# A Double Inequality for the Trigamma Function and Its Applications

Authors: Zhen-Hang Yang; Yu-Ming Chu; Xiao-Jing Tao
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We prove thatp=1 and q=2 are the best possible
parameters in the interval (0,∞) such that the double inequality ep/x+1-e-p/x/2p<ψ′x+1<eq/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∞-1n/(2n+1)2 are given. Here, ψ′ is the trigamma function.


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## Body

## 1. Introduction

For real and positive values ofx, the classical Euler’s gamma function Γ and its logarithmic derivative ψ, the so-called psi function, are defined as
(1)Γ(x)=∫0∞‍tx-1e-tdt,ψ(x)=Γ′(x)Γ(x).
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
(2)ψ′(x)<e-ψ(x)
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
(3)(x+a*)e-2ψ(x+1)<ψ′(x+1)<(x+b*)e-2ψ(x+1)
holds for all x>0, where γ is Euler’s constant. Batir [4] also showed that
(4)12(2x2-1+e2/(x+1)-e-2ψ(x+1))<ψ′(x+1)<12(2x2+1-e-2/x+e-2ψ(x+1))
for all x>0. In [5, (1.11)], Guo and Qi established that
(5)ϑeϑ/xx2(eϑ/x-1)<ψ′(x)<θeθ/xx2(eθ/x-1)
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:
(6)ψ′(x)<e1/x-1
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 e1/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 thatm∈(0,∞) and g and gm 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)>(<)gm(x) holds for all x>0 if gλ(x)≥(≤)gμ(x) on (0,∞), or gλ(x) and gμ(x) are not comparable on (0,∞), and lim⁡x→∞(g(x)-gλ(x))/(g(x)-gμ(x))=0 for any μ∈(0,∞) satisfies g(x)>(<)gμ(x) on the interval (0,∞).The main purpose of this paper is to find the best possible constantsp,q∈(0,∞) such that the double inequality
(7)θ(x,p)<ψ′(x+1)<θ(x,q)
or equivalently
(8)1x2+θ(x,p)<ψ′(x)<1x2+θ(x,q)
holds for all x>0, where
(9)θ(x,m)=em/(x+1)-e-m/x2m,m>0.
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 Theorem1, we clearly see the following.Corollary 2.
The double inequality(10)sinh⁡1x+1<ψ′(x+1)<12sinh⁡2x
holds for all x>0.


