A Double Inequality for the Trigamma Function and Its Applications

Abstract

We prove that 𝑝=1 and 𝑞=2 are the best possible parameters in the interval (0,∞) such that the double inequality (𝑒^{𝑝/(𝑥+1)}−𝑒^−𝑝/𝑥)/2𝑝<𝜓^′(𝑥+1)<(𝑒^{𝑞/(𝑥+1)}−𝑒^−𝑞/𝑥)/2𝑞 holds for 𝑥>0. As applications, some new approximation algorithms for the circumference ratio 𝜋 and Catalan constant 𝐺=∑^∞_𝑛=0((−1)^𝑛/(2𝑛+1)²) are given. Here, 𝜓^′ is the trigamma function.

1. Introduction

For real and positive values of 𝑥, the classical Euler’s gamma function Γ and its logarithmic derivative 𝜓, the so-called psi function, are defined as Γ(𝑥)=^∞₀𝑡^𝑥−1𝑒^−𝑡(𝑥)=Γ𝑑𝑡,𝜓^′(𝑥)Γ(𝑥). (1) For extension of these functions to complex variables and for basic properties, see [1]. The derivatives 𝜓^′,𝜓^′′,𝜓^′′′,… are known as polygamma functions (see [2]). In particular, 𝜓^′ is called trigamma function.

Recently, the bounds for the trigamma function using exponential functions have attracted the attention of many researchers. For example, Elezović et al. [3] proved that the inequality 𝜓^′(𝑥)<𝑒^{−𝜓(𝑥)} (2) holds for all 𝑥>0. In [4, Theorem 2.7], Batir proved that 𝑎^∗=1/2 and 𝑏^∗=𝜋²𝑒^−2𝛾/6 are the best possible constants such that the double inequality 𝑥+𝑎^∗𝑒^{−2𝜓(𝑥+1)}<𝜓^′(𝑥+1)<𝑥+𝑏^∗𝑒^{−2𝜓(𝑥+1)} (3) holds for all 𝑥>0, where 𝛾 is Euler’s constant. Batir [4] also showed that 122𝑥²−1+𝑒^2/(𝑥+1)−𝑒^{−2𝜓(𝑥+1)}<𝜓^′()<1𝑥+122𝑥²+1−𝑒^−2/𝑥+𝑒^{−2𝜓(𝑥+1)} (4) for all 𝑥>0. In [5, (1.11)], Guo and Qi established that 𝜗𝑒^𝜗/𝑥𝑥²𝑒^𝜗/𝑥−1<𝜓^′(𝑥)<𝜃𝑒^𝜃/𝑥𝑥²𝑒^𝜃/𝑥−1 (5) if 𝑥>0 and 0<𝜗≤1, 𝜃≥2. They [6, Lemma 2] found a very simple upper bound for trigamma function in terms of exponential function as follows: 𝜓^′(𝑥)<𝑒^1/𝑥−1 (6) for all 𝑥>0. The inequality (6) was generalized in [7, Theorem 3.1], [8, Theorem 1.1], and [9, Theorem 1.1] to a complete monotonicity which reads that the difference 𝑒^1/𝑥−𝜓^′(𝑥) is completely monotonic on (0,∞). Many other new results involving the psi and trigamma functions can be found in the literature [10, 11].

Suppose that 𝑚∈(0,∞) and 𝑔 and 𝑔_𝑚 are the real functions defined on (0,∞). 𝑚=𝜆 is said to be the best possible constant in (0,∞) such that the inequality 𝑔(𝑥) >(<) 𝑔_𝑚(𝑥) holds for all 𝑥>0 if 𝑔_𝜆(𝑥) ≥(≤) 𝑔_𝜇(𝑥) on (0,∞), or 𝑔_𝜆(𝑥) and 𝑔_𝜇(𝑥) are not comparable on (0,∞), and lim_𝑥→∞(𝑔(𝑥)−𝑔_𝜆(𝑥))/(𝑔(𝑥)−𝑔_𝜇(𝑥))=0 for any 𝜇∈(0,∞) satisfies 𝑔(𝑥) >(<) 𝑔_𝜇(𝑥) on the interval (0,∞).

The main purpose of this paper is to find the best possible constants 𝑝,𝑞∈(0,∞) such that the double inequality 𝜃𝑥,𝑝<𝜓^′(𝑥+1)<𝜃𝑥,𝑞 (7) or equivalently 1𝑥²+𝜃𝑥,𝑝<𝜓^′(𝑥)<1𝑥²+𝜃𝑥,𝑞 (8) holds for all 𝑥>0, where 𝜃()=𝑒𝑥,𝑚^{𝑚/(𝑥+1)}−𝑒^−𝑚/𝑥2𝑚,𝑚>0. (9) Our main result is the following Theorem 1.

Theorem 1. 𝑝=1 and 𝑞=2 are the best possible constants in the interval (0,∞) such that the double inequality (7) or (8) holds for all 𝑥>0.

From Theorem 1, we clearly see the following.

Corollary 2. The double inequality 1sinh𝑥+1<𝜓^′()<1𝑥+122sinh𝑥 (10) holds for all 𝑥>0.

2. Lemmas

Lemma 3. Let the function 𝜃 be defined on (0,∞)² by (9). Then the function 𝜃 is strictly decreasing with respect to 𝑚 on (0,1] and strictly increasing on [3/2,+∞).

