# New Improved Inequalities Associated with Ratios of Gamma Functions

Authors: Jenica Cringanu
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we improve the estimate for the products∏k=1n4k−3/4k and ∏k=1n4k−1/4k, obtained by Cringanu “inequalities associated with ratios of gamma functions.”


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## Body

## 1. Introduction

Cringanu [1] proved the following inequalities for the Wallis ratio:(1)an34≤1⋅5,…,4n−34⋅8,…,4n<bn34,(2)cn4≤3⋅7,…,4n−14⋅8,…,4n<dn4,for all n≥1, with the best constants a=1/4=0.25,b=1/Γ1/4=0.2758…,c=3/4=0.75, and d=1/Γ3/4=0.8160,….These inequalities are a consequence of the complete monotonicity on0,∞ of the functions(3)x⟶ln1/Γ1/44x3Γx+1/4/Γx+1Γ1/44,x⟶ln1/Γ3/44xΓx+3/4/Γx+1Γ3/44,where Γx=∫0∞tx−1e−xdt for x>0 is the gamma function (see, for example, [2–5]).Since(4)limn⟶∞1⋅5,…,4n−34⋅8,…,4nn34=1Γ1/4,limn⟶∞3⋅7,…,4n−14⋅8,…,4nn4=1Γ3/4,(see [1]), the following approximations as n⟶∞ are true:(5)1⋅5,…,4n−34⋅8,…,4n≈1Γ1/4n34,3⋅7,…,4n−14⋅8,…,4n≈1Γ3/4n4.First, we concentrate on the approximation of1⋅5,…,4n−3/4⋅8,…,4n and introduce the relative error sequence wn by the following relation for every integer n≥1:(6)1⋅5,…,4n−34⋅8,…,4n=1Γ1/4n34expwn.A tool for measuring speed of convergence is a result by Mortici (lemma, [6]), who states that a sequence xn converging to zero is the fastest possible when the difference xn−xn+1 is the fastest possible. More precisely, if there exists limn⟶∞nkxn−xn+1=l, then limn⟶∞nk−1xn=l/k−1. Recent results using this lemma have been given in [7–12]. In our case of wn, we have(7)wn−wn+1=ln4n+44n+1−34lnn+1n,and using the Maclaurin series of the function wn−wn+1 in the variable 1/n, we obtain(8)wn−wn+1=−332n2+O1n3.Then,limn⟶∞n2wn−wn+1=−3/32 and so limn⟶∞nwn=−3/32.Consequently, the following approximation is true:wn≈−3/32n as n⟶∞.Similarly, if3⋅7,…,4n−1/4⋅8,…,4n=1/Γ3/4n4expvn, then vn−vn+1=−3/32n2+O1/n3 and so the following approximation is true: vn≈−3/32n as n⟶∞.