## 2. Lemmas

Lemma 3.
Let the functionθ 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
(11)m2x2(m+x)em/x∂θ∂m=(1-x(x-m+1)(x+1)(m+x)exp⁡(mx+mx+1))=:hm(x),∂hm∂x=m2xexp⁡((m/x)((2x+1)/(x+1)))(x+1)3(m+x)2×((3-2m)x2+(3-2m)x+(1-m)).
If m∈(0,1], then ∂hm/∂x>0; that is, hm is strictly increasing with respect to x>0. Therefore,
(12)hm(x)<lim⁡x→∞hm(x)=0,
which implies that ∂θ/∂m<0.
Ifm≥3/2, then ∂hm/∂x<0; that is, hm is strictly decreasing with respect to x>0, which leads to the conclusion that
(13)hm(x)>lim⁡x→∞hm(x)=0.Lemma 4.
Let the functionθ be defined on (0,∞)2 by (9) and
(14)Fm(x)=ψ′(x+1)-θ(x,m).
Then the equation
(15)Fm(0+)=ψ′(1)-em2m=0
has two roots
(16)m1=0.5023⋯,m2=1.7510⋯
such that Fm(0+)>0 for m∈(m1,m2) and Fm(0+)<0 for m∈(0,m1)∪(m2,∞).Proof.
Differentiation yields(17)ddmFm(0+)=-12m2em(m-1),
which reveals that the function m↦Fm(0+) is strictly increasing on (0,1) and strictly decreasing on (1,∞). Therefore, Lemma 4 follows from the piecewise monotonicity of the function m↦Fm(0+) and the numerical computations results:
(18)F0.5023(0+)=16π2-e0.50231.0046=-1.685⋯×10-5<0,F0.5024(0+)=16π2-e0.50241.0048=1.46⋯×10-4>0,F1.751(0+)=16π2-e1.7513.502=5.68⋯×10-5>0,F1.7511(0+)=16π2-e1.75113.5022=-1.374⋯×10-5<0.Lemma 5.
Letm≥0, m1=0.5023⋯, and m2=1.7510⋯ and let Fm(x) be defined as in Lemma 4. Then the following statements are true:(i)
if the inequalityFm(x)≥0 holds for all x>0, then m∈[1,m2];(ii)
if the inequalityFm(x)≤0 holds for all x>0, then m∈(0,m1]∪[2,∞).Proof.
It follows from the series formulas that(19)ψ′(x)=1x+12x2+16x3-130x5+⋯,(20)em/(x+1)=1+mx+1+12(mx+1)2+16(mx+1)3+124(mx+1)5+⋯,(21)e-m/x=1-mx+12(-mx)2+16(-mx)3+124(-mx)5+⋯,
and we get
(22)lim⁡x→∞Fm(x)x-3=lim⁡x→∞((((1-mx+12(-mx)2+16(-mx)3)1+mx+1+12(mx+1)2+16(mx+1)3((((1-mx+12(-mx)2+16(-mx)3)1+mx+1+12(mx+1)2+16(mx+1)31(x+1)+12(x+1)2+16(x+1)3mmmmnmmm-(((1-mx+12(-mx)2+16(-mx)3)1+mx+1+12(mx+1)2+16(mx+1)3mmmmmmmmmmnm-(1-mx+12(-mx)2+16(-mx)3))mmmmmnmmmmmm×(2m)-1((1-mx+12(-mx)2+16(-mx)3)1+mx+1+12(mx+1)2+16(mx+1)3)(((1-mx+12(-mx)2+16(-mx)3)1+mx+1+12(mx+1)2+16(mx+1)3)iiiiiiiiiii×(x-3)-1(((1-mx+12(-mx)2+16(-mx)3)1+mx+1+12(mx+1)2+16(mx+1)3)=-112lim⁡x→∞(((x)3)-1((2m2-6m+4)x3+(3m2-9m+6)x2mmmmmmmmii+(3m2-3m)x+m2)iiiiiiiiiiiiiiiii×((x+1)3)-1)=-112(2m2-6m+4)=-16(m-1)(m-2).(i)
If inequalityFm(x)≥0 holds for all x>0, then, from
(23)Fm(0+)=ψ′(1)-em2m≥0,lim⁡x→∞Fm(x)x-3=-16(m-1)(m-2)≥0and Lemma 4, we clearly see that m∈[m1,m2]∩[1,2]=[1,m2].(ii)
If inequalityFm(x)≤0 holds for all x>0, then
(24)Fm(0+)=ψ′(1)-em2m≤0,lim⁡x→∞Fm(x)x-3=-16(m-1)(m-2)≤0and Lemma 4 lead to the conclusion that
