Journal of Inequalities and Applications
Volume 2010 (2010), Article ID 931275, 4 pages
doi:10.1155/2010/931275
Research Article

Sharpening the Becker-Stark Inequalities

Ling Zhu and Jiukun Hua

Department of Mathematics, Zhejiang Gongshang University, Hangzhou, Zhejiang 310018, China

Received 3 April 2009; Accepted 14 January 2010

Academic Editor: Sever Silvestru Dragomir

Copyright © 2010 Ling Zhu and Jiukun Hua. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we establish a general refinement of the Becker-Stark inequalities by using the power series expansion of the tangent function via Bernoulli numbers and the property of a function involving Riemann's zeta one.

1. Introduction

Steckin [1] (or see Mitrinovic [2, 3.4.19, page 246]) gives us a result as follows.

Theorem 1.1 (see [1, Lemma 2 . 1 ]). If 0 < 𝑥 < 𝜋 / 2 , then 4 𝜋 𝑥 𝜋 − 2 𝑥 < t a n 𝑥 . ( 1 . 1 )

Later, Becker and Stark [3] (or see Kuang [4, 5.1.102, page 248]) obtain the following two-sided rational approximation for ( t a n 𝑥 ) / 𝑥 .

Theorem 1.2. Let 0 < 𝑥 < 𝜋 / 2 , then 8 𝜋 2 − 4 𝑥 2 < t a n 𝑥 𝑥 < 𝜋 2 𝜋 2 − 4 𝑥 2 . ( 1 . 2 ) Furthermore, 8 and 𝜋 2 are the best constants in (1.2).

In fact, we can obtain the following further results.

Theorem 1.3. Let 0 < 𝑥 < 𝜋 / 2 , then 𝜋 2 + 4 8 − 𝜋 2 / 𝜋 2 𝑥 2 𝜋 2 − 4 𝑥 2 < t a n 𝑥 𝑥 < 𝜋 2 + 𝜋 2 𝑥 / 3 − 4 2 𝜋 2 − 4 𝑥 2 . ( 1 . 3 ) Furthermore, 𝛼 = ( 4 ( 8 − 𝜋 2 ) ) / 𝜋 2 and 𝛽 = 𝜋 2 / 3 − 4 are the best constants in (1.3).

In this paper, in the form of (1.2) and (1.3) we shall show a general refinement of the Becker-Stark inequalities as follows.

Theorem 1.4. Let 0 < 𝑥 < 𝜋 / 2 , and let 𝑁 ≥ 0 be a natural number. Then 𝑃 2 𝑁 ( 𝑥 ) + 𝛼 𝑥 2 𝑁 + 2 𝜋 2 − 4 𝑥 2 < t a n 𝑥 𝑥 < 𝑃 2 𝑁 ( 𝑥 ) + 𝛽 𝑥 2 𝑁 + 2 𝜋 2 − 4 𝑥 2 ( 1 . 4 ) holds, where 𝑃 2 𝑁 ( 𝑥 ) = 𝑎 0 + 𝑎 1 𝑥 2 + ⋯ + 𝑎 𝑁 𝑥 2 𝑁 , and 𝑎 𝑛 = 2 2 𝑛 + 2 2 2 𝑛 + 2 𝜋 − 1 2 | | 𝐵 ( 2 𝑛 + 2 ) ! 2 𝑛 + 2 | | − 4 ⋅ 2 2 𝑛 2 2 𝑛 − 1 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | , 𝑛 = 0 , 1 , 2 , … , ( 1 . 5 ) where 𝐵 2 𝑛 are the even-indexed Bernoulli numbers.
Furthermore, 𝛼 = ( 8 − 𝑎 0 − 𝑎 1 ( 𝜋 / 2 ) 2 − ⋯ − 𝑎 𝑁 ( 𝜋 / 2 ) 2 𝑁 ) / ( 𝜋 / 2 ) 2 𝑁 + 2 and 𝛽 = 𝑎 𝑁 + 1 are the best constants in (1.4).

2. Four Lemmas

Lemma 2.1. The function ( 1 − 1 / 2 𝑛 ) 𝜁 ( 𝑛 ) ( 𝑛 = 1 , 2 , … ) is decreasing, where 𝜁 ( 𝑛 ) is Riemann's zeta function.

Proof. ( 1 − 1 / 2 𝑛 ) 𝜁 ( 𝑛 ) = 𝜁 ( 𝑛 ) − 𝜁 ( 𝑛 ) / 2 𝑛 is equivalent to the function ∑ 𝜆 ( 𝑛 ) = ∞ 𝑘 = 0 1 / ( 2 𝑘 + 1 ) 𝑛 , which is decreasing.