Proof. It follows from (9) that 𝑚²𝑥2()𝑒𝑚+𝑥^𝑚/𝑥𝜕𝜃=𝑥()𝜕𝑚1−𝑥−𝑚+1()𝑚𝑥+1)(𝑚+𝑥exp𝑥+𝑚𝑥+1=∶ℎ_𝑚(𝑥),𝜕ℎ_𝑚=𝑚𝜕𝑥𝑥²exp((𝑚/𝑥)((2𝑥+1)/(𝑥+1)))(𝑥+1)³(𝑚+𝑥)²×(3−2𝑚)𝑥²+(3−2𝑚)𝑥+(1−𝑚). (11) If 𝑚∈(0,1], then 𝜕ℎ_𝑚/𝜕𝑥>0; that is, ℎ_𝑚 is strictly increasing with respect to 𝑥>0. Therefore, ℎ_𝑚(𝑥)<lim_𝑥→∞ℎ_𝑚(𝑥)=0, (12) which implies that 𝜕𝜃/𝜕𝑚<0.
If 𝑚≥3/2, then 𝜕ℎ_𝑚/𝜕𝑥<0; that is, ℎ_𝑚 is strictly decreasing with respect to 𝑥>0, which leads to the conclusion that ℎ_𝑚(𝑥)>lim_𝑥→∞ℎ_𝑚(𝑥)=0. (13)

Lemma 4. Let the function 𝜃 be defined on (0,∞)² by (9) and 𝐹_𝑚(𝑥)=𝜓^′(𝑥+1)−𝜃(𝑥,𝑚). (14) Then the equation 𝐹_𝑚0⁺=𝜓^′(1)−𝑒^𝑚2𝑚=0 (15) has two roots 𝑚₁=0.5023⋯,𝑚₂=1.7510⋯ (16) such that 𝐹_𝑚(0⁺)>0 for 𝑚∈(𝑚₁,𝑚₂) and 𝐹_𝑚(0⁺)<0 for 𝑚∈(0,𝑚₁)∪(𝑚₂,∞).

Proof. Differentiation yields 𝑑𝐹𝑑𝑚^𝑚0⁺1=−2𝑚²𝑒^𝑚(𝑚−1), (17) which reveals that the function 𝑚↦𝐹_𝑚(0⁺) is strictly increasing on (0,1) and strictly decreasing on (1,∞). Therefore, Lemma 4 follows from the piecewise monotonicity of the function 𝑚↦𝐹_𝑚(0⁺) and the numerical computations results: 𝐹_0.50230⁺=16𝜋²−𝑒^0.50231.0046=−1.685⋯×10⁻⁵𝐹<0,_0.50240⁺=16𝜋²−𝑒^0.50241.0048=1.46⋯×10⁻⁴𝐹>0,_1.7510⁺=16𝜋²−𝑒^1.7513.502=5.68⋯×10⁻⁵𝐹>0,_1.75110⁺=16𝜋²−𝑒^1.75113.5022=−1.374⋯×10⁻⁵<0. (18)

Lemma 5. Let 𝑚≥0, 𝑚₁=0.5023⋯, and 𝑚₂=1.7510⋯ and let 𝐹_𝑚(𝑥) be defined as in Lemma 4. Then the following statements are true:(i)if the inequality 𝐹_𝑚(𝑥)≥0 holds for all 𝑥>0, then 𝑚∈[1,𝑚₂];(ii)if the inequality 𝐹_𝑚(𝑥)≤0 holds for all 𝑥>0, then 𝑚∈(0,𝑚₁]∪[2,∞).

Proof. It follows from the series formulas that 𝜓^′(𝑥)=1𝑥+12𝑥²+16𝑥³−130𝑥⁵+⋯,𝑒^{𝑚/(𝑥+1)}𝑚=1++1𝑥+12𝑚𝑥+1²+16𝑚𝑥+1³+1𝑚24𝑥+1⁵+⋯,𝑒^−𝑚/𝑥𝑚=1−𝑥+12−𝑚𝑥²+16−𝑚𝑥³+1−𝑚24𝑥⁵+⋯, (19)(20)(21) and we get lim_𝑥→∞𝐹_𝑚(𝑥)𝑥⁻³=lim_𝑥→∞1()+1𝑥+12(𝑥+1)²+16(𝑥+1)³−𝑚1++1𝑥+12𝑚𝑥+1²+16𝑚𝑥+1³−𝑚1−𝑥+12−𝑚𝑥²+16−𝑚𝑥³×(2𝑚)⁻¹×𝑥⁻³⁻¹1=−12lim_𝑥→∞2𝑚²−6𝑚+4𝑥³+3𝑚²−9𝑚+6𝑥²+3𝑚²−3𝑚𝑥+𝑚²×(𝑥+1)³⁻¹1=−122𝑚²1−6𝑚+4=−6(𝑚−1)(𝑚−2). (22)(i)If inequality 𝐹_𝑚(𝑥)≥0 holds for all 𝑥>0, then, from 𝐹_𝑚0⁺=𝜓^′(1)−𝑒^𝑚2𝑚≥0,lim_𝑥→∞𝐹_𝑚(𝑥)𝑥⁻³1=−6(𝑚−1)(𝑚−2)≥0 (23)and Lemma 4, we clearly see that 𝑚∈[𝑚₁,𝑚₂]∩[1,2]=[1,𝑚₂]. (ii)If inequality 𝐹_𝑚(𝑥)≤0 holds for all 𝑥>0, then 𝐹_𝑚0⁺=𝜓^′(1)−𝑒^𝑚2𝑚≤0,lim_𝑥→∞𝐹_𝑚(𝑥)𝑥⁻³1=−6(𝑚−1)(𝑚−2)≤0 (24)and Lemma 4 lead to the conclusion that 𝑚∈0,𝑚₁∪𝑚₂∩]∪[=,∞((0,12,∞))0,𝑚₁∪[2,∞). (25)

Lemma 6. Let the function 𝜃 be defined on (0,∞)² by (9). Then 𝜃(𝑥,1) and 𝜃(𝑥,𝑚) are not comparable for all 𝑥>0 if 𝑚∈(1,2).

