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

Authors: Hua-feng Ge
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We obtain new sharp bounds for the Bernoulli numbers:2(2n)!/(π2n(22n−1))<|B2n|≤(2(22k−1)/22k)ζ(2k)(2n)!/(π2n(22n−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.


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## Body

## 1. Introduction

The classical Bernoulli numbersBn(n=1,2,…) can be defined by (see [1])
(1.1)xex-1=∑n=0∞Bnxnn!,|x|<2π.Reference [2] shows a upper bound for |B2n|=(-1)n+1B2n(1.2)|B2n|<2(2n)!(2π)2n11-21-2n,n=1,2,….
On the other hand, [3] presents a lower bound for |B2n| as follows:(1.3)|B2n|>2(2n)!(2π)2n11-2-2n,n=1,2,….On the basis of (1.2) and (1.3), Alzer [4] obtains the further results.Theorem A.
For all integersn≥1 one has
(1.4)2(2n)!(2π)2n11-2α-2n<|B2n|<2(2n)!(2π)2n11-2β-2n,
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.
Letk∈N+, n=k,k+1,…, then
(1.5)2(2n)!π2n(22n-1)<|B2n|≤2(22k-1)22kζ(2k)(2n)!π2n(22n-1).
The equality holds in (1.5) if and only if n=k. Furthermore, 2 and (2(22k-1)/22k)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.
If0<x<π/2, then
(1.6)4πxπ-2x<tanx.
Now, we show a upper bound fortanx and obtain the following sharp Steckin's inequalities.Theorem 1.2.
If0<x<π/2, then
(1.7)4πxπ-2x<tanx<πxπ-2x
or
(1.8)4π1π-2x<tanxx<π1π-2x.
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 inequalitiesTheorem C.
Let0<t<1, then
(1.9)4πt1-t2<tanπ2t<π2t1-t2.
Furthermore, 4/π and π/2 are the best constants in (1.9).Letx=(π/2)t in (1.9), then Theorem C is equivalent to.Theorem D.
Let0<x<π/2, then
(1.10)8π2-4x2<tanxx<π2π2-4x2.
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.
Let0≤x<π/2, then
(1.11)xsec2x-tanx≤8π2x3(π2-4x2)2. Bach [8] prove Theorem E and obtain a further result.Theorem F.
Let0≤x<π/2, then
(1.12)xsec2x-tanx≤2π43x3(π2-4x2)2.In this section, we first obtain sharp Papenfuss-Bach inequalities described as Theorem1.3.Theorem 1.3.
Let0<x<π/2, then
(1.13)64x3(π2-4x2)2<xsec2x-tanx<2π43x3(π2-4x2)2.
Furthermore, 64 and 2π4/3 are the best constants in (1.13).The inequalities (1.13) are equivalent to(1.14)64x(π2-4x2)2<xsec2x-tanxx2<2π43x(π2-4x2)2,x∈(0,π2).
That is,(1.15)64x(π2-4x2)2<(tanxx)′<2π43x(π2-4x2)2,x∈(0,π2).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).
Let0<x<π/2, then
(1.16)8π2-4x2+(1-8π2)<tanxx<π4121π2-4x2+(1-π212).
An application of Theorem1.4 leads to Theorem 1.5 ( the refinement of Steckin's inequalities).Theorem 1.5  (Refinement of Steckin's Inequalities).
If0<x<π/2, then
(1.17)4π1π-2x+(1-8π2)<tanxx<π3121π-2x+(1-π212).Finally, we will show a new proof of Ruehr-Shafer inequality.Theorem G (Ruehr-Shafer Inequality, see [8]).
Let0≤x<π/2, then
(1.18)xsec2x-tanx≤2π2tanx-xπ2-4x2.


## 2. Two Lemmas

Lemma 2.1 (see [9, Lemma  2.1]).
The function(1-(1/2n))ζ(n)(n=1,2,…) is decreasing, where ζ(n) is the Riemann's zeta function.Lemma 2.2 (see [10]).
Letln and mn(n=1,2,…) be real numbers, and let the power series L(x)=∑n=1∞lnxn and M(x)=∑n=1∞mnxn be convergent for |x|<R. If mn>0 for n=1,2,… and if ln/mn is strictly decreasing for n=1,2,…, then the function L(x)/M(x) is strictly decreasing on (0,R).


## 3. Proof of Theorem1.1

Using the representation(3.1)ζ(2n)=(2π)2n2(2n)!|B2n|,n=1,2,…,
(cf. [11, page 266]), we have (3.2)G(n)=|B2n|π2n(22n-1)(2n)!=2(1-122n)ζ(2n),n=k,k+1,…;k∈N+.
From Lemma 2.1, we know that G(n) is decreasing and G(k)=(2(22k-1)/22k)ζ(2k),G(+∞)=2limn→∞ζ(2n)=2. Then, the proof of Theorem 1.1 is complete.