Lemma 2.2 (see [5, Theorem 3 . 4 ]). Let 𝜁 ( 𝑛 ) be Riemann's zeta function and 𝐵 2 𝑛 the even-indexed Bernoulli numbers. Then 𝜁 ( 2 𝑛 ) = ( 2 𝜋 ) 2 𝑛 2 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | , 𝑛 = 1 , 2 , … . ( 2 . 1 )

Lemma 2.3 (see [6, 1.3.1.4 (1.3)]). Let | 𝑥 | < 𝜋 / 2 . Then t a n 𝑥 = ∞ 𝑛 = 1 2 2 𝑛 2 2 𝑛 − 1 ( ( 2 𝑛 ) ! − 1 ) 𝑛 − 1 𝐵 2 𝑛 𝑥 2 𝑛 − 1 = ∞ 𝑛 = 1 2 2 𝑛 2 2 𝑛 − 1 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | 𝑥 2 𝑛 − 1 . ( 2 . 2 )

Lemma 2.4. Let 𝐹 ( 𝑥 ) = ( 𝜋 2 − 4 𝑥 2 ) ( t a n 𝑥 / 𝑥 ) and | 𝑥 | < 𝜋 / 2 . Then 𝐹 ( 𝑥 ) = 𝜋 2 + ∑ + ∞ 𝑛 = 1 𝑎 𝑛 𝑥 2 𝑛 , where 𝑎 𝑛 = 2 2 𝑛 + 2 2 2 𝑛 + 2 𝜋 − 1 2 | | 𝐵 ( 2 𝑛 + 2 ) ! 2 𝑛 + 2 | | − 4 ⋅ 2 2 𝑛 2 2 𝑛 − 1 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | < 0 , 𝑛 = 1 , 2 , … . ( 2 . 3 )

Proof. By Lemma 2.3, we have 𝜋 𝐹 ( 𝑥 ) = 2 − 4 𝑥 2 ∞ 𝑛 = 1 2 2 𝑛 2 2 𝑛 − 1 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | 𝑥 2 𝑛 − 2 = 𝜋 2 + + ∞ 𝑛 = 1 2 2 𝑛 + 2 2 2 𝑛 + 2 𝜋 − 1 2 | | 𝐵 ( 2 𝑛 + 2 ) ! 2 𝑛 + 2 | | − 4 ⋅ 2 2 𝑛 2 2 𝑛 − 1 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | 𝑥 2 𝑛 ∶ = 𝜋 2 + + ∞ 𝑛 = 1 𝑎 𝑛 𝑥 2 𝑛 . ( 2 . 4 ) Since ( 1 − ( 1 / 2 2 𝑛 ) ) 𝜁 ( 2 𝑛 ) is decreasing by Lemma 2.1, it follows that 2 2 𝑛 + 2 − 1 4 2 𝜁 ( 2 𝑛 + 2 ) < 2 𝑛 − 1 𝜁 ( 2 𝑛 ) . ( 2 . 5 ) From Lemma 2.2, we get 𝜋 2 2 2 𝑛 + 2 − 1 | | 𝐵 ( 2 𝑛 + 2 ) ! 2 𝑛 + 2 | | < 2 2 𝑛 − 1 | | 𝐵 ( 2 𝑛 ) ! 2 𝑛 | | , ( 2 . 6 ) which implies that 𝑎 𝑛 < 0 for 𝑛 = 1 , 2 , … .

3. Proofs of Theorems

Proof of Theorem 1.4. Let 𝐺 𝜋 ( 𝑥 ) = ( ( t a n 𝑥 ) / 𝑥 ) 2 − 4 𝑥 2 − 𝑎 0 + 𝑎 1 𝑥 2 + ⋯ + 𝑎 𝑁 𝑥 2 𝑁 𝑥 2 𝑁 + 2 . ( 3 . 1 ) Then 𝑎 𝐺 ( 𝑥 ) = 𝐹 ( 𝑥 ) − 0 + 𝑎 1 𝑥 2 + ⋅ ⋅ ⋅ + 𝑎 𝑁 𝑥 2 𝑁 𝑥 2 𝑁 + 2 = ∑ + ∞ 𝑛 = 𝑁 + 1 𝑎 𝑛 𝑥 2 𝑛 𝑥 2 𝑁 + 2 = + ∞ 𝑘 = 0 𝑎 𝑁 + 1 + 𝑘 𝑥 2 𝑘 . ( 3 . 2 ) By Lemma 2.4, we have 𝑎 𝑛 < 0 for 𝑛 ∈ ℕ + , and 𝐺 ( 𝑥 ) is decreasing on ( 0 , 𝜋 / 2 ) .
At the same time, 𝛼 = l i m 𝑥 → ( 𝜋 / 2 ) − 𝐺 ( 𝑥 ) = ( 8 − 𝑎 0 − 𝑎 1 ( 𝜋 / 2 ) 2 − ⋯ − 𝑎 𝑁 ( 𝜋 / 2 ) 2 𝑁 ) / ( 𝜋 / 2 ) 2 𝑁 + 2 by (3.1), and 𝛽 = l i m 𝑥 → 0 + 𝐺 ( 𝑥 ) = 𝑎 𝑁 + 1 by (3.2), so 𝛼 and 𝛽 are the best constants in (1.4).

Proof of Theorem 1.3. Let 𝑁 = 0 in Theorem 1.4; we obtain that 𝛼 = ( 4 ( 8 − 𝜋 2 ) ) / 𝜋 2 and 𝛽 = 𝜋 2 / 3 − 4 . Then the proof of Theorem 1.3 is complete.

References

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  3. M. Becker and E. L. Strak, “On a hierarchy of quolynomial inequalities for tanx,” University of Beograd Publikacije Elektrotehnicki Fakultet. Serija Matematika i fizika, no. 602–633, pp. 133–138, 1978.
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