Research Article | Open Access

Zhen-Hang Yang, Yu-Ming Chu, Xiao-Jing Tao, "A Double Inequality for the Trigamma Function and Its Applications", *Abstract and Applied Analysis*, vol. 2014, Article ID 702718, 9 pages, 2014. https://doi.org/10.1155/2014/702718

# A Double Inequality for the Trigamma Function and Its Applications

**Academic Editor:**Wing-Sum Cheung

#### Abstract

We prove that and are the best possible parameters in the interval such that the double inequality holds for . As applications, some new approximation algorithms for the circumference ratio and Catalan constant are given. Here, is the trigamma function.

#### 1. Introduction

For real and positive values of , the classical Eulerâ€™s gamma function and its logarithmic derivative , the so-called psi function, are defined as For extension of these functions to complex variables and for basic properties, see [1]. The derivatives are known as polygamma functions (see [2]). In particular, is called trigamma function.

Recently, the bounds for the trigamma function using exponential functions have attracted the attention of many researchers. For example, ElezoviÄ‡ et al. [3] proved that the inequality holds for all . In [4, Theorem 2.7], Batir proved that and are the best possible constants such that the double inequality holds for all , where is Eulerâ€™s constant. Batir [4] also showed that for all . In [5, (1.11)], Guo and Qi established that if and , . They [6, Lemma 2] found a very simple upper bound for trigamma function in terms of exponential function as follows: for all . The inequality (6) was generalized in [7, Theorem 3.1], [8, Theorem 1.1], and [9, Theorem 1.1] to a complete monotonicity which reads that the difference is completely monotonic on . Many other new results involving the psi and trigamma functions can be found in the literature [10, 11].

Suppose that and and are the real functions defined on . is said to be the best possible constant in such that the inequality holds for all if on , or and are not comparable on , and for any satisfies on the interval .

The main purpose of this paper is to find the best possible constants such that the double inequality or equivalently holds for all , where Our main result is the following Theorem 1.

Theorem 1. * and are the best possible constants in the interval such that the double inequality (7) or (8) holds for all .*

From Theorem 1, we clearly see the following.

Corollary 2. *The double inequality
**
holds for all .*

#### 2. Lemmas

Lemma 3. *Let the function be defined on by (9). Then the function is strictly decreasing with respect to on and strictly increasing on .*

*Proof. *It follows from (9) that
If , then ; that is, is strictly increasing with respect to . Therefore,
which implies that .

If , then ; that is, is strictly decreasing with respect to , which leads to the conclusion that

Lemma 4. *Let the function be defined on by (9) and
**
Then the equation
**
has two roots
**
such that for and for .*

*Proof. *Differentiation yields
which reveals that the function is strictly increasing on and strictly decreasing on . Therefore, Lemma 4 follows from the piecewise monotonicity of the function and the numerical computations results:

Lemma 5. *Let , , and and let be defined as in Lemma 4. Then the following statements are true:*(i)*if the inequality holds for all , then ;*(ii)*if the inequality holds for all , then .*

*Proof. *It follows from the series formulas that
and we get
(i)If inequality holds for all , then, from
and Lemma 4, we clearly see that .â€‰(ii)If inequality holds for all , then
and Lemma 4 lead to the conclusion that

Lemma 6. *Let the function be defined on by (9). Then and are not comparable for all if .*

*Proof. *For and , let
Then simple computation leads to
From (20) and (21), we have
Differentiation yields
which shows that is strictly decreasing on . Therefore,
if . On the other hand, we clearly see that for .

Lemma 7. *Let and with and let be the polynomial of degree defined by
**
where and for with . Then, there exists such that and for and for .*

*Proof. *Differentiating gives
Note that
for .

From , we clearly see that is strictly increasing on . Then and lead to the conclusion that there exists such that and for and for . Therefore, is strictly decreasing on and strictly increasing on .

It follows from the piecewise monotonicity of and that for , and there exists such that and for and for . Therefore, is strictly decreasing on and strictly increasing on .

