AZUMA Hiroki

Fujita Yasuhiro

University of Toyama Repository

Toyama Math. J.Vol. 43(2022), 77-90The best constant for an inequality related tothe Mathieu seriesHiroki Azuma and Yasuhiro FujitaAbstract. Let S(r) be the Mathieu series defined byS(r) =∞∑n=12n(n2 + r2)2, r ∈ (0,∞),which converges to 2ζ(3) as r → 0+. Here ζ is the Riemann zetafunction. Let b ∈ (0,∞) be a constant. Our aim of this paper is toshow that the inequalityS(r) <2ζ(3)br2 + 1, r ∈ (0,∞)holds if and only if b ≤ b∗ := (2ζ(5))/ζ(3). Hence b∗ is the bestconstant satisfying this inequality. This best constant is related tothe Taylor expansion of S at r = 0.1. IntroductionIn 1890, French mathematician Émile Léonard Mathieu [5] first intro-duced the series S(r), which is now called the Mathieu series, defined as(1.1) S(r) =∞∑n=12n(n2 + r2)2, r ∈ (0,∞).2020 Mathematics Subject Classification. Primary 26D15; Secondary 26A09.Key words and phrases. The Mathieu series, the best constant.The second author is supported in part by JSPS KAKENHI No. 21K03322.7778 Hiroki Azuma and Yasuhiro FujitaHe conjectured that S(r) < 1/r2 for all r ∈ (0,∞), and used this inequalityin his work on elasticity of solid bodies. However, he could not prove thisconjecture.In 1952, Berg [2] proved this inequality, although his proof was diffi-cult enough. It was Makai [4] who gave in 1957 an elementary proof andobtained the following estimates:(1.2)1r2 + 12< S(r) <1r2, r ∈ (0,∞).In 1998, Alzer, Brenner and Ruehr [1] showed that the best constants k1and k2 in the two–sided estimations(1.3)1r2 + k1< S(r) <1r2 + k2, r ∈ (0,∞)are given by k1 = 1/(2ζ(3)) and k2 = 1/6, where ζ(s) is the Riemann zetafunction defined by(1.4) ζ(s) =∞∑n=11ns, s ∈ (1,∞).In 2007, Hoorfar and Qi [3] gave the inequality(1.5) S(r) <1√(r2 + 1)2 + 1− 1, r ∈ (0,∞).By a simple calculation, the inequality1√(r2 + 1)2 + 1− 1<1r2 + 16holds if and only if 0 < r <√2312 = 1.38443 · · · . Thus inequality (1.5) isbetter than the right-hand side inequality in (1.3) with k2 = 1/6 only when0 < r <√2312 .Till now, the Mathieu series has attracted increasing interest and at-tention. However, as far as the authors know, the upper bounds obtainedfor S(r) never reflect the fact such that S(r) → 2ζ(3) = 2.40411 · · · asr → 0+. Indeed, the upper bound in (1.3) with k2 = 1/6 converges to 6which is greater than 2ζ(3) as r → 0+; the upper bound in (1.5) convergesThe best constant for an inequality related to the Mathieu series 79to√2 + 1 = 2.41421 · · · which is also greater than 2ζ(3) as r → 0+. Thisis contrary to lower bounds for S(r), since the lower bound in (1.3) withk1 = 1/(2ζ(3)) converges to 2ζ(3) as r → 0+.In this paper, we would like to present an inequality for S(r), whichreflects the fact such that S(r) → 2ζ(3) as r → 0+. For this purpose, for agiven constant b ∈ (0,∞), we consider the inequality of the form(1.6) S(r) <2ζ(3)br2 + 1, r ∈(0,∞).Note that the right–hand side of inequality (1.6) converges to 2ζ(3) as r →0+. In this sense, inequality (1.6) reflects the fact such that S(r) → 2ζ(3)as r → 0+.Furthermore, the denominator of the right–hand side of (1.6) comes from(1.3). Indeed, the right–hand side of (1.6) can be rewritten as(2ζ(3)/b)r2 + (1/b)whose denominator has the same form as those of (1.3).Our aim of this paper is to determine completely the range of b satisfyinginequality (1.6). Let us define the constant b∗ by(1.7) b∗ =2ζ(5)ζ(3)= 1.72525 · · · .Our main result isTheorem 1.1. Let b ∈ (0,∞) be a constant. Inequality (1.6) holds if andonly if b ≤ b∗.This theorem shows that the best constant b satisfying inequality (1.6)is given by b∗. As indicated by Lemma 2.1 and the proof of Theorem 1.1below, this best constant is related to the Taylor expansion of S at r = 0.We emphasize that inequality (1.6) with b = b∗ is important for a smallr. Indeed, as shown in Lemma 2.6 below, there exists a constant r0 ∈ (0, 12)such that the inequality1√(r2 + 1)2 + 1− 1<2ζ(3)b∗r2 + 1holds if and only if r ∈ (r0,∞). Reflecting inequality (1.5), the objectiveinequality (1.6) with b = b∗ has an advantage for only r ∈ (0, r0].80 Hiroki Azuma and Yasuhiro FujitaThe paper is organized as follow: In Section 2, we give some preliminar-ies. In Section 3, we prove Theorem 1.1. In Section 4, we give a concludingremark to compare the three inequalities (1.3) with k2 = 1/6, (1.5) and(1.6) with b = b∗.2. PreliminariesIn this section, we give some preliminaries to prove Theorem 1.1. Firstwe give the Taylor expansion of S at r = 0.Lemma 2.1.limr→0+S(r)− 2ζ(3)r2= −4ζ(5).Proof. This is clear, sinceS(r)− 2ζ(3)r2= −2∞∑n=12n2 + r2n3(n2 + r2)2→ −2∞∑n=12n2n7= −4ζ(5) (r → 0+).This completes the proof.Lemma 2.2. Let a ∈ (0,∞) be a constant. If a function g : [a,∞) → R isstrictly convex on [a,∞), then we haveg(x) <∫ x+ 12x− 12g(t) dt, x ∈ [a+ 12,∞).Proof. Let x ∈ [a + 12 ,∞) and y ∈ [−12 ,12 ]. Note that a ≤ x + y, x − y.Since g is strictly convex on [a,∞) and x+ y ̸= x− y for y ̸= 0, we haveg(x) = g(12(x+y)+12(x−y))<12g(x+y)+12g(x−y), y ∈ [−12,12]\{0}.Integrating this inequality with respect to y over [−12 ,12 ], we conclude thatg(x) <12∫ 12− 12g(x+ y) dy +12∫ 12− 12g(x− y) dy =∫ x+ 12x− 12g(t) dt.This completes the proof.The best constant for an inequality related to the Mathieu series 81Lemma 2.3. We haveS(r) <1r2 + 14, r ∈ (0, 12].Remark 2.4. Milovanović and Pogány [6] showed this inequality for r ∈(0,√32 ]. In this paper, it is sufficient to show this inequality for r ∈ (0,12 ].In order to make this paper self–contained, we give a proof to this lemma.Proof. Fix r ∈ (0, 12 ]. Letf(x) =2x(x2 + r2)2, x ∈ (0,∞).Sincef ′′(x) = 24xx2 − r2(x2 + r2)4, x ∈ (0,∞),we see that f is strictly convex on [r,∞). Note that r+ 12 ≤ 1 ≤ n for eachpositive integer n. By Lemma 2.2, we havef(n) <∫ n+ 12n− 12f(t) dt.ThusS(r) =∞∑n=1f(n) <∫ ∞12f(t) dt =[−(t2 + r2)−1]∞12=1r2 + 14.This completes the proof.Lemma 2.5. We haveS′(r) < − 8ζ(5)r(1 + r2)3, r ∈ (0,∞).Proof. Note thatS′(r) = −8r∞∑n=1n(n2 + r2)3, r ∈ (0,∞).82 Hiroki Azuma and Yasuhiro