New Sharp Bounds for the Bernoulli Numbers and Refinement of Becker-Stark Inequalities

Abstract

We obtain new sharp bounds for the Bernoulli numbers: 2(2𝑛)!/(𝜋^2𝑛(2^2𝑛−1))<|𝐵_2𝑛|≤(2(2^2𝑘−1)/2^2𝑘)𝜁(2𝑘)(2𝑛)!/(𝜋^2𝑛(2^2𝑛−1)), 𝑛=𝑘,𝑘+1,…,𝑘∈𝑁⁺, and establish sharpening of Papenfuss's inequalities, the refinements of Becker-Stark, and Steckin's inequalities. Finally, we show a new simple proof of Ruehr-Shafer inequality.

1. Introduction

The classical Bernoulli numbers 𝐵_𝑛(𝑛=1,2,…) can be defined by (see [1]) 𝑥𝑒^𝑥=−1^∞_𝑛=0𝐵_𝑛𝑥^𝑛,|𝑥|𝑛!<2𝜋. (1.1)Reference [2] shows a upper bound for |𝐵_2𝑛|=(−1)^𝑛+1𝐵_2𝑛𝐵_2𝑛<2(2𝑛)!(2𝜋)^2𝑛11−2^1−2𝑛,𝑛=1,2,…. (1.2) On the other hand, [3] presents a lower bound for |𝐵_2𝑛| as follows:𝐵_2𝑛>2(2𝑛)!(2𝜋)^2𝑛11−2^−2𝑛,𝑛=1,2,…. (1.3)

On the basis of (1.2) and (1.3), Alzer [4] obtains the further results.

Theorem A. For all integers 𝑛≥1 one has 2(2𝑛)!(2𝜋)^2𝑛11−2^𝛼−2𝑛<𝐵_2𝑛<2(2𝑛)!(2𝜋)^2𝑛11−2^𝛽−2𝑛, (1.4) with the best possible constants 𝛼=0 and 𝛽=2+ln(1−6/𝜋²)/ln2≈0.6491⋯.

In this paper, we obtain new bounds for the Bernoulli numbers as follows.

Theorem 1.1. Let 𝑘∈𝐍⁺, 𝑛=𝑘,𝑘+1,…, then 2(2𝑛)!𝜋^2𝑛2^2𝑛<𝐵−1^2𝑛≤22^2𝑘−12^2𝑘𝜁()()!2𝑘2𝑛𝜋^2𝑛2^2𝑛−1. (1.5) The equality holds in (1.5) if and only if 𝑛=𝑘. Furthermore, 2 and (2(2^2𝑘−1)/2^2𝑘)𝜁(2𝑘) are the best constants in (1.5).

In the following, we study on some trigonometric inequalities.

Mitrinovic [5] gives us a result which belongs to Steckin.

Theorem B. If 0<𝑥<𝜋/2, then 4𝜋𝑥𝜋−2𝑥<tan𝑥. (1.6)
Now, we show a upper bound for tan𝑥 and obtain the following sharp Steckin's inequalities.

Theorem 1.2. If 0<𝑥<𝜋/2, then 4𝜋𝑥𝑥𝜋−2𝑥<tan𝑥<𝜋𝜋−2𝑥 (1.7) or 4𝜋1<𝜋−2𝑥tan𝑥𝑥1<𝜋.𝜋−2𝑥 (1.8) Furthermore, 4/𝜋 and 𝜋 are the best constants in (1.7) and (1.8).

Kuang [6] gives us the further results described as Becker-Stark inequalities

Theorem C. Let 0<𝑡<1, then 4𝜋𝑡1−𝑡²𝜋<𝑡𝑎𝑛2𝜋𝑡<2𝑡1−𝑡². (1.9) Furthermore, 4/𝜋 and 𝜋/2 are the best constants in (1.9).

Let 𝑥=(𝜋/2)𝑡 in (1.9), then Theorem 𝐶 is equivalent to.

Theorem D. Let 0<𝑥<𝜋/2, then 8𝜋²−4𝑥²<tan𝑥𝑥<𝜋²𝜋²−4𝑥². (1.10) Furthermore, 8 and 𝜋² are the best constants in (1.10).

Clearly, Becker-Stark inequalities (1.10) are the generalization of the strengthened Steckin's inequalities (1.8).

On the other hand, Papenfuss [7] proposes an open problem described as the following statement.