After repeating the same steps as above times, we deduce that there exists such that and for and for .

Lemma 8. *Let the function be defined on by (9). Then there exists such that and are not comparable for all if , and for all if .*

*Proof. *For and , let
Then simple computation leads to

(i) We prove that there exists such that and are not comparable for all if . For this end, it suffices to prove that there exists such that and if .

Indeed, it follows from
that the function is strictly increasing on . Numerical computations show that

Therefore, there exists such that
and for and for .

On the other hand, it follows from together with (20) and (21) that

(ii)â€‰â€‰We prove that for all if . From Lemma 3, we know that is strictly decreasing on , so it suffices to prove that for all .

Let
Then
Differentiation yields
where
Note that
where the second equation in (46) follows from (38).

Differentiating leads to
where
We assert that there exists a unique such that for and for , which leads to the conclusion that is strictly decreasing on and strictly increasing on . To this end, it is enough to verify that the coefficients of satisfy the conditions of Lemma 7. In fact, it follows from that we have
From the piecewise monotonicity of together with (46) and (47), we clearly see that
that is, for . Then (43) and (45) lead to the conclusion that for , which implies that is strictly increasing on and for .

Therefore, follows easily from (40) and .

Lemma 9 (see [12, pp. 258â€“260]). *Let and . Then
*

From the proof of [4, Theorem 2.6], we get the following.

Lemma 10. *The inequality
**
holds for .*

The following lemma can be derived immediately from the proof of [4, Theorem 2.1].

Lemma 11 (see [4, Theorem 2.1]). *Let be the function defined on by
**
Then for .*

The well-known Hermite-Hadamard inequality for convex function can be stated as follows.

Lemma 12 (see [13]). *Let be an interval, with , and let be a convex function. Then
*

#### 3. Proofs of Theorem 1

*Proof of Theorem 1. *We divide the proof into four parts.

(I) We prove the first inequality in (7); that is,
where is defined by (14).

It follows from Lemma 10 that
where

We clearly see that it is enough to prove that for .

Differentiating gives
where denotes the logarithmic mean of positive numbers and , and
Differentiating leads to
for , which means that is strictly increasing on and . It in turn implies that is strictly decreasing on . Therefore, for .

(II) We prove that is the best possible constant such that for all .

From Lemma 5, we know that if for all , where . It follows from Lemma 6 that and are not comparable for all if ; that is to say, is not a better lower bound of than even if there exists such that .

For any , (22) leads to

It follows from (19), (20), and (21) that we get

(III) We prove the second inequality (7); that is,
for , where is defined by (14).

Lemma 11 implies that the function is strictly convex on . Then, making use of Lemma 12, we get
for . That is,

Therefore, inequality (64) follows easily from (52) and the last inequality above.

(IV) We prove that is the best possible constant such that for all .

From Lemma 5, we know that if for all , where .

It follows from Lemma 8 that there exists such that and are not comparable for all if and for all if . Lemma 3 leads to the conclusion that for all if .

If there exists such that for all , then, from (19), (20), (21), and (22), we get

From the above proof and Lemma 3 we get the following.

Corollary 13. *Let the function be defined on by (9) and let be the root of (38) on . Then the inequalities
**
or equivalently
**
hold for all .*

#### 4. Remarks

*Remark 14. *It follows from (67) and the facts that
that we clearly see that the upper bound in Theorem 1 for the trigamma function is better than the upper bounds given in (2), (3), (4), (5), and (6) if is large enough.

Lemma 15. *One has
*

*Proof. *Differentiation leads to
If , then, from , we get , which leads to
If , then the first inequality in (71) still holds by making a change of variable . If , since , we see that for and for . Hence,

*Remark 16. *Using inequality (71), one has
for , which shows that the upper bound in (8) is better than the upper bound in (6).

*Remark 17. *The conclusion that the difference is completely monotonic on given in [6, Lemma 2] implies that
or
It is easy to check that the lower bound and for given in (10) are not comparable due to

*Remark 18. *Guo et al. [14] proved that
for if