(25)m∈((0,m1]∪[m2,∞])∩((0,1]∪[2,∞))=(0,m1]∪[2,∞).Lemma 6.
Let the functionθ 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.
Forx>0 and m>0, let
(26)G1,m(x)=θ(x,1)-θ(x,m)=e1/(x+1)-e-1/x2-em/(x+1)-e-m/x2m.
Then simple computation leads to
(27)G1,m(0+)=e2-em2m.
From (20) and (21), we have
(28)lim⁡x→∞G1,m(x)x-3=-16(m-1)(m-2).
Differentiation yields
(29)ddmG1,m(0+)=-em2m2(m-1),
which shows that m↦G1,m(0+) is strictly decreasing on (1,∞). Therefore,
(30)G1,m(0+)<G1,1(0+)=0
if m∈(1,2). On the other hand, we clearly see that lim⁡x→∞[x3G1,m(x)]>0 for m∈(1,2).Lemma 7.
Letn∈N={1,2,3,…} and m∈N∪{0} with n>m and let Pn(t) be the polynomial of degree n defined by
(31)Pn(t)=∑i=m+1n‍aiti-∑i=0m‍aiti,
where an,am>0 and ai≥0 for 0≤i≤n-1 with i≠m. Then, there exists tm+1∈(0,∞) such that Pn(tm+1)=0 and Pn(t)<0 for t∈(0,tm+1) and Pn(t)>0 for t∈(tm+1,∞).Proof.
DifferentiatingPn(t) gives
(32)Pn(k)(t)=∑i=m+1n‍i!(i-k)!aiti-k-∑i=km‍i!(i-k)!aiti-kfor1≤k≤m,Pn(m+1)(t)=∑i=m+1n‍i!(i-m-1)!aiti-m-1>0.
Note that
(33)Pn(k)(∞)=∞,Pn(k)(0+)=-ak
for 1≤k≤m.
FromPn(m+1)(t)>0, we clearly see that Pn(m)(t) is strictly increasing on (0,∞). Then Pn(m)(∞)>0 and Pn(m)(0+)=-am<0 lead to the conclusion that there exists t1∈(0,∞) such that Pn(m)(t1)=0 and Pn(m)(t)<0 for t∈(0,t1) and Pn(m)(t)>0 for t∈(t1,∞). Therefore, Pn(m-1)(t) is strictly decreasing on (0,t1) and strictly increasing on (t1,∞).
It follows from the piecewise monotonicity ofPn(m-1)(t) and Pn(m-1)(∞)=∞ that Pn(m-1)(t)<Pn(m-1)(0+)=-am-1≤0 for t∈(0,t1), and there exists t2∈(t1,∞) such that Pn(m-1)(t2)=0 and Pn(m-1)(t)<0 for t∈(0,t2) and Pn(m-1)(t)>0 for t∈(t2,∞). Therefore, Pn(m-2)(t) is strictly decreasing on (0,t2) and strictly increasing on (t2,∞).
After repeating the same steps as abovem+1 times, we deduce that there exists tm+1∈(tm,∞)⊂(0,∞) such that Pn(tm+1)=0 and Pn(t)<0 for t∈(0,tm+1) and Pn(t)>0 for t∈(tm+1,∞).Lemma 8.
Let the functionθ 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.
Forx>0 and m>0, let
(34)G2,m(x)=θ(x,2)-θ(x,m)=e2/(x+1)-e-2/x4-em/(x+1)-e-m/x2m.
Then simple computation leads to
(35)G2,m(0+)=e24-em2m.
(i) We prove that there existsm0∈(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 G2,m(0+)>0 and lim⁡x→∞x3G2,m(x)<0 if m∈(m0,1).
Indeed, it follows from(36)ddmG2,m(0+)=-em2m2(m-1)
that the function m↦G2,m(0+) is strictly increasing on (0,1). Numerical computations show that
(37)G2,2/5(0+)=14e2-54e2/5<0,G2,9/20(0+)=14e2-109e9/20>0.
Therefore, there existsm0∈(2/5,9/20) such that