Proof. For 𝑥>0 and 𝑚>0, let 𝐺_1,𝑚(𝑥)()()=𝑒=𝜃𝑥,1−𝜃𝑥,𝑚_1/(𝑥+1)−𝑒^−1/𝑥2−𝑒^{𝑚/(𝑥+1)}−𝑒^−𝑚/𝑥.2𝑚 (26) Then simple computation leads to 𝐺_1,𝑚0⁺=𝑒2−𝑒^𝑚.2𝑚 (27) From (20) and (21), we have lim_𝑥→∞𝐺_1,𝑚(𝑥)𝑥⁻³1=−6(𝑚−1)(𝑚−2). (28) Differentiation yields 𝑑𝐺𝑑𝑚^1,𝑚0⁺𝑒=−^𝑚2𝑚²(𝑚−1), (29) which shows that 𝑚↦𝐺_1,𝑚(0⁺) is strictly decreasing on (1,∞). Therefore, 𝐺_1,𝑚0⁺<𝐺_1,10⁺=0 (30) if 𝑚∈(1,2). On the other hand, we clearly see that lim_𝑥→∞[𝑥³𝐺_1,𝑚(𝑥)]>0 for 𝑚∈(1,2).

Lemma 7. Let 𝑛∈ℕ={1,2,3,…} and 𝑚∈ℕ∪{0} with 𝑛>𝑚 and let 𝑃_𝑛(𝑡) be the polynomial of degree 𝑛 defined by 𝑃_𝑛(𝑡)=^𝑛_𝑖=𝑚+1𝑎_𝑖𝑡^𝑖−^𝑚_𝑖=0𝑎_𝑖𝑡^𝑖, (31) where 𝑎_𝑛,𝑎_𝑚>0 and 𝑎_𝑖≥0 for 0≤𝑖≤𝑛−1 with 𝑖≠𝑚. Then, there exists 𝑡_𝑚+1∈(0,∞) such that 𝑃_𝑛(𝑡_𝑚+1)=0 and 𝑃_𝑛(𝑡)<0 for 𝑡∈(0,𝑡_𝑚+1) and 𝑃_𝑛(𝑡)>0 for 𝑡∈(𝑡_𝑚+1,∞).

Proof. Differentiating 𝑃_𝑛(𝑡) gives 𝑃_𝑛^(𝑘)(𝑡)=^𝑛_𝑖=𝑚+1𝑖!()!𝑎𝑖−𝑘^𝑖𝑡^𝑖−𝑘−^𝑚_𝑖=𝑘𝑖!()!𝑎𝑖−𝑘^𝑖𝑡^𝑖−𝑘𝑃for1≤𝑘≤𝑚,_𝑛(𝑚+1)(𝑡)=^𝑛_𝑖=𝑚+1𝑖!()!𝑎𝑖−𝑚−1^𝑖𝑡^{𝑖−𝑚−1}>0. (32) Note that 𝑃_𝑛^(𝑘)(∞)=∞,𝑃_𝑛^(𝑘)0⁺=−𝑎_𝑘 (33) for 1≤𝑘≤𝑚.
From 𝑃_𝑛^(𝑚+1)(𝑡)>0, we clearly see that 𝑃_𝑛^(𝑚)(𝑡) is strictly increasing on (0,∞). Then 𝑃_𝑛^(𝑚)(∞)>0 and 𝑃_𝑛^(𝑚)(0⁺)=−𝑎_𝑚<0 lead to the conclusion that there exists 𝑡₁∈(0,∞) such that 𝑃_𝑛^(𝑚)(𝑡₁)=0 and 𝑃_𝑛^(𝑚)(𝑡)<0 for 𝑡∈(0,𝑡₁) and 𝑃_𝑛^(𝑚)(𝑡)>0 for 𝑡∈(𝑡₁,∞). Therefore, 𝑃_𝑛^(𝑚−1)(𝑡) is strictly decreasing on (0,𝑡₁) and strictly increasing on (𝑡₁,∞).
It follows from the piecewise monotonicity of 𝑃_𝑛^(𝑚−1)(𝑡) and 𝑃_𝑛^(𝑚−1)(∞)=∞ that 𝑃_𝑛^(𝑚−1)(𝑡)<𝑃_𝑛^(𝑚−1)(0⁺)=−𝑎_𝑚−1≤0 for 𝑡∈(0,𝑡₁), and there exists 𝑡₂∈(𝑡₁,∞) such that 𝑃_𝑛^(𝑚−1)(𝑡₂)=0 and 𝑃_𝑛^(𝑚−1)(𝑡)<0 for 𝑡∈(0,𝑡₂) and 𝑃_𝑛^(𝑚−1)(𝑡)>0 for 𝑡∈(𝑡₂,∞). Therefore, 𝑃_𝑛^(𝑚−2)(𝑡) is strictly decreasing on (0,𝑡₂) and strictly increasing on (𝑡₂,∞).
After repeating the same steps as above 𝑚+1 times, we deduce that there exists 𝑡_𝑚+1∈(𝑡_𝑚,∞)⊂(0,∞) such that 𝑃_𝑛(𝑡_𝑚+1)=0 and 𝑃_𝑛(𝑡)<0 for 𝑡∈(0,𝑡_𝑚+1) and 𝑃_𝑛(𝑡)>0 for 𝑡∈(𝑡_𝑚+1,∞).

