#### Abstract

A lower bound on the sinc function is given. Application for the sequence which related to Carleman inequality is given as well.

#### 1. Introduction

The sinc function is defined to be

This function plays a key role in many areas of mathematics and its applications [1–6].

The following result that provides a lower bound for the sinc is well known as Jordan inequality [7]: where equality holds if and only if .

This inequality has been further refined by many authors in the past few years [8–35].

In [36], it was proposed that

We noticed that the lower bound in (3) is the fractional function. Similar result has been reported as follows [1]:

To the best of the authors’ knowledge, few results have been obtained on fractional lower bound for the sinc function. It is the first aim of the present paper to establish the following fractional lower bound for the sinc function.

Theorem 1. *For any , one has
*

In [37], Yang proved that for any positive integer , the following Carleman type inequality holds: whenever , , with , where

From a mathematical point of view, the sequence has very interesting properties. Yang [38] and Gyllenberg and Ping [39] have proved that, for any positive integer ,

In [40], the authors proved that where

As an application of Theorem 1, it is the second aim of the present paper to give a better upper bound on the sequence .

Theorem 2. *For any positive integer , one has
*

#### 2. The Proof of Theorem 1

The proof is not based on (3). We first prove the following result.

Lemma 3. *For any , one has
*

*Proof. *Set , . Then inequality (13) is equivalent to
To prove (14) by (4), it is enough to prove that
namely,
Next we prove (16). Let
We need only to prove that . Elementary calculations reveal that
Noting that, for , we have
Thus, from (19) and (18), we get
This completes the proof. Now we prove Theorem 1.

*Proof. *By using the power series expansions of and , we find that
where
Set , . Consider the function defined by
From (21), we get and . Lemma 3 implies
where
Elementary calculations reveal that for ,
Hence, for , we have
Therefore,
If we set
then we have
The intermediate value theorem implies that there must be at least one root with such that . Using Maple, we find that on the open interval the equation has a unique real root .

Hence, from (28) we get
By (21), (24), and (31), Theorem 1 follows.

#### 3. The Proof of Theorem 2

First, we need an auxiliary result.

Lemma 4. *For any , one has
*

*Proof. *By letting , , the requested inequality can be equivalently written as
so it suffices to show that the function
is negative on . Theorem 1 implies
Hence,
The required inequality follows. Now we prove Theorem 2.

*Proof. *Let
We first consider the case .

Taking the natural log gives
Taking the second derivative of both sides of (38), we have
By Lemma 4, it follows that
Thus,
and therefore for , we have
For the case , since , , and is concave up, it follows that
Using (10) from (42) and (43), we have
This proves Theorem 2.

#### Conflict of Interests

There is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors are very grateful to the anonymous referees and the editor for their insightful comments and suggestions. The work of the second author was supported by a Grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI project no. PN-II-ID-PCE-2011-3-0087.