
and found that it satisfies the double inequalities
and
for all \(n\in\mathbb{N}\).
Villarino ([14], Theorem 1.7) proved that the double inequality
holds for all \(n\in\mathbb{N}\) with the best possible constants \(21/5\) and \(1/(1-\ln3+\ln2-\gamma)-54=3.739\ldots\) .
In [18], Chen proved that the double inequality
holds for all \(n\in\mathbb{N}\) with the best possible constants
and \(1/2\).
Mortici ([28], Theorem 2.1) presented the bounds for \(R_{n}-\gamma\) as follows:
In [29–31], the authors established the inequality
which is equivalent to
Karatsuba [32] proved that the sequence
is strictly increasing with respect to all \(n\in\mathbb{N}\).
In [33], the authors pointed out that \((1+1/n)^{2}H(n)\) is a strictly decreasing and convex sequence by use of computer experiments. Chen ([15], Theorem 2) proved that both \(H(n)\) and \(( ( n+1/2 ) /n)^{2}H(n)\) are strictly increasing and concave sequences, while \(( (n+1)/n ) ^{2}H(n)\) is a strictly decreasing and convex sequence, and conjectured that:
-
(i)
The two functions \(H(x)=[\psi(x+1)-\ln(x+1/2)]x^{2}\) and \([ (x+1/2 )/x]^{2}H(x)\) are so-called Bernstein functions on \((0,\infty)\). That is,
$$\begin{aligned}& H(x) > 0,\quad ( -1 ) ^{n} \bigl[ H(x) \bigr] ^{ ( n+1 ) }>0, \\& \bigl( ( x+1/2 ) /x\bigr)^{2}H(x) > 0,\quad ( -1 ) ^{n} \bigl[ \bigl( ( x+1/2 ) /x\bigr)^{2}H(x) \bigr] ^{ ( n+1 ) }>0 \end{aligned}$$for \(x>0\) and \(n\in\mathbb{N}\).
-
(ii)
The function \(( (x+1)/x ) ^{2}H(x)\) is strictly completely monotonic on \((0,\infty)\).
It is not difficult to verify that
Therefore, the function \(H(x)\) is not a Bernstein function on \((0,\infty)\).
The main purpose of this paper is to give a positive answer to the conjecture (ii) and present several necessary and sufficient conditions such that the functions involving
with a parameter are strictly completely monotone on \(( 0,\infty )\).
2 Lemmas
In order to prove our results we need several lemmas, which we present in this section.
Lemma 1
Let \(R(x)\) be defined by (1.10) and \(Q(t)\) be defined on \((0, \infty)\) by
Then the following identities are valid:
Proof
Making use of the integral representations [34], p.259
we get
Integration by parts leads to
where the last equality holds due to \(\lim_{t\rightarrow\infty} ( e^{-xt}Q ( t ) ) =\lim_{t\rightarrow0} ( e^{-xt}Q ( t ) ) =0\).
Integration by parts again together with
leads to
Similarly, we have
due to \(\lim_{t\rightarrow\infty} ( e^{-xt}Q^{\prime\prime} ( t ) ) =\lim_{t\rightarrow0} ( e^{-xt}Q^{\prime\prime } ( t ) ) =0\), and
due to
□
Lemma 2
([35], Lemma 7)
Let \(P ( t ) \) be a power series which is convergent on \(( 0,\infty ) \) defined by
where \(a_{i}\geq0\) and \(a_{j}\geq0\) for \(i\geq m+1\) and \(0\leq j\leq m-1\), \(a_{m}>0\), and \(\sum_{i=m+1}^{\infty}a_{i}>0\). Then there exists \(t_{0}\in ( 0,\infty ) \) such that \(P (t_{0} ) =0\), \(P ( t ) <0\) for \(t\in ( 0,t_{0} ) \) and \(P ( t ) >0\) for \(t\in ( t_{0},\infty )\).
Lemma 3
Let \(Q(t)\) be defined by (2.1). Then \(Q^{\prime } ( t ) \geq c_{0}Q ( t ) \) for \(t>0\), where
Proof
Simple computations lead to
where
Using the ‘product into sum’ formulas and Taylor expansion we get
where
It is not difficult to verify that \(u_{3}=0\), \(u_{n}<0\) for \(4\leq n\leq10\) and \(u_{11}=1\text{,}636\text{,}643\text{,}754\text{,}240>0\). Note that
for \(n\geq11\). Therefore, \(u_{n}\geq0\) for \(n\geq11\).