Lemma 8. Let the function 𝜃 be defined on (0,∞)² by (9). Then there exists 𝑚₀∈(2/5,9/20) such that 𝜃(𝑥,2) and 𝜃(𝑥,𝑚) are not comparable for all 𝑥>0 if 𝑚∈(𝑚₀,1), and 𝜃(𝑥,2)<𝜃(𝑥,𝑚) for all 𝑥>0 if 𝑚∈(0,𝑚₀].

Proof. For 𝑥>0 and 𝑚>0, let 𝐺_2,𝑚(𝑥)()()=𝑒=𝜃𝑥,2−𝜃𝑥,𝑚_2/(𝑥+1)−𝑒^−2/𝑥4−𝑒^{𝑚/(𝑥+1)}−𝑒^−𝑚/𝑥.2𝑚 (34) Then simple computation leads to 𝐺_2,𝑚0⁺=𝑒²4−𝑒^𝑚.2𝑚 (35)
(i) We prove that there exists 𝑚₀∈(2/5,9/20) such that 𝜃(𝑥,2) and 𝜃(𝑥,𝑚) are not comparable for all 𝑥>0 if 𝑚∈(𝑚₀,1). For this end, it suffices to prove that there exists 𝑚₀∈(2/5,9/20) such that 𝐺_2,𝑚(0⁺)>0 and lim_𝑥→∞𝑥³𝐺_2,𝑚(𝑥)<0 if 𝑚∈(𝑚₀,1).
Indeed, it follows from 𝑑𝐺𝑑𝑚^2,𝑚0⁺𝑒=−^𝑚2𝑚²(𝑚−1) (36) that the function 𝑚↦𝐺_2,𝑚(0⁺) is strictly increasing on (0,1). Numerical computations show that 𝐺_2,2/50⁺=14𝑒²−54𝑒^2/5𝐺<0,_2,9/200⁺=14𝑒²−109𝑒^9/20>0. (37)
Therefore, there exists 𝑚₀∈(2/5,9/20) such that 𝐺_2,𝑚0⁺=𝑒²4−𝑒^𝑚2𝑚=0 (38) and 𝐺_2,𝑚(0⁺)<0 for 𝑚∈(0,𝑚₀) and 𝐺_2,𝑚(0⁺)>0 for 𝑚∈(𝑚₀,1).
On the other hand, it follows from 𝑚∈(𝑚₀,1) together with (20) and (21) that lim_𝑥→∞𝐺_2,𝑚(𝑥)𝑥⁻³1=−6(𝑚−1)(𝑚−2)<0. (39)
(ii)  We prove that 𝜃(𝑥,2)<𝜃(𝑥,𝑚) for all 𝑥>0 if 𝑚∈(0,𝑚₀]. From Lemma 3, we know that 𝑚↦𝜃(𝑥,𝑚) is strictly decreasing on (0,1], so it suffices to prove that 𝜃(𝑥,2)<𝜃(𝑥,𝑚₀) for all 𝑥>0.
Let 𝑔₁(𝑥)=𝑒^𝑚0/𝑥𝐺_2,𝑚0(𝑥)=14𝑚exp⁰𝑥+2−1𝑥+14𝑚exp⁰𝑥−2𝑥+12𝑚₀𝑚1−exp⁰𝑥+𝑚₀.𝑥+1 (40) Then lim_𝑥→∞𝑔₁(𝑥)=0. (41) Differentiation yields 42−𝑚₀𝑥²𝑒^{(2−𝑚0)/𝑥}×𝑔^′₁(𝑥)=22𝑥²+2𝑥+12−𝑚₀(𝑥+1)²2exp𝑥+𝑚₀−𝑚𝑥+1−1₀+2𝑥²+2𝑚₀𝑥+𝑚₀2−𝑚₀(𝑥+1)²2exp𝑥+2𝑥+1∶=𝑔₂(𝑥),𝑔^′₂(𝑥)=22𝑚₀+2𝑥⁴+2𝑚₀+10𝑥³+𝑚₀+14𝑥²×+8𝑥+22−𝑚₀𝑥²(𝑥+1)⁴⁻¹𝑚×exp⁰+2𝑥+1𝑥𝑔₃(𝑥), (42)(43) where 𝑔₃(𝑥)=2𝑚₀+2𝑥⁴+6𝑚₀+2𝑥³+7𝑚₀+2𝑥²+4𝑚₀𝑥+𝑚₀×2𝑚₀+2𝑥⁴+2𝑚₀+10𝑥³+𝑚₀+14𝑥²+8𝑥+2⁻¹×𝑒^{(2−𝑚0)/(𝑥+1)}−1. (44) Note that lim_𝑥→∞𝑔₂(𝑥)=0,lim_𝑥→0+𝑔₃(𝑥)=12𝑚₀𝑒^2−𝑚0−1=0,lim_𝑥→∞𝑔₃(𝑥)=0, (45)(46)(47) where the second equation in (46) follows from (38).