## 4. Proofs of Theorem1.3 and G

### 4.1. Proof of Theorem1.3

The following power series expansion can be found in [12]:(4.1)tanx=∑n=1∞22n(22n-1)(2n)!(-1)n-1B2nx2n-1=∑n=1∞22n(22n-1)(2n)!|B2n|x2n-1,|x|<π2.
Then(4.2)sec2x=(tanx)′=∑n=1∞22n(22n-1)(2n)!(2n-1)|B2n|x2n-2,|x|<π2,xsec2x-tanx=∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1,|x|<π2.Let(4.3)A(x)=xsec2x-tanxx3/(π2-4x2)2=L(x)M(x),
where(4.4)L(x)=xsec2x-tanx=∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1,|x|<π2,M(x)=x3(π2-4x2)2=∑n=2∞2n-232(2π)2nx2n-1,|x|<π2.
Then, ln=((22n(22n-1)/(2n)!)(2n-2)|B2n|, mn=((2n-2)/32)(2/π)2n>0(n≥2) and ln/mn=64(1-1/22n)ζ(2n). So, ln/mn 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, limx→0+A(x)=2π4/3 and limx→(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 (4.5)(xsec2x-tanx)(π2-4x2)=2π2∑n=2∞22n(22n-1)(2n)!(n-1)|B2n|x2n-1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1,2π2(tanx-x)=2π2∑n=2∞22n(22n-1)(2n)!|B2n|x2n-1,
so, (1.18) is equivalent to (4.6)2π2∑n=2∞22n(22n-1)(2n)!(n-2)|B2n|x2n-1≤4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1,
that is,(4.7)2π2∑n=3∞22n(22n-1)(2n)!(n-2)|B2n|x2n-1≤4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1,
or (4.8)π2∑n=2∞22n+2(22n+2-1)(2n+2)!(n-1)|B2n+2|x2n+1≤4∑n=2∞22n(22n-1)(2n)!(n-1)|B2n|x2n+1.From Lemma2.1, we have that (1-1/22n)ζ(2n) is decreasing or (4.9)22n+2-14ζ(2n+2)<(22n-1)ζ(2n)
holds. By (3.1), we get (4.10)π2(22n+2-1)(2n+2)!|B2n+2|<(22n-1)(2n)!|B2n|,
so, (4.8) holds.


## 4.1. Proof of Theorem1.3

The following power series expansion can be found in [12]:(4.1)tanx=∑n=1∞22n(22n-1)(2n)!(-1)n-1B2nx2n-1=∑n=1∞22n(22n-1)(2n)!|B2n|x2n-1,|x|<π2.
Then(4.2)sec2x=(tanx)′=∑n=1∞22n(22n-1)(2n)!(2n-1)|B2n|x2n-2,|x|<π2,xsec2x-tanx=∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1,|x|<π2.Let(4.3)A(x)=xsec2x-tanxx3/(π2-4x2)2=L(x)M(x),
where(4.4)L(x)=xsec2x-tanx=∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n-1,|x|<π2,M(x)=x3(π2-4x2)2=∑n=2∞2n-232(2π)2nx2n-1,|x|<π2.
Then, ln=((22n(22n-1)/(2n)!)(2n-2)|B2n|, mn=((2n-2)/32)(2/π)2n>0(n≥2) and ln/mn=64(1-1/22n)ζ(2n). So, ln/mn 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, limx→0+A(x)=2π4/3 and limx→(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 (4.5)(xsec2x-tanx)(π2-4x2)=2π2∑n=2∞22n(22n-1)(2n)!(n-1)|B2n|x2n-1-4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1,2π2(tanx-x)=2π2∑n=2∞22n(22n-1)(2n)!|B2n|x2n-1,
so, (1.18) is equivalent to (4.6)2π2∑n=2∞22n(22n-1)(2n)!(n-2)|B2n|x2n-1≤4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1,
that is,(4.7)2π2∑n=3∞22n(22n-1)(2n)!(n-2)|B2n|x2n-1≤4∑n=2∞22n(22n-1)(2n)!(2n-2)|B2n|x2n+1,
or (4.8)π2∑n=2∞22n+2(22n+2-1)(2n+2)!(n-1)|B2n+2|x2n+1≤4∑n=2∞22n(22n-1)(2n)!(n-1)|B2n|x2n+1.From Lemma2.1, we have that (1-1/22n)ζ(2n) is decreasing or (4.9)22n+2-14ζ(2n+2)<(22n-1)ζ(2n)
holds. By (3.1), we get (4.10)π2(22n+2-1)(2n+2)!|B2n+2|<(22n-1)(2n)!|B2n|,
so, (4.8) holds.


## 5. Remark

In 2010, Zhu and Hua [9] proved for x∈(0,π/2)that(5.1)π2+((4(8-π2))/π2)x2π2-4x2<tanxx<π2+(π2/3-4)x2π2-4x2.Now, we can compare the results of (5.1) with (1.16). In fact, we can easy check that (5.2)8π2-4x2+(1-8π2)=π2+((4(8-π2))/π2)x2π2-4x2,π4121π2-4x2+(1-π212)=π2+(π2/3-4)x2π2-4x2.
So, (5.1) is equivalent to (1.16).



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*Source: 137507-2011-09-08.xml*