(38)G2,m(0+)=e24-em2m=0
and G2,m(0+)<0 for m∈(0,m0) and G2,m(0+)>0 for m∈(m0,1).
On the other hand, it follows fromm∈(m0,1) together with (20) and (21) that
(39)lim⁡x→∞G2,m(x)x-3=-16(m-1)(m-2)<0.
(ii)  We prove thatθ(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(40)g1(x)=em0/xG2,m0(x)=14exp⁡(m0x+2x+1)-14exp⁡(m0x-2x)+12m0(1-exp⁡(m0x+m0x+1)).
Then
(41)lim⁡x→∞g1(x)=0.
Differentiation yields
(42)42-m0x2e(2-m0)/x×g1′(x)=2(2x2+2x+1)(2-m0)(x+1)2exp⁡(2x+m0x+1)-1-(m0+2)x2+2m0x+m0(2-m0)(x+1)2exp⁡(2x+2x+1):=g2(x),(43)g2′(x)=2((2m0+2)x4+(2m0+10)x3iiiiiiii+(m0+14)x2+8x+2)×((2-m0)x2(x+1)4)-1×exp⁡(m0x+1+2x)g3(x),
where
(44)g3(x)=((2m0+2)x4+(6m0+2)x3+(7m0+2)x2+4m0x+m0(2m0+2)x4+(6m0+2)x3)×((2m0+2)x4+(2m0+10)x3iiiiiiii+(m0+14)x2+8x+2(2m0+2)x4+(2m0+10)x3)-1×e(2-m0)/(x+1)-1.
Note that
(45)lim⁡x→∞g2(x)=0,(46)lim⁡x→0+g3(x)=12m0e2-m0-1=0,(47)lim⁡x→∞g3(x)=0,
where the second equation in (46) follows from (38).
Differentiatingg3(x) leads to
(48)g3′(x)=((2-m0)e(2-m0)/(x+1))×(+2m0x4+14x2+10x3+2x4+2)2(x+1)2(8x+m0x2+2m0x3iiiiiiiiiiiiiiiiiiii+2m0x4+14x2+10x3+2x4+2)2)-1×g4(x),
where
(49)g4(x)=(4-4m02)x8+(16-16m02)x7+(-28m02-24m0+32)x6+(-28m02-72m0+40)x5-(17m02+104m0-42)x4-(6m02+88m0-36)x3-(m02+46m0-18)x2-(14m0-4)x-2m0≔a8x8+a7x7+a6x6+a5x5-a4x4-a3x3-a2x2-a1x-a0.
We assert that there exists a unique x4*∈(0,∞) such that g4(x)<0 for x∈(0,x4*) and g4(x)>0 for x∈(x4*,∞), which leads to the conclusion that g3(x) is strictly decreasing on (0,x4*] and strictly increasing on [x4*,∞). 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
(50)a8=4-4m02=4(1-m02)>0,a7=16-16m02=16(1-m02)>0,a6=-28m02-24m0+32>-28m0+2-24m0++32=1553100>0,a5=-28m02-72m0+40>-28m0+2-72m0++40=193100>0,a4=(17m02+104m0-42)>17m0-2+104m0--42=5825>0,a3=(6m02+88m0-36)>6m0-2+88m0--36=425>0,a2=(m02+46m0-18)>m0-2+46m0--18=1425>0,a1=(14m0-4)>14m0--4=85>0,a0=2m0>0.
From the piecewise monotonicity of g3(x) together with (46) and (47), we clearly see that
(51)g3(x)<lim⁡x→0+g3(x)=0forx∈(0,x4*],g3(x)<lim⁡x→∞g3(x)=0forx∈[x4*,∞);
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]).
Letx>0 and n∈{0,1,2,…}. Then
(52)ψ(n)(x+1)-ψ(n)(x)=(-1)nn!xn+1.From the proof of [4, Theorem 2.6], we get the following.Lemma 10.
The inequality(53)ψ′(x+1)>2x+12x2+2x+2/3
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]).
Lety be the function defined on (0,∞) by
(54)y(x)=e2ψ(x).
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]).
LetI⊆R be an interval, a,b∈I with a<b, and let f:I→R be a convex function. Then
(55)f(a+b2)≤∫abf(x)dxb-a≤f(a)+f(b)2.