Theorem E. Let 0≤𝑥<𝜋/2, then 𝑥sec²𝑥−tan𝑥≤8𝜋²𝑥³𝜋²−4𝑥²². (1.11) Bach [8] prove Theorem 𝐸 and obtain a further result.

Theorem F. Let 0≤𝑥<𝜋/2, then 𝑥sec²𝑥−tan𝑥≤2𝜋⁴3𝑥³𝜋²−4𝑥²². (1.12)

In this section, we first obtain sharp Papenfuss-Bach inequalities described as Theorem 1.3.

Theorem 1.3. Let 0<𝑥<𝜋/2, then 64𝑥³𝜋²−4𝑥²²<𝑥sec²𝑥−tan𝑥<2𝜋⁴3𝑥³𝜋²−4𝑥²². (1.13) Furthermore, 64 and 2𝜋⁴/3 are the best constants in (1.13).

The inequalities (1.13) are equivalent to64𝑥𝜋²−4𝑥²²<𝑥sec²𝑥−tan𝑥𝑥²<2𝜋⁴3𝑥𝜋²−4𝑥²²𝜋,𝑥∈0,2. (1.14) That is,64𝑥𝜋²−4𝑥²²<tan𝑥𝑥^′<2𝜋⁴3𝑥𝜋²−4𝑥²²𝜋,𝑥∈0,2. (1.15)Then, integrating the three functions in (1.15) from 0 to 𝑥, where 𝑥∈(0,𝜋/2), we obtain the following refinement of Becker-Stark inequalities.

Theorem 1.4 (Refinement of Becker-Stark Inequalities). Let 0<𝑥<𝜋/2, then 8𝜋²−4𝑥²+81−𝜋²<tan𝑥𝑥<𝜋⁴112𝜋²−4𝑥²+𝜋1−².12 (1.16)
An application of Theorem 1.4 leads to Theorem 1.5 ( the refinement of Steckin's inequalities).

Theorem 1.5 (Refinement of Steckin's Inequalities). If 0<𝑥<𝜋/2, then 4𝜋1+8𝜋−2𝑥1−𝜋²<tan𝑥𝑥<𝜋³112+𝜋𝜋−2𝑥1−².12 (1.17)Finally, we will show a new proof of Ruehr-Shafer inequality.

Theorem G (Ruehr-Shafer Inequality, see [8]). Let 0≤𝑥<𝜋/2, then 𝑥sec²𝑥−tan𝑥≤2𝜋²tan𝑥−𝑥𝜋²−4𝑥². (1.18)

2. Two Lemmas

Lemma 2.1 (see [9, Lemma  2.1]). The function (1−(1/2^𝑛))𝜁(𝑛)(𝑛=1,2,…) is decreasing, where 𝜁(𝑛) is the Riemann's zeta function.

Lemma 2.2 (see [10]). Let 𝑙_𝑛 and 𝑚_𝑛(𝑛=1,2,…) be real numbers, and let the power series 𝐿(𝑥)=∑^∞_𝑛=1𝑙_𝑛𝑥^𝑛 and 𝑀(𝑥)=∑^∞_𝑛=1𝑚_𝑛𝑥^𝑛 be convergent for |𝑥|<𝑅. If 𝑚_𝑛>0 for 𝑛=1,2,… and if 𝑙_𝑛/𝑚_𝑛 is strictly decreasing for 𝑛=1,2,…, then the function 𝐿(𝑥)/𝑀(𝑥) is strictly decreasing on (0,𝑅).

3. Proof of Theorem 1.1

Using the representation𝜁()=()2𝑛2𝜋^2𝑛2()!𝐵2𝑛^2𝑛,𝑛=1,2,…, (3.1) (cf. [11, page 266]), we have 𝐺(𝑛)=𝐵_2𝑛𝜋^2𝑛2^2𝑛−1()!12𝑛=21−2^2𝑛𝜁(2𝑛),𝑛=𝑘,𝑘+1,…;𝑘∈𝐍⁺. (3.2) From Lemma 2.1, we know that 𝐺(𝑛) is decreasing and 𝐺(𝑘)=(2(2^2𝑘−1)/2^2𝑘)𝜁(2𝑘),𝐺(+∞)=2lim_𝑛→∞𝜁(2𝑛)=2. Then, the proof of Theorem 1.1 is complete.