Differentiating 𝑔₃(𝑥) leads to 𝑔^′₃(𝑥)=2−𝑚₀𝑒^{(2−𝑚0)/(𝑥+1)}×(𝑥+1)²8𝑥+𝑚₀𝑥²+2𝑚₀𝑥³+2𝑚₀𝑥⁴+14𝑥²+10𝑥³+2𝑥⁴+2²⁻¹×𝑔₄(𝑥), (48) where 𝑔₄(𝑥)=4−4𝑚²₀𝑥⁸+16−16𝑚²₀𝑥⁷+−28𝑚²₀−24𝑚₀+32𝑥⁶+−28𝑚²₀−72𝑚₀+40𝑥⁵−17𝑚²₀+104𝑚₀−42𝑥⁴−6𝑚²₀+88𝑚₀−36𝑥³−𝑚²₀+46𝑚₀−18𝑥²−14𝑚₀−4𝑥−2𝑚₀∶=𝑎₈𝑥⁸+𝑎₇𝑥⁷+𝑎₆𝑥⁶+𝑎₅𝑥⁵−𝑎₄𝑥⁴−𝑎₃𝑥³−𝑎₂𝑥²−𝑎₁𝑥−𝑎₀. (49) We assert that there exists a unique 𝑥^∗₄∈(0,∞) such that 𝑔₄(𝑥)<0 for 𝑥∈(0,𝑥^∗₄) and 𝑔₄(𝑥)>0 for 𝑥∈(𝑥^∗₄,∞), which leads to the conclusion that 𝑔₃(𝑥) is strictly decreasing on (0,𝑥^∗₄] and strictly increasing on [𝑥^∗₄,∞). To this end, it is enough to verify that the coefficients of 𝑔₄(𝑥) satisfy the conditions of Lemma 7. In fact, it follows from 𝑚₀∈(2/5,9/20)∶=(𝑚₀₋,𝑚₀₊) that we have 𝑎₈=4−4𝑚²₀=41−𝑚²₀𝑎>0,₇=16−16𝑚²₀=161−𝑚²₀𝑎>0,₆=−28𝑚²₀−24𝑚₀+32>−28𝑚²₀₊−24𝑚₀₊+32=1553𝑎100>0,₅=−28𝑚²₀−72𝑚₀+40>−28𝑚²₀₊−72𝑚₀₊+40=193𝑎100>0,₄=17𝑚²₀+104𝑚₀−42>17𝑚²₀₋+104𝑚₀₋−42=58𝑎25>0,₃=6𝑚²₀+88𝑚₀−36>6𝑚²₀₋+88𝑚₀₋4−36=𝑎25>0,₂=𝑚²₀+46𝑚₀−18>𝑚²₀₋+46𝑚₀₋−18=14𝑎25>0,₁=14𝑚₀−4>14𝑚₀₋8−4=5𝑎>0,₀=2𝑚₀>0. (50) From the piecewise monotonicity of 𝑔₃(𝑥) together with (46) and (47), we clearly see that 𝑔₃(𝑥)<lim_𝑥→0+𝑔₃(𝑥)=0for𝑥∈0,𝑥^∗₄,𝑔₃(𝑥)<lim_𝑥→∞𝑔₃(𝑥)=0for𝑥∈𝑥^∗₄,∞; (51) that is, 𝑔₃(𝑥)<0 for 𝑥∈(0,∞). Then (43) and (45) lead to the conclusion that 𝑔₂(𝑥)>lim_𝑥→∞𝑔₂(𝑥)=0 for 𝑥∈(0,∞), which implies that 𝑔₁(𝑥) is strictly increasing on (0,∞) and 𝑔₁(𝑥)<lim_𝑥→∞𝑔₁(𝑥)=0 for 𝑥∈(0,∞).
Therefore, 𝜃(𝑥,2)<𝜃(𝑥,𝑚₀) follows easily from (40) and 𝑔₁(𝑥)<0.