## 3. Proofs of Theorem1

Proof of Theorem1.
We divide the proof into four parts.
(I) We prove the first inequality in (7); that is,
(56)F1(x)=ψ′(x+1)-e1/(x+1)-e-1/x2>0,
where Fm(x) is defined by (14).
It follows from Lemma10 that
(57)F1(x)=ψ′(x+1)-e1/(x+1)-e-1/x2>2x+12x2+2x+2/3-e1/(x+1)-e-1/x2:=e-1/xH(x),
where
(58)H(x)=2x+12x2+2x+2/3e1/x-exp⁡((1/(x+1))+(1/x))-12.
We clearly see that it is enough to prove thatH(x)>0 for x>0.
DifferentiatingH(x) gives
(59)H′(x)=12x2e1/x×(3(x+1)(6x3+6x2+4x+1)(3x2+3x+1)22x2+2x+1(x+1)2e1/(x+1)iiiiiiiii-3(x+1)(6x3+6x2+4x+1)(3x2+3x+1)2)=12x2e1/xL×(3(x+1)(6x3+6x2+4x+1)(3x2+3x+1)22x2+2x+1(x+1)2e1/(x+1),iiiiiiii3(x+1)(6x3+6x2+4x+1)(3x2+3x+1)2)×h(x),
where L(a,b)=(b-a)/(ln⁡a-ln⁡b) denotes the logarithmic mean of positive numbers a and b, and
(60)h(x)=1x+1+ln⁡2x2+2x+1(x+1)2-ln⁡3(x+1)(6x3+6x2+4x+1)(3x2+3x+1)2.
Differentiating h(x) leads to
(61)h′(x)=(x(2x+1)(7x2+7x+3))×((x+1)2(3x2+3x+1)(2x2+2x+1)iiiiiii×(6x3+6x2+4x+1))-1>0
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 thatm=1 is the best possible constant such that ψ′(x+1)>θ(x,m) for all x>0.
From Lemma5, 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 anym*∈(1,m2], (22) leads to
(62)lim⁡x→∞[x3Fm*(x)]=16(m*-1)(2-m*).
It follows from (19), (20), and (21) that we get
(63)lim⁡x→∞[x5F1(x)]=124.
(III) We prove the second inequality (7); that is,
(64)F2(x)=ψ′(x+1)-e2/(x+1)-e-2/x4<0
for x>0, where Fm(x) is defined by (14).
Lemma11 implies that the function x↦y′(x)=2ψ′(x)e2ψ(x) is strictly convex on (0,∞). Then, making use of Lemma 12, we get
(65)y′(x+x+22)<∫xx+2y′(t)dtx+2-x
for x>0. That is,
(66)2ψ′(x+1)e2ψ(x+1)<e2ψ(x+2)-e2ψ(x)2,ψ′(x+1)<14[e2(ψ(x+2)-ψ(x+1))-e2(ψ(x)-ψ(x+1))].
Therefore, inequality (64) follows easily from (52) and the last inequality above.
(IV) We prove thatm=2 is the best possible constant such that ψ′(x+1)<θ(x,m) for all x>0.
From Lemma5, 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 Lemma8 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 existsm*∈(m0,m1] such that Fm*(x)=ψ′(x+1)-θ(x,m*)<0 for all x>0, then, from (19), (20), (21), and (22), we get
(67)lim⁡x→∞[x3Fm*(x)]=-16(1-m*)(2-m*),lim⁡x→∞[x7F2(x)]=-145.From the above proof and Lemma3 we get the following.Corollary 13.
Let the functionθ 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
(68)ψ′(x+1)<θ(x,2)<θ(x,m0)<lim⁡m→0θ(x,m)=12(x+1)+12x
or equivalently
(69)ψ′(x)<12(x+1)+12x+1x2
hold for all x>0.