From Lemma 2 we clearly see that there exists \(t_{0}\in(0, \infty)\) such that \(Q^{\prime}(t)/Q(t)\) is strictly decreasing on \((0, t_{0}) \) and strictly increasing on \((t_{0}, \infty)\). Therefore, Lemma 3 follows from the piecewise monotonicity of \(Q^{\prime}/Q\) and the numerical computations results \(t_{0}=15.4015\ldots\) and \(Q^{\prime}(t_{0})/Q(t_{0})=-0.06187\ldots\) . □
Lemma 4
The inequalities
hold for \(t>0\).
Proof
Inequality (2.8) can be found in [36], Theorem 18.
To prove (2.9), it suffices to show that for \(t>0\),
Simple computations lead to
which implies that \(p_{1} ( t ) >p_{1} ( 0 ) =0\).
Similarly, inequality (2.10) is equivalent to
Differentiating \(p_{2}(t)\) yields
where \(x=\cosh t>1\). Therefore, \(p_{2} ( t ) < p_{2} ( 0 ) =0\). □
Lemma 5
Let \(Q(t)\) be defined by (2.1). Then the inequality
holds for all \(t>0\).
Proof
Simple computations lead to
Making use of inequality (2.8) we get
□
Lemma 6
Let \(Q(t)\) be defined by (2.1). Then
for all \(t>0\).
Proof
Simple computations lead to
Let
Then it suffices to prove that \(U ( ( \sinh t ) /t ) >0\) for \(t>0\).
It follows from \(U^{\prime} ( y ) =31\sinh^{4}t-2\text{,}520y^{4}\) that U is strictly increasing with respect to y on \((1, \sqrt [4]{31/2\text{,}520}\sinh t ]\) and strictly decreasing with respect to y on \([\sqrt [4]{31/2\text{,}520}\sinh t,\infty)\). We divide the proof into two cases.
Case 1: \(t\in ( \sqrt[4]{2\text{,}520/31},\infty ) \). Then inequality (2.8) leads to
that is,
and so
Let \(\cosh t=x\), then \(\sinh^{2}t=x^{2}-1\), and
where
It is not difficult to verify that \(U_{1} ( x ) >U_{1} ( 1 ) =7\text{,}290\text{,} 000>0\), which implies that \(U ( ( \sinh t ) /t ) >0 \) for \(t\in ( \sqrt[4]{2\text{,}520/31},\infty ) \).
Case 2: \(t\in(0,\sqrt[4]{2\text{,}520/31}]\). Then it follows from (2.10) and the piecewise monotonicity of U that
Let \(\cosh t=x\), then
where
□
Lemma 7
Let \(Q(t)\) be defined by (2.1). Then
for all \(t>0\).
Proof
It follows from (2.1), (2.11), and (2.12) that
Let \(\cosh t=x>1\), then (2.9) leads to
where the last inequality holds due to
□
3 Main results
Theorem 1
Let \(R(x)\) be defined on \(( 0,\infty ) \) by (1.10). Then the function
is strictly completely monotonic on \((0,\infty)\) if \(a\geq a_{0}=\sqrt{ c_{0}^{2}+7/40}-c_{0}=0.4847\ldots\) , where \(c_{0}=-0.06187\ldots\) is given by (2.7).
Proof
It follows from (2.2)-(2.4) that
We clearly see that \(Q ( t ) >0\) for \(t>0\) and Lemmas 3 and 5 imply that
if \(a\geq\sqrt{c_{0}^{2}+7/40}-c_{0}\). □
Taking \(a=1/2\) and replacing x by \(( x+1/2 ) \) in Theorem 1, we have the following.
Corollary 1
The function \(( (x+1)/x ) ^{2}H(x)\) is strictly completely monotonic on \((-1/2,\infty)\).
Remark 1
Corollary 1 gives a positive answer to the conjecture (ii) posed by Chen in [15].
Theorem 2
Let \(R(x)\) be defined on \(( 0,\infty ) \) by (1.10). Then the function
is strictly completely monotonic on \(( 0,\infty ) \) if and only if \(a\geq a_{1}=7/40\).
Proof
The necessity follows from
It follows from (2.2) and (2.4) that
From Lemma 5 we clearly see that
for \(t>0\) if \(a\geq7/40\). □
Note that
Therefore, we have the following.