Lemma 9 (see [12, pp. 258–260]). Let 𝑥>0 and 𝑛∈{0,1,2,…}. Then 𝜓^(𝑛)(𝑥+1)−𝜓^(𝑛)(𝑥)=(−1)^𝑛𝑛!𝑥^𝑛+1. (52)

From the proof of [4, Theorem 2.6], we get the following.

Lemma 10. The inequality 𝜓^′(𝑥+1)>2𝑥+12𝑥²+2𝑥+2/3 (53) holds for 𝑥>−1.

The following lemma can be derived immediately from the proof of [4, Theorem 2.1].

Lemma 11 (see [4, Theorem 2.1]). Let 𝑦 be the function defined on (0,∞) by 𝑦(𝑥)=𝑒^2𝜓(𝑥). (54) Then 𝑦^′′′(𝑥)>0 for 𝑥∈(0,∞).

The well-known Hermite-Hadamard inequality for convex function can be stated as follows.

Lemma 12 (see [13]). Let 𝐼⊆ℝ be an interval, 𝑎,𝑏∈𝐼 with 𝑎<𝑏, and let 𝑓∶𝐼→ℝ be a convex function. Then 𝑓𝑎+𝑏2≤∫^𝑏_𝑎𝑓(𝑥)𝑑𝑥≤𝑓(𝑎)(𝑏)𝑏−𝑎+𝑓2. (55)

3. Proofs of Theorem 1

Proof of Theorem 1. We divide the proof into four parts.
(I) We prove the first inequality in (7); that is, 𝐹₁(𝑥)=𝜓^′()−𝑒𝑥+1^1/(𝑥+1)−𝑒^−1/𝑥2>0, (56) where 𝐹_𝑚(𝑥) is defined by (14).
It follows from Lemma 10 that 𝐹₁(𝑥)=𝜓^′()−𝑒𝑥+1^1/(𝑥+1)−𝑒^−1/𝑥2>2𝑥+12𝑥²−𝑒+2𝑥+2/3^1/(𝑥+1)−𝑒^−1/𝑥2∶=𝑒^−1/𝑥𝐻(𝑥), (57) where 𝐻(𝑥)=2𝑥+12𝑥²𝑒+2𝑥+2/3^1/𝑥−exp((1/(𝑥+1))+(1/𝑥))−12. (58)
We clearly see that it is enough to prove that 𝐻(𝑥)>0 for 𝑥>0.
Differentiating 𝐻(𝑥) gives 𝐻^′(𝑥)=12𝑥²𝑒^1/𝑥×2𝑥²+2𝑥+1(𝑥+1)²𝑒^1/(𝑥+1)−3(𝑥+1)6𝑥³+6𝑥²+4𝑥+13𝑥²+3𝑥+1²=12𝑥²𝑒^1/𝑥𝐿×2𝑥²+2𝑥+1(𝑥+1)²𝑒^1/(𝑥+1),3(𝑥+1)6𝑥³+6𝑥²+4𝑥+13𝑥²+3𝑥+1²×ℎ(𝑥), (59) where 𝐿(𝑎,𝑏)=(𝑏−𝑎)/(ln𝑎−ln𝑏) denotes the logarithmic mean of positive numbers 𝑎 and 𝑏, and ℎ(𝑥)=1𝑥+1+ln2𝑥²+2𝑥+1(𝑥+1)²3()−ln𝑥+16𝑥³+6𝑥²+4𝑥+13𝑥²+3𝑥+1². (60) Differentiating ℎ(𝑥) leads to ℎ^′(𝑥)=𝑥(2𝑥+1)7𝑥²×()+7𝑥+3𝑥+1²3𝑥²+3𝑥+12𝑥²×+2𝑥+16𝑥³+6𝑥²+4𝑥+1⁻¹>0 (61) for 𝑥>0, which means that ℎ is strictly increasing on (0,∞) and ℎ(𝑥)<lim_𝑥→∞ℎ(𝑥)=0. It in turn implies that 𝐻 is strictly decreasing on (0,∞). Therefore, 𝐻(𝑥)>lim_𝑥→∞𝐻(𝑥)=0 for 𝑥>0.
(II) We prove that 𝑚=1 is the best possible constant such that 𝜓^′(𝑥+1)>𝜃(𝑥,𝑚) for all 𝑥>0.
From Lemma 5, we know that 𝑚∈[1,𝑚₂] if 𝜓^′(𝑥+1)>𝜃(𝑥,𝑚) for all 𝑥>0, where 𝑚₂=1.7510…. It follows from Lemma 6 that 𝜃(𝑥,1) and 𝜃(𝑥,𝑚) are not comparable for all 𝑥>0 if 𝑚∈(1,2); that is to say, 𝜃(𝑥,𝑚^∗) is not a better lower bound of 𝜓^′(𝑥+1) than 𝜃(𝑥,1) even if there exists 𝑚^∗∈(1,𝑚₂] such that 𝜓^′(𝑥+1)>𝜃(𝑥,𝑚^∗).
For any 𝑚^∗∈(1,𝑚₂], (22) leads to lim_𝑥→∞𝑥³𝐹_𝑚∗(𝑥=1)6𝑚^∗−12−𝑚^∗. (62)
It follows from (19), (20), and (21) that we get lim_𝑥→∞𝑥⁵𝐹₁(𝑥=1).24 (63)
(III) We prove the second inequality (7); that is, 𝐹₂(𝑥)=𝜓^′()−𝑒𝑥+1^2/(𝑥+1)−𝑒^−2/𝑥4<0 (64) for 𝑥>0, where 𝐹_𝑚(𝑥) is defined by (14).
Lemma 11 implies that the function 𝑥↦𝑦^′(𝑥)=2𝜓^′(𝑥)𝑒^2𝜓(𝑥) is strictly convex on (0,∞). Then, making use of Lemma 12, we get 𝑦^′𝑥+𝑥+22<∫_𝑥^𝑥+2𝑦^′(𝑡)𝑑𝑡𝑥+2−𝑥 (65) for 𝑥>0. That is, 2𝜓^′(𝑥+1)𝑒^{2𝜓(𝑥+1)}<𝑒^{2𝜓(𝑥+2)}−𝑒^2𝜓(𝑥)2,𝜓^′()<1𝑥+14𝑒^{2(𝜓(𝑥+2)−𝜓(𝑥+1))}−𝑒^{2(𝜓(𝑥)−𝜓(𝑥+1))}. (66)
Therefore, inequality (64) follows easily from (52) and the last inequality above.
(IV) We prove that 𝑚=2 is the best possible constant such that 𝜓^′(𝑥+1)<𝜃(𝑥,𝑚) for all 𝑥>0.
From Lemma 5, we know that 𝑚∈(0,𝑚₁]∪[2,∞) if 𝜓^′(𝑥+1)<𝜃(𝑥,𝑚) for all 𝑥>0, where 𝑚₁=0.5023⋯.
It follows from Lemma 8 that there exists 𝑚₀∈(2/5,9/20)⊂(0,𝑚₁) such that 𝜃(𝑥,2) and 𝜃(𝑥,𝑚) are not comparable for all 𝑥>0 if 𝑚∈(𝑚₀,𝑚₁] and 𝜃(𝑥,2)<𝜃(𝑥,𝑚) for all 𝑥>0 if 𝑚∈(0,𝑚₀]. Lemma 3 leads to the conclusion that 𝜃(𝑥,2)<𝜃(𝑥,𝑚) for all 𝑥>0 if 𝑚∈(2,∞).
If there exists 𝑚^∗∈(𝑚₀,𝑚₁] such that 𝐹_𝑚∗(𝑥)=𝜓^′(𝑥+1)−𝜃(𝑥,𝑚^∗)<0 for all 𝑥>0, then, from (19), (20), (21), and (22), we get lim_𝑥→∞𝑥³𝐹_𝑚∗(𝑥1)=−61−𝑚^∗2−𝑚^∗,lim_𝑥→∞𝑥⁷𝐹₂(𝑥1)=−.45 (67)