## 4. Remarks

Remark 14.
It follows from (67) and the facts that
(70)lim⁡x→∞ψ′(x)-e-ψ(x)x-3=-124,lim⁡x→∞ψ′(x+1)-(x+π2e-2γ/6)e-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-6=-160,lim⁡x→∞ψ′(x)-(e1/x-1)x-4=-124,lim⁡x→∞ψ′(x)-(2e2/x/x2(e2/x-1))x-2=-12
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(71)e1/(x+1)+e-1/x>2ifx∈(-∞,-1)∪(0,∞),e1/(x+1)+e-1/x≥2e2ifx∈(-1,0).Proof.
Differentiation leads to(72)(e1/(x+1)+e-1/x)′=e-1/xx2(1-x2(x+1)2exp⁡(1x+1+1x)):=e-1/xx2V(x),V′(x)=exp⁡(1/(x+1)+(1/x))(x+1)4>0.
If x∈(0,∞), then, from V′(x)>0, we get V(x)<lim⁡x→∞V(x)=0, which leads to
(73)e1/(x+1)+e-1/x>lim⁡x→∞(e1/(x+1)+e-1/x)=2.
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,
(74)e1/(x+1)+e-1/x≥[e1/(x+1)+e-1/x]x=-1/2=2e2.Remark 16.
Using inequality (71), one has
(75)1x2+θ(x,2)-(e1/x-1)=1x2+e2/(x+1)-e-2/x4-(e1/x-1)<1x2+e2/(x+1)-(2-e1/(x+1))24-(e1/x-1)=1x2+e1/(x+1)-e1/x=1x2+∑n=0∞‍1n!(1(x+1)n-1xn)<1x2+∑n=03‍1n!(1(x+1)n-1xn)=-16x3(x+1)3<0
for x>0, which shows that the upper bound x-2+θ(x,2) in (8) is better than the upper bound (e1/x-1) in (6).Remark 17.
The conclusion that the differencee1/x-ψ′(x) is completely monotonic on (0,∞) given in [6, Lemma 2] implies that
(76)e-ψ′(1)>e1/(x+1)-ψ′(x+1)>1
or
(77)e1/(x+1)-e+π26<ψ′(x+1)<e1/(x+1)-1<12sinh⁡2x.
It is easy to check that the lower bound e1/(x+1)-e+π2/6 and sinh⁡1/(x+1) for ψ′(x+1) given in (10) are not comparable due to
(78)e1/(x+1)-e+π26>(<)sinh⁡1x+1ifx<(>)1.62670⋯.Remark 18.
Guo et al. [14] proved that
(79)(k-1)!xk+k!2xk+1<(-1)k+1ψ(k)(x)<(k-1)!xk+k!xk+1
for x>0 if k∈N. In particular, if k=1, one has
(80)1x+12x2<ψ′(x)<1x+1x2.
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 constantsπ and G (Catalan constant).Remark 19.
It is well known that(81)∑k=1∞‍1k2=π26.
Let Qn=∑k=1n‍(1/k2), and then ψ′(n+1)=(π2/6)-Qn. From Theorem 1, we clearly see that the double inequality
(82)e1/(n+1)-e-1/n2<π26-Qn<e2/(n+1)-e-2/n4
holds for all n∈N.
Let(83)ln=Qn+e1/(n+1)-e-1/n2,Ln=Qn+e2/(n+1)-e-2/n4.
Then inequalities (82) can be rewritten as
(84)ln<π26<Ln
or
(85)6ln<π<6Ln.
It follows from (63) and (67) that
(86)lim⁡n→∞n5(ln-π26)=-124,lim⁡n→∞n7(Ln-π26)=145.
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 6Qn (see Table 1).
More estimate methods forπ can be found in [15–19].Table 1
n
|
6
Q
n
-
π
|
|
6
l
n
-
π
|
|
6
L
n
-
π
|
1
0.6921
4.3127
×
10
-
3
7.663
×
10
-
4
2
0.40298
3.7549
×
10
-
4
2.8250
×
10
-
5
5
0.1782
7.7596
×
10
-
6
1.3276
×
10
-
7
10
9.2231
×
10
-
2
3.1013
×
10
-
7
1.4875
×
10
-
9
50
1.8966
×
10
-
2
1.2112
×
10
-
10
2.5320
×
10
-
14
100
9.5161
×
10
-
3
3.8807
×
10
-
12
2.0489
×
10
-
16
200
4.7663
×
10
-
3
1.2280
×
10
-
13
1.6291
×
10
-
18
500
1.9085
×
10
-
3
1.2669
×
10
-
15
2.6973
×
10
-
21Remark 20.
The Catalan constant(87)G=∑n=0∞‍(-1)n(2n+1)2=0.9159655941772190⋯
is a mysterious constant in mathematics and physics. From ψ′(1/4)=π2+8G and ψ′(n+(1/4)+1)=ψ′(1/4)-∑k=0n‍(1/(k+1/4)2) (see [17, 20]) together with Theorem 1, we get
(88)e1/(n+5/4)-e-1/(n+1/4)16+18∑k=0n‍1(k+1/4)2-π28<G<e2/(n+5/4)-e-2/(n+1/4)32+18∑k=0n‍1(k+1/4)2-π28
for n∈N.
Let(89)un=e1/(n+5/4)-e-1/(n+1/4)16+18∑k=0n‍1(k+1/4)2-π28,Un=e2/(n+5/4)-e-2/(n+1/4)32+18∑k=0n‍1(k+1/4)2-π28.
Then inequalities (88) can be rewritten as
(90)un<G<Un.
It follows from (63) and (67) that we have
(91)lim⁡n→∞n5(un-G)=-124,lim⁡n→∞n7(Un-G)=145.
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 ∑k=0n‍((-1)k/(2k+1)2) (see Table 2).Table 2
n
|
∑
k
=
0
n
‍
(
-
1
)
k
/
(
2
k
+
1
)
2
-
G
|
|
u
n
-
G
|
|
U
n
-
G
|
1
2.7077
×
10
-
2
2.7274
×
10
-
4
3.7885
×
10
-
5
2
1.2923
×
10
-
2
3.0912
×
10
-
5
1.9591
×
10
-
6
5
3.4037
×
10
-
3
8.1456
×
10
-
7
1.2782
×
10
-
8
10
1.0268
×
10
-
3
3.6098
×
10
-
8
1.6524
×
10
-
10
50
4.8045
×
10
-
5
1.5468
×
10
-
11
3.2017
×
10
-
15
100
1.2253
×
10
-
5
5.0171
×
10
-
13
2.6358
×
10
-
17
200
3.0939
×
10
-
6
1.5974
×
10
-
14
2.1139
×
10
-
19
500
4.98
×
10
-
7
1.6542
×
10
-
16
3.5184
×
10
-
22



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*Source: 702718-2014-07-16.xml*