Corollary 2
Let \(R_{n}\) be defined by (1.3). Then the double inequality
holds for \(n\in\mathbb{N}\) with the best possible constants \(\lambda _{1}=F_{7/40} ( 3/2 ) =0.00797\ldots\) .
Theorem 3
Let \(R(x)\) be defined on \(( 0,\infty )\) by (1.10). Then the function
is strictly completely monotonic on \(( 0,\infty ) \) if and only if \(a\leq a_{2}=-31/336\).
Proof
The necessity can be deduced by
It follows from (2.2), (2.4), and (2.6) that
From Lemma 6 we clearly see that
if \(a\leq a_{2}=-31/336\). □
Making use of the monotonicity of \(f_{a_{2}}\) and the facts that
we have the following.
Corollary 3
Let \(R_{n}\) be defined by (1.3). Then the double inequality
holds for \(n\in\mathbb{N}\) with the best possible constant \(\lambda _{2}=f_{a_{2}} ( 3/2 ) =0.009063\ldots\) .
Remark 2
The upper bound for \(R_{n}-\gamma\) given in (3.2) is better than that given in (1.5). Indeed, simple computations show that
for all \(n\in\mathbb{N}\).
Theorem 4
Let \(R(x)\) be defined on \(( 0,\infty ) \) by (1.10). Then the function
is strictly completely monotonic on \(( 0,\infty ) \) if and only if \(a\geq a_{3}=11\text{,}165/8\text{,}284\).
Proof
The necessity can be derived from
It follows from (2.2), (2.4), and (2.6) that
From Lemmas 5 and 7 we clearly see that
if \(a\geq a_{3}=11\text{,}165/8\text{,}284\). □
The monotonicity of \(G_{a_{3}}\) and the facts that
lead to the following.
Corollary 4
Let \(R_{n}\) be defined by (1.3). Then the double inequality
holds for all \(n\in\mathbb{N}\) with the best possible constant \(\lambda _{3}=G_{a_{3}} ( 3/2 ) =0.001690\ldots\) .
4 Remarks
Remark 3
The function \(G_{a}\) defined by (3.3) can be rewritten as
Theorem 4 leads to the conclusion that
Moreover, we can prove that
It suffices to prove the function
is increasing on \(( 0,\infty ) \). Differentiation gives
From \(\psi^{\prime} ( x+1 ) -\psi^{\prime} ( x ) =-1/x^{2}\) we get
where
for \(x>0\). Therefore,
for all \(x>0\).
In addition, (4.1) implies that the necessary condition such that the function \(-G_{a}\) is completely monotone on \(( 0,\infty ) \) is
Motivated by inequalities (4.2) and (4.3) we propose two conjectures.
Conjecture 1
Let \(R(x)\) be defined on \(( 0,\infty ) \) by (1.10). Then we conjecture that
-
(i)
the function
$$ x\mapsto\frac{-24 ( x^{4}-\frac{31}{336} ) R ( x ) +x^{2}-\frac{7}{40}}{24 ( x^{2}+\frac{7}{40} ) R ( x ) -1} $$is increasing on \(( 0,\infty ) \);
-
(ii)
the function \(-G_{a}\) is completely monotone on \(( 0,\infty ) \) if and only if \(a\leq155/294\).
Remark 4
The monotonicity of the function V proved in Remark 3 and the facts that
lead to the conclusion that the double inequality
holds with the best possible constant \(\lambda_{4}=-0.00003238\ldots\) .
The upper bound for \(R_{n}-\gamma\) in (4.4) is better than that in (3.2) because of
Remark 5
Let
Then Theorems 3 and 4 together with Remark 4 lead to
and simple computations show that
Lastly, inspired by Theorems 2-4, we propose an open problem as follows.
Problem 1
We wonder what the sequences \(\{a_{k}\}\) and \(\{b_{k}\}\) are such that the function
is completely monotone on \(( 0,\infty ) \) and
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Acknowledgements
This research was supported by the Natural Science Foundation of China under Grants 11401191 and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.
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Yang, ZH., Chu, YM. & Zhang, XH. Necessary and sufficient conditions for functions involving the psi function to be completely monotonic. J Inequal Appl 2015, 157 (2015). https://doi.org/10.1186/s13660-015-0674-8
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DOI: https://doi.org/10.1186/s13660-015-0674-8