From the above proof and Lemma 3 we get the following.

Corollary 13. Let the function 𝜃 be defined on (0,∞)² by (9) and let 𝑚₀∈(2/5,9/20) be the root of (38) on (0,1). Then the inequalities 𝜓^′(𝑥+1)<𝜃(𝑥,2)<𝜃𝑥,𝑚₀<lim_𝑚→0𝜃()=1𝑥,𝑚2()+1𝑥+12𝑥 (68) or equivalently 𝜓^′(𝑥)<12()+1𝑥+1+12𝑥𝑥² (69) hold for all 𝑥>0.

4. Remarks

Remark 14. It follows from (67) and the facts that lim_𝑥→∞𝜓^′(𝑥)−𝑒^{−𝜓(𝑥)}𝑥⁻³1=−,24lim_𝑥→∞𝜓^′(𝑥+1)−𝑥+𝜋²𝑒^−2𝛾/6𝑒^{−2𝜓(𝑥+1)}𝑥⁻²=12−16𝜋²𝑒^−2𝛾,lim_𝑥→∞𝜓^′(𝑥+1)−(1/2)2/𝑥²+1−𝑒^−2/𝑥+𝑒^{−2𝜓(𝑥+1)}𝑥⁻⁶1=−,60lim_𝑥→∞𝜓^′(𝑥)−𝑒^1/𝑥−1𝑥⁻⁴1=−,24lim_𝑥→∞𝜓^′(𝑥)−2𝑒^2/𝑥/𝑥²𝑒^2/𝑥−1𝑥⁻²1=−2 (70) that we clearly see that the upper bound in Theorem 1 for the trigamma function 𝜓^′ is better than the upper bounds given in (2), (3), (4), (5), and (6) if 𝑥 is large enough.

Lemma 15. One has 𝑒^1/(𝑥+1)+𝑒^−1/𝑥()∪(),𝑒>2𝑖𝑓𝑥∈−∞,−10,∞_1/(𝑥+1)+𝑒^−1/𝑥≥2𝑒²𝑖𝑓𝑥∈(−1,0). (71)

Proof. Differentiation leads to 𝑒^1/(𝑥+1)+𝑒^−1/𝑥′=𝑒^−1/𝑥𝑥²𝑥1−²(𝑥+1)²1exp+1𝑥+1𝑥𝑒∶=^−1/𝑥𝑥²𝑉(𝑥),𝑉^′(𝑥)=exp(1/(𝑥+1)+(1/𝑥))(𝑥+1)⁴>0. (72) If 𝑥∈(0,∞), then, from 𝑉^′(𝑥)>0, we get 𝑉(𝑥)<lim_𝑥→∞𝑉(𝑥)=0, which leads to 𝑒^1/(𝑥+1)+𝑒^−1/𝑥>lim_𝑥→∞𝑒^1/(𝑥+1)+𝑒^−1/𝑥=2. (73) If 𝑥∈(−∞,−1), then the first inequality in (71) still holds by making a change of variable 𝑦=−(𝑥+1). If 𝑥∈(−1,0), since 𝑉(−1/2)=0, we see that 𝑉(𝑥)<𝑉(−1/2)=0 for 𝑥∈(−1,−1/2) and 𝑉(𝑥)>𝑉(−1/2)=0 for 𝑥∈(−1/2,0). Hence, 𝑒^1/(𝑥+1)+𝑒^−1/𝑥≥𝑒^1/(𝑥+1)+𝑒^−1/𝑥_𝑥=−1/2=2𝑒². (74)

Remark 16. Using inequality (71), one has 1𝑥²+𝜃(𝑥,2)−𝑒^1/𝑥=1−1𝑥²+𝑒^2/(𝑥+1)−𝑒^−2/𝑥4−𝑒^1/𝑥<1−1𝑥²+𝑒^2/(𝑥+1)−2−𝑒^1/(𝑥+1)24−𝑒^1/𝑥=1−1𝑥²+𝑒^1/(𝑥+1)−𝑒^1/𝑥=1𝑥²+^∞_𝑛=011𝑛!(𝑥+1)^𝑛−1𝑥^𝑛<1𝑥²+³_𝑛=011𝑛!(𝑥+1)^𝑛−1𝑥^𝑛1=−6𝑥³(𝑥+1)³<0 (75) for 𝑥>0, which shows that the upper bound 𝑥⁻²+𝜃(𝑥,2) in (8) is better than the upper bound (𝑒^1/𝑥−1) in (6).