## 2. The Main Results

Theorem 1.
(i)
For all integern≥1, we have(9)1Γ1/4n34exp−332n−an2<1⋅5,…,4n−34⋅8,…,4n,with the best constanta=1/128.(ii)
For everya<1/128, there exists na∈N,n≥1, such that(10)1⋅5,…,4n−34⋅8,…,4n<1Γ1/4n34exp−332n−an2,for all n≥na.Proof.
We define the sequencean for n≥1,(11)an=ln1⋅5,…,4n−34⋅8,…,4n+lnΓ14+34lnn+332n+an2,and so an+1−an=fn, where(12)fn=ln4n+14n+4+34lnn+1−34lnn+332n+1−332n+an+12−an2.
The derivative of the functionf is equal to(13)f′n=Pn32n3n+134n+1,where Pn=6n3128a−1+3n2320a−1+n448a+3+64a.(i)
Ifa=1/128, then Pn=1/29n2+13n+1>0, for all n≥1, and then f is strictly increasing.Sincef∞=0, it follows that fn<0, for all n≥1, so that an is strictly decreasing. The sequence an converges to zero, and then it follows that an>0, for all n≥1, so that(14)1Γ1/4n34exp−332n−1128n2<1⋅5,…,4n−34⋅8,…,4n,for alln≥1.(ii)
Ifa<1/128, then there exists na∈N,n≥1, such that Pn<0 for all n≥na, and then f is strictly decreasing on na,∞. Since f∞=0, it follows that fn>0, for all n≥na, so that ann≥na is strictly increasing. The sequence an converges to zero, and then it follows that an<0, for all n≥na, so that(15)1⋅5,…,4n−34⋅8,…,4n<1Γ1/4n34exp−332n−an2,for all n≥na.
Now, we find the constantna in some particular cases.
For example, ifa=1/256<1/128, then Pn=−3n3+3/4n2+19/4n+1/4<0 for all n≥2 and so(16)1Γ1/4n34exp−332n−1128n2<1⋅5,…,4n−34⋅8,…,4n<1Γ1/4n34exp−332n−1256n2,for all n≥2.
Let us remark that a direct calculus shows that these inequalities hold forn=1, and then(17)1Γ1/4n34exp−332n−1128n2<1⋅5,…,4n−34⋅8,…,4n<1Γ1/4n34exp−332n−1256n2,for all n≥1.
Now, ifa=1/160<1/128, then Pn=−6/5n3+3n2+29/5n+2/5<0 for all n≥4 and so(18)1Γ1/4n34exp−332n−1128n2<1⋅5,…,4n−34⋅8,…,4n<1Γ1/4n34exp−332n−1160n2,for all n≥4.
Let us remark that a direct calculus shows that these inequalities hold forn=3, and then(19)1Γ1/4n34exp−332n−1128n2<1⋅5,…,4n−34⋅8,…,4n<1Γ1/4n34exp−332n−1160n2,for all n≥3.Remark 1.
Let us remark that1/4n34<1/Γ1/4n34exp−3/32n−1/128n2, for all n≥2, and for a∈0,1/128,(20)1Γ1/4n34exp−332n−an2<1Γ1/4n34,for alln≥1,which is shown to improve inequality (1).Theorem 2.
(i)
For all integern≥1, we have(21)1Γ3/4n4exp−332n+an2<3⋅7,…,4n−14⋅8,…,4n,with the best constanta=1/128.(ii)
For everya>1/128, there exists na∈N,na≥1, such that(22)3⋅7,…,4n−14⋅8,…,4n<1Γ3/4n4exp−332n+an2,for all n≥na.Proof.
We define the sequencean for n≥1,(23)an=ln3⋅7,…,4n−14⋅8,…,4n+lnΓ34+14lnn+332n−an2,and so an+1−an=fn, where(24)fn=ln4n+34n+4+14lnn+1−14lnn+332n+1−332n−an+12+an2.
The derivative of the functionf is equal to(25)f′n=Qn32n3n+134n+3,where Qn=6n31−128a+3n25−448a+n9−832a−192a.(i)
Ifa=1/128, then Qn=1/29n2+5n−3>0, for all n≥1, and then f is strictly increasing.Sincef∞=0, it follows that fn<0, for all n≥1, so that an is strictly decreasing. The sequence an converges to zero, and then it follows that an>0, for all n≥1, so that(26)1Γ3/4n4exp−332n+1128n2<3⋅7,…,4n−14⋅8,…,4n,for alln≥1.(ii)
Ifa>1/128, then there exists na∈N,n≥1, such that Qn<0 for all n≥na, and then f is strictly decreasing on na,∞. Since f∞=0, it follows that fn>0, for all n≥na, so that ann≥na is strictly increasing. The sequence an converges to zero, and then it follows that an<0, for all n≥na, so that(27)3⋅7,…,4n−14⋅8,…,4n<1Γ3/4n4exp−332n+an2,for all n≥na.
Now, we find the constantna in some particular cases.
For example, ifa=1/64>1/128, then(28)Qn=−6n3−6n2−4n−3<0,for all n≥1, and so(29)1Γ3/4n4exp−332n+1128n2<3⋅7,…,4n−14⋅8,…,4n<1Γ3/4n4exp−332n+164n2,for all n≥1.
Now, ifa=1/120>1/128, then Qn=−1/156n3−57n2−31n+24<0 for all n≥10 and so(30)1Γ3/4n4exp−332n+1128n2<1⋅5,…,4n−34⋅8,…,4n<1Γ3/4n4exp−332n+1120n2,for all n≥10.
Let us remark that a direct calculus shows that these inequalities hold forn∈1,2,…,9, and then(31)1Γ3/4n4exp−332n+1128n2<1⋅5,…,4n−34⋅8,…,4n<1Γ3/4n4exp−332n+1120n2,for all n≥1.Remark 2.
Let us remark that3/4n4<1/Γ3/4n4exp−3/32n+1/128n2, for all n≥1, and for a>1/128,(32)1Γ3/4n4exp−332n+an2<1Γ3/4n4,for alln>32a3,which is shown to improve inequality (2).



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*Source: 2008466-2021-09-22.xml*