4. Proofs of Theorem 1.3 and G

4.1. Proof of Theorem 1.3

The following power series expansion can be found in [12]:tan𝑥=^∞_𝑛=12^2𝑛2^2𝑛−1()!()2𝑛−1^𝑛−1𝐵_2𝑛𝑥^2𝑛−1=^∞_𝑛=12^2𝑛2^2𝑛−1()!𝐵2𝑛^2𝑛𝑥^2𝑛−1,|𝑥|<𝜋2. (4.1) Thensec²𝑥=(tan𝑥)^′=^∞_𝑛=12^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−1)_2𝑛𝑥^2𝑛−2,|𝑥|<𝜋2,𝑥sec²𝑥−tan𝑥=^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛−1,|𝑥|<𝜋2. (4.2)

Let 𝐴(𝑥)=𝑥sec²𝑥−tan𝑥𝑥³/𝜋²−4𝑥²²=𝐿(𝑥)𝑀(𝑥), (4.3) where𝐿(𝑥)=𝑥sec²𝑥−tan𝑥=^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛−1,|𝑥|<𝜋2,𝑀(𝑥)=𝑥³𝜋²−4𝑥²²=^∞_𝑛=22𝑛−2232𝜋^2𝑛𝑥^2𝑛−1,|𝑥|<𝜋2. (4.4) Then, 𝑙_𝑛=((2^2𝑛(2^2𝑛−1)/(2𝑛)!)(2𝑛−2)|𝐵_2𝑛|, 𝑚_𝑛=((2𝑛−2)/32)(2/𝜋)^2𝑛>0(𝑛≥2) and 𝑙_𝑛/𝑚_𝑛=64(1−1/2^2𝑛)𝜁(2𝑛). So, 𝑙_𝑛/𝑚_𝑛 is decreasing by Lemma 2.1. Therefore, 𝐴(𝑥)=𝐿(𝑥)/𝑀(𝑥)is decreasing on (0,𝜋/2) by Lemma 2.2. At the same time, lim_𝑥→0+𝐴(𝑥)=2𝜋⁴/3 and lim_{𝑥→(𝜋/2)}⁻𝐴(𝑥)=64, so 64 and 2𝜋⁴/3 are the best constants in (1.13).

4.2. Proof of Theorem G

By (4.1) and (4.2), we have 𝑥sec²𝑥−tan𝑥𝜋²−4𝑥²=2𝜋∑^2∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛𝑛−1)_2𝑛𝑥^2𝑛−1−4∑^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1,2𝜋²(tan𝑥−𝑥)=2𝜋^2∞_𝑛=22^2𝑛2^2𝑛−1()!𝐵2𝑛^2𝑛𝑥^2𝑛−1, (4.5) so, (1.18) is equivalent to 2𝜋^2∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛𝑛−2)_2𝑛𝑥^2𝑛−1≤4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1, (4.6) that is,2𝜋^2∞_𝑛=32^2𝑛2^2𝑛−1()!(𝐵2𝑛𝑛−2)_2𝑛𝑥^2𝑛−1≤4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛2𝑛−2)_2𝑛𝑥^2𝑛+1, (4.7) or 𝜋^2∞_𝑛=22^2𝑛+22^2𝑛+2−1()!(𝐵2𝑛+2𝑛−1)_2𝑛+2𝑥^2𝑛+1≤4^∞_𝑛=22^2𝑛2^2𝑛−1()!(𝐵2𝑛𝑛−1)_2𝑛𝑥^2𝑛+1. (4.8)

From Lemma 2.1, we have that (1−1/2^2𝑛)𝜁(2𝑛) is decreasing or 2^2𝑛+2−14𝜁(2𝑛+2)<2^2𝑛−1𝜁(2𝑛) (4.9) holds. By (3.1), we get 𝜋²2^2𝑛+2−1()!𝐵2𝑛+2^2𝑛+2<2^2𝑛−1()!𝐵2𝑛^2𝑛, (4.10) so, (4.8) holds.

5. Remark

In 2010, Zhu and Hua [9] proved for 𝑥∈(0,𝜋/2)that𝜋²+48−𝜋²/𝜋²𝑥²𝜋²−4𝑥²<tan𝑥𝑥<𝜋²+𝜋²/3−4𝑥²𝜋²−4𝑥². (5.1)

Now, we can compare the results of (5.1) with (1.16). In fact, we can easy check that 8𝜋²−4𝑥²+81−𝜋²=𝜋²+48−𝜋²/𝜋²𝑥²𝜋²−4𝑥²,𝜋⁴112𝜋²−4𝑥²+𝜋1−²=𝜋12²+𝜋²/3−4𝑥²𝜋²−4𝑥². (5.2) So, (5.1) is equivalent to (1.16).