FujitaSincen(n2 + r2)3>1n5(1 + r2)3, n ∈ {2, 3, 4, · · · }, r ∈ (0,∞),we haveS′(r) = −8r∞∑n=1n(n2 + r2)3< −8r∞∑n=11n51(1 + r2)3= − 8ζ(5)r(1 + r2)3, r ∈ (0,∞).This completes the proof.In the following, we often use the following estimates:(2.1) 1.20205 < ζ(3) < 1.20206, 1.03692 < ζ(5) < 1.03693.Lemma 2.6. There exists a constant r0 ∈ (0, 12) such that the inequality1√(r2 + 1)2 + 1− 1<2ζ(3)b∗r2 + 1holds if and only if r ∈ (r0,∞).Proof. Let r ∈ (0,∞). By (1.7), we have2ζ(3)b∗r2 + 1− 1√(r2 + 1)2 + 1− 1=2ζ(3)2√(r2 + 1)2 + 1− 2ζ(5)r2 − 2ζ(3)2 − ζ(3)(2ζ(5)r2 + ζ(3))(√(r2 + 1)2 + 1− 1).Let t = r2 ∈ (0,∞). We setf(t) = 2ζ(3)2√(t+ 1)2 + 1− 2ζ(5)t− 2ζ(3)2 − ζ(3), t ∈ (0,∞).We consider the range of t ∈ (0,∞) such that f(t) > 0. We see thatf(t) > 0 if and only if the inequality4ζ(3)4[(t+ 1)2 + 1] > [2ζ(5)t+ 2ζ(3)2 + ζ(3)]2holds. This inequality is equivalent to the oneat2 + bt+ c > 0,The best constant for an inequality related to the Mathieu series 83where a, b, c are the constants given bya = 4(ζ(3)4 − ζ(5)2),b = 4ζ(3)(2ζ(3)3 − 2ζ(5)ζ(3)− ζ(5)),c = ζ(3)2(4ζ(3)2 − 4ζ(3)− 1).By (2.1), we have ζ(3)2 − ζ(5) > 1.202052 − 1.03693 > 0.40799, so thata > 0. Since 0 < ζ(3) < 1.20206 < 1+√22 =: γ and γ is a solution of theequation 4s2 − 4s− 1 = 0, we see that c < 0; (although we can determinethe sign of b, it is not necessary in this proof.) Hence, reflecting t > 0, wesee that the inequality at2 + bt+ c > 0 holds if and only ift >−b+√b2 − 4ac2a> 0.Lettingr0 =(−b+√b2 − 4ac2a)1/2,we conclude that f(r2) > 0 if and only if r ∈ (r0,∞).Finally, when r = 12 , we have t =14 . Note that√41 > 6.40312. By (2.1),we havef(14)=(√412− 2)ζ(3)2 − ζ(5)2− ζ(3)>(6.403122− 2)· 1.202052 − 1.036932− 1.20206 > 0.01563.This implies that r0 <12 . This completes the proof.Proposition 2.7. Let b ∈ (0, ζ(5)]. Then the inequalityS(r) <2ζ(3)br2 + 1, r ∈(0,12]holds.84 Hiroki Azuma and Yasuhiro FujitaProof. Let b ∈ (0, ζ(5)]. By Lemmas 2.3 and 2.5, we have, for r ∈ (0, 12),ddr[(br2 + 1)S(r)]= 2brS(r) + (br2 + 1)S′(r)< 2br1r2 + 14+ (br2 + 1)(− 8ζ(5)r(1 + r2)3)=8r(4r2 + 1)(1 + r2)3[b(1 + r2)3 − ζ(5)(br2 + 1)(4r2 + 1)]=8r(4r2 + 1)(1 + r2)3g(r),whereg(r) := br6 + b(3− 4ζ(5))r4 + [3b− ζ(5)(b+ 4)]r2 + b− ζ(5), r ∈ R.We show that g(r) ≤ 0 for r ∈ [0, 12 ]. Then, sinceddr[(br2 + 1)S(r)]< 0, r ∈ (0, 12),we conclude that (br2 + 1)S(r) < S(0+) = 2ζ(3) for r ∈ (0, 12 ].It remains to show that g(r) ≤ 0 for r ∈ [0, 12 ]. Seth(t) := bt3 + b(3− 4ζ(5))t2 + [3b− ζ(5)(b+ 4)]t+ b− ζ(5), t ∈ R.Thenh′(t) = 3bt2 + 2b(3− 4ζ(5))t+ 3b− ζ(5)(b+ 4)= 3b(t− 4ζ(5)− 33)2 − 3b(4ζ(5)− 33)2+ 3b− ζ(5)(b+ 4).By (2.1), note that4ζ(5)− 33> 0.38256 >14.Thus, h′(t) ≤ h′(0) for t ∈ [0, 14 ]. Sinceh′(0) = 3b− ζ(5)(b+ 4) = b(3− ζ(5))− 4ζ(5)≤ ζ(5)(3− ζ(5))− 4ζ(5) = −ζ(5)(1 + ζ(5)) < 0,we have h′(t) < 0 for t ∈ [0, 14 ], and h is strictly decreasing on [0,14 ].Therefore,h(t) ≤ h(0) = b− ζ(5) ≤ 0.This implies that g(r) ≤ 0 for r ∈ [0, 12 ]. This completes the proof.The best constant for an inequality related to the Mathieu series 85Recall that b∗ is the constant of (1.7). By (2.1), we have(2.2) b∗ <2 · 1.036931.20205< 1.72527.Thusζ(5)8b∗ +ζ(5)2=ζ(5)8(b∗ + 4) <1.036938· 5.72527 < 0.74209.Since12564ζ(3) > 2.34775,we have12564ζ(3)− ζ(5)8b∗ − ζ(5)2> 0.Hence,(2.3) η(a) :=ζ(5)(a2+ 2)12564ζ(3)− ζ(5)8a− ζ(5)2> 0, a ∈ (0, b∗].This inequality is important in the following proposition.Proposition 2.8. Let a ∈ (0, b∗]. Assume that the inequalityS(r) <2ζ(3)ar2 + 1, r ∈ (0, 12]holds. Then, for a constant b with(2.4) 0 < b ≤ min{b∗, η(a)},the inequalityS(r) <2ζ(3)br2 + 1, r ∈ (0, 12]holds.Proof. Let b be a constant with (2.4). By Lemma 2.5, we have, for r ∈(0, 12),ddr[(br2 + 1)S(r)]= 2brS(r) + (br2 + 1)S′(r)< 2br2ζ(3)ar2 + 1+ (br2 + 1)(− 8ζ(5)r(1 + r2)3)=4r(ar2 + 1)(1 + r2)3[bζ(3)(1 + r2)3 − 2ζ(5)(ar2 + 1)(br2 + 1)]=4r(ar2 + 1)(1 + r2)3g(r),86 Hiroki Azuma and Yasuhiro Fujitawhereg(r) := bζ(3)r6+b(3ζ(3)−2aζ(5))r4+[3bζ(3)−2ζ(5)(a+b)]r2+bζ(3)−2ζ(5)for r ∈ R. We show that g(r) ≤ 0 for r ∈ [0, 12 ]. Then, sinceddr[(br2 + 1)S(r)]< 0, r ∈ (0, 12),we conclude that (br2 + 1)S(r) < S(0+) = 2ζ(3) for r ∈ (0, 12 ].It remains to show that g(r) ≤ 0 for r ∈ [0, 12 ]. Seth(t) := bζ(3)t3+b(3ζ(3)−2aζ(5))t2+[3bζ(3)−2ζ(5)(a+b)]t+bζ(3)−2ζ(5)for t ∈ R. Thenh′′(t) = 6bζ(3)t+ 2b(3ζ(3)− 2aζ(5)).Since a ≤ b∗, we note by (2.1) and (2.2) that3ζ(3)− 2aζ(5) ≥ 3ζ(3)− 2b∗ζ(5) > 3 · 1.20205− 2 · 1.72527 · 1.03693 > 0.Thus h is convex on [0, 14 ]. Hence,h(t) ≤ max{h(0), h(14)}, t ∈ [0, 14].Note thath(0) = bζ(3)− 2ζ(5) ≤ 0,h(14) = b(12564ζ(3)− ζ(5)8a− ζ(5)2)− ζ(5)(a2+ 2)=(12564ζ(3)− ζ(5)8a− ζ(5)2)(b− η(a)) ≤ 0by (2.3) and (2.4). Hence, h(t) ≤ 0 for t ∈ [0, 14 ]. This implies that g(r) ≤ 0for r ∈ [0, 12 ]. This completes the proof.3. Proof of Theorem 1.1Proof of Theorem 1.1. First, we show only if part. Let b ∈ (0,∞) be aconstant. Assume that inequality (1.6) holds. From inequality (1.6), wehaveS(r)− 2ζ(3) < 2ζ(3)br2 + 1− 2ζ(3) = −2ζ(3)br2br2 + 1, r ∈ (0,∞).The best constant for an inequality related to the Mathieu series 87Thus we haveS(r)− 2ζ(3)r2< − 2ζ(3)bbr2 + 1r ∈ (0,∞).Letting r → 0+ and using Lemma 2.1, we have−4ζ(5) ≤ −2ζ(3)b.Thus we conclude that b ≤ (2ζ(5))/ζ(3) = b∗. Hence we have shown onlyif part.Next we prove if part. By (2.1), we have(3.1)12564ζ(3) < 2.34778,ζ(5)2> 0.51846.On the other hand, by (2.1) and Proposition 2.7, we see that the assumptionof Proposition 2.8 is fulfilled for a = 1.03692. For this a, we have, by (2.1),ζ(5)(a2+ 2) > 2.61144,ζ(5)8a > 0.13440.Thus, by (3.1), we obtainη(a) =ζ(5)(a2+ 2)12564ζ(3)− ζ(5)8a− ζ(5)2>2.611442.34778− 0.13440− 0.51846> 1.54074.Since 1.54074 < b∗ by (2.2), Proposition 2.8 implies that the inequalityS(r) <2ζ(3)br2 + 1, r ∈ (0, 12]holds for b ∈ (0, 1.54074].Now, let â = 1.54074. For this â, we have, by (2.1),ζ(5)(â2+ 2) > 2.87265,ζ(5)8â > 0.19970.Thus, by (3.1), we obtainη(â) =ζ(5)(â2+ 2)12564ζ(3)− ζ(5)8â− ζ(5)2>2.872652.34778− 0.19970− 0.51846> 1.76277.88 Hiroki Azuma and Yasuhiro FujitaSince 1.76277 > b∗ by (2.2), Proposition 2.8 implies that the inequalityS(r) <2ζ(3)br2 + 1, r ∈ (0, 12]holds for b ∈ (0, b∗]. Hence, reflecting Lemma 2.6 and inequality (1.5), wehave shown if part. This completes the proof.4. A concluding remarkIn this section, we give a concluding remark for the three inequalities(1.3) with k2 = 1/6, (1.5) and (1.6) with b = b∗. Let us define the threefunctions f, g, h on (0,∞) as follows:f(r) =1r2 + 16,g(r) =1√(r2 + 1)2 + 1− 1,h(r) =2ζ(3)b∗r2 + 1.Note that these functions are different from ones used in Section 2. Then,we have considered the following inequalities:S(r) < f(r),(4.1)S(r) < g(r),(4.2)S(r) < h(r).(4.3)Inequalities (4.1), (4.2) and (4.3) appear, respectively, in (1.3) with k2 =1/6, (1.5) and (1.6) with b = b∗. Dividing the range of r ∈ (0,∞), we givea table to show which inequality is best among (4.1), (4.2) and (4.3). Letr0 be the constant of Lemma 2.6, andr1 =√ζ(3)(3− ζ(3))6(ζ(3)2 − ζ(5))= 0.93958 · · · , r2 =√2312.Note that 0 < r0 < r1 < r2 and thatr0 is a unique solution of the equation g(r) = h(r),r1 is a unique solution of the equation f(r) = h(r),r2 is a unique solution of the equation f(r) = g(r).The best constant for an inequality related to the Mathieu series 89Then we have the following table:The range of r Inequality The best one among (4.1), (4.2) and (4.3)0 < r < r0 h(r) < g(r) < f(r) (4.3)r0 < r < r1 g(r) < h(r) < f(r) (4.2)r1 < r < r2 g(r) < f(r) < h(r) (4.2)r2 < r f(r) < g(r) < h(r) (4.1)Acknowledgment. The authors would like to thank the referees forvaluable comments.References[1] H. Alzer, J. L. Brenner and O. G. Ruehr, On Mathieu’s inequality, J.Math. Anal. Appl. 218 (1998), 607–610.[2] L. Berg, Über eine Abschätzung von Mathieu, Math, Nachr. German7 (1952), 257–259.[3] A. Hoorfar and F. Qi, Some new bounds for Mathieu’s series, Abstractand Appl. Anal., 2007, DOI: 10.1155/2007/94854.[4] E. Makai, On the inequality of Mathieu, Publ. Math. Debrecen 5(1957), 204–205.[5] É. L. Mathieu, Traité de Physique Mathématique, vol. VI-VII: Théoriede l’élasticité des corps solides (Part2). Gauthier-Villars. Paris. 1890.[6] G. V. Milovanović and T. K. Pogány, New integral forms of generalizedMathieu series and related applications, Appl. Anal. Discrete Math. 7(2013), 180–192.[7] D. S. Mitrinović, J. E. Pečarić and A. M. Fink, Classical and New In-equalities in Analysis, Mathematics and its Applications (East Euro-pean Series) 61, Kluwer Academic Publishers Group, Dordrecht, 1993.90 Hiroki Azuma and Yasuhiro FujitaHiroki AzumaGraduate School of Science and Engineering for EducationToyama UniversityGofuku, Toyama 930-8555, JAPANe-mail: e-math@outlook.jpYasuhiro FujitaDepartment of MathematicsFaculty of ScienceToyama UniversityGofuku, Toyama 930-8555, JAPANe-mail: yfujita@sci.u-toyama.ac.jp(Received November 11, 2021)(Revised September 28, 2022)

The best constant for an inequality related to the Mathieu series

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The best constant for an inequality related to the Mathieu series

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