Remark 17. The conclusion that the difference 𝑒^1/𝑥−𝜓^′(𝑥) is completely monotonic on (0,∞) given in [6, Lemma 2] implies that 𝑒−𝜓^′(1)>𝑒^1/(𝑥+1)−𝜓^′(𝑥+1)>1 (76) or 𝑒^1/(𝑥+1)𝜋−𝑒+²6<𝜓^′(𝑥+1)<𝑒^1/(𝑥+1)1−1<22sinh𝑥. (77) It is easy to check that the lower bound 𝑒^1/(𝑥+1)−𝑒+𝜋²/6 and sinh1/(𝑥+1) for 𝜓^′(𝑥+1) given in (10) are not comparable due to 𝑒^1/(𝑥+1)𝜋−𝑒+²6>(<)1sinh(>)𝑥+1if𝑥<1.62670⋯. (78)

Remark 18. Guo et al. [14] proved that (𝑘−1)!𝑥^𝑘+𝑘!2𝑥^𝑘+1<(−1)^𝑘+1𝜓^(𝑘)(𝑥)<(𝑘−1)!𝑥^𝑘+𝑘!𝑥^𝑘+1 (79) for 𝑥>0 if 𝑘∈ℕ. In particular, if 𝑘=1, one has 1𝑥+12𝑥²<𝜓^′(𝑥)<1𝑥+1𝑥². (80)
We clearly see that the upper bound given in (69) is better than that in (80) for the trigamma function 𝜓^′(𝑥).

Finally, we give remarks on two mathematical constants 𝜋 and 𝐺 (Catalan constant).

Remark 19. It is well known that ^∞_𝑘=11𝑘²=𝜋²6. (81) Let 𝑄_𝑛=∑^𝑛_𝑘=1(1/𝑘²), and then 𝜓^′(𝑛+1)=(𝜋²/6)−𝑄_𝑛. From Theorem 1, we clearly see that the double inequality 𝑒^1/(𝑛+1)−𝑒^−1/𝑛2<𝜋²6−𝑄_𝑛<𝑒^2/(𝑛+1)−𝑒^−2/𝑛4 (82) holds for all 𝑛∈ℕ.
Let 𝑙_𝑛=𝑄_𝑛+𝑒^1/(𝑛+1)−𝑒^−1/𝑛2,𝐿_𝑛=𝑄_𝑛+𝑒^2/(𝑛+1)−𝑒^−2/𝑛4. (83) Then inequalities (82) can be rewritten as 𝑙_𝑛<𝜋²6<𝐿_𝑛 (84) or 6𝑙_𝑛<𝜋<6𝐿_𝑛. (85) It follows from (63) and (67) that lim_𝑛→∞𝑛⁵𝑙_𝑛−𝜋²61=−,24lim_𝑛→∞𝑛⁷𝐿_𝑛−𝜋²6=1.45 (86) Therefore, (85) provides a new approximation algorithm for 𝜋. Numerical simulations results carried out with mathematical software show that the given algorithm is more accurate than 6𝑄_𝑛 (see Table 1).
More estimate methods for 𝜋 can be found in [15–19].

Table 1 

Remark 20. The Catalan constant 𝐺=^∞_𝑛=0(−1)^𝑛(2𝑛+1)²=0.9159655941772190⋯ (87) is a mysterious constant in mathematics and physics. From 𝜓^′(1/4)=𝜋²+8𝐺 and 𝜓^′(𝑛+(1/4)+1)=𝜓^′(1/4)−∑^𝑛_𝑘=0(1/(𝑘+1/4)²) (see [17, 20]) together with Theorem 1, we get 𝑒^1/(𝑛+5/4)−𝑒^{−1/(𝑛+1/4)}+1168^𝑛_𝑘=01(𝑘+1/4)²−𝜋²8<𝑒<𝐺_2/(𝑛+5/4)−𝑒^{−2/(𝑛+1/4)}+1328^𝑛_𝑘=01(𝑘+1/4)²−𝜋²8 (88) for 𝑛∈ℕ.
Let 𝑢_𝑛=𝑒^1/(𝑛+5/4)−𝑒^{−1/(𝑛+1/4)}+1168^𝑛_𝑘=01(𝑘+1/4)²−𝜋²8,𝑈_𝑛=𝑒^2/(𝑛+5/4)−𝑒^{−2/(𝑛+1/4)}+1328^𝑛_𝑘=01(𝑘+1/4)²−𝜋²8. (89) Then inequalities (88) can be rewritten as 𝑢_𝑛<𝐺<𝑈_𝑛. (90) It follows from (63) and (67) that we have lim_𝑛→∞𝑛⁵𝑢_𝑛1−𝐺=−,24lim_𝑛→∞𝑛⁷𝑈_𝑛=1−𝐺.45 (91) Therefore, (90) provides a new approximation algorithm for the Catalan constant 𝐺. Numerical simulations results carried out with mathematical software show that the given algorithm is more accurate than ∑^𝑛_𝑘=0((−1)^𝑘/(2𝑘+1)²) (see Table 2).

Table 2 